help-28

[lime ether]

how would i use integration here to find the equation of a curve

do i integrate the dy/dx?

y is the coordinate significant

oh

ok

let me try that

wojld i have to

first

separate the variables

like y dy

and x dx

but

wouldn’t both sides have a plus c

that are different

[lilac pagoda]

yeah

but C1 - C2 = C3

they are all constants

like it doesnt matter

you can just subtract to one side and call it + C

[lime ether]

so would this be right then

and then i go back and solve the equation for y?

[lilac pagoda]

yes

[lilac pagoda]

wait wdym?

[lime ether]

well it asks for the equation of the curve

and how to get there

ik it’s an ellipse

i’m just confused how to get there

like after integrating

when i have 1/2 y^2

do i solve for y there?

or no

no

i put it in terms of an ellipse

[lilac pagoda]

you have $\frac{y^2}{2} = \frac{-2x^2}{9} + 4$ right now correct?

[glossy valve]

Abyss

[lime ether]

the equation of an ellipse

yes

[lilac pagoda]

$y^2 = \frac{-4x^2}{9} + 8$

[glossy valve]

Abyss

[lilac pagoda]

$\frac{y^2}{8} = \frac{-x^2}{18} + 1$

[glossy valve]

Abyss

[lilac pagoda]

$\frac{y^2}{8} + \frac{x^2}{18} = 1$

[glossy valve]

Abyss

[lime ether]

ohhh so u divide by 8

[lilac pagoda]

yeah just manipluate until u get the equation it says

[lime ether]

yea i got there by doing this

so c2-c1 =4

and that’s c3

so move everything to one side

and divide by 4

which is the same thing u did essentially

[lilac pagoda]

yea

[lime ether]

ok thanks

one more thing

if i were to integrate this equation to find the area

what would my limits of integration be?

like integrating a circle to find the area is easy because u can just find the area of one quadrant

and multiply by four

like the limits would be from o the the radius

how does it work for ellipses?

[lilac pagoda]

have u taken calc 3?

[lime ether]

no

[lilac pagoda]

just calc 2?

hmm

[lime ether]

i’m just self studying this book called calculus for the practical man

i’m still in high school lol calc bc is just too easy and i study on my own

[lilac pagoda]

well tbh i forgot calc 2, id just take the double integral using eliptical coordinates

[lime ether]

only calc 3 ik is really just partial derivatives

[lilac pagoda]

im sure u could solve using a single integral, i just dont know how

[lime ether]

and ik how to do multiple integrals

idk any of the applications of multiple integrals

a book i worked through had some

they were simple tho

really? so i don’t have to just do the three algebraic equations

ig those r really just two dimensional tho

v= v_o +at or the v^2

yea i’ve heard of volume of solids

or density too right?

is that a triple integral

i actually bought a workbook for multi variable just because it’ll help me self study e and m

so do u know how to use a single integral to find the area of an ellipse?

i’m gonna bet it’s not the same approach as it is for circles

[lilac pagoda]

x = root(18) * r * cos(theta)

y = root(8) * r * sin(theta)

integral from 0 to 2 pi integral from 0 to 1 of root(18) * root(8) * r dr dtheta

single integral idk how i'd actually set it up

[lime ether]

what’s annoying is that there are two radii so do i account for the negative area beneath the x axis

i can’t

can?

sorry mistype

so just find it in one quadrant

and multiply by 4?

[lilac pagoda]

solving in cartesian coords is just going to be ugly

[lime ether]

hold on

so it would just be this

and then simple trig sub

x=sqrt18 sin theta

u mean the y=?

i just simplified the sqrt

common denominator of 18

then factor out the 1/rad 18

and rad 18 is 3rad2

rad 2 cancels

ok but i got 4/3 pi

it says the answer is 12 pi

because the radius is sqrt 18

right the x axis radius

that was my only concern

since the y axis has a different radius

is there a better name for the two radii?

the answer in the back says 12 pi

i got 4/3 pi

so maybe it’s the limits of integration that’s wrong?

which step

right it simplifies to dtheta

yea so i solved for theta

which was arcsin x/rad18

and plugged in the limits

if u plug in zero u get zero

if u plug in rad 18 u get pi/2 times 8/3

which should be 4/3 pi no?

or wait

i’m dumb

i think ik

i have to integrate cos

no no i think

dx = rad 18 cos dtheta

so i should integrate cos

and get - sin

then solve for - sun

and then plug in the limits

idk y i canceled the rad 18 cos

no

wait

that’s still wrong

what am i doing wrong

let me put it in an integral calculator to make sure

ok yea so i did something wrong

the answer is 12 pi

yea i see

i skipped a step

that’s y i shouldn’t jump ahead in my head

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[torn jolt]

i just wanted to confirm if this was right

[torn jolt]

would this be 3√4

like cube root of 4

[shut thorn]

No WHAT

take the 1\3 out side

[forest blade]

16=2^4

use log rules

[shut thorn]

Now evaluate log 16 to the base 2

[torn jolt]

i did this:

log2 (16)^1/3
= log 2 (2^4) 1/3
= 4^1/3
= 3√4

idk if that's right 😍

[narrow ermine]

It's right, it is cbrt(4)

[torn jolt]

YAYY

[narrow ermine]

Assuming we interpret it as $(\log_2(16))^{\frac{1}{3}}$.

[glossy valve]

Azyrashacorki

[torn jolt]

yes

that was it

thank u sm @narrow ermine 💌

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[limpid glen]

is there a way to get the actual value of log? or is it purely calculator? (esp if it is a decimal)

[fast peak]

well depends on what you mean by "actual value"

most of them are irrational so you cant ever write them down fully

you could do by hand what the calculator does

[merry thicket]

Yeah, there's two — exact (fraction) or approximate (decimal)

[limpid glen]

example, log base 7 of 35 is 1.827-

[lilac pagoda]

get a book of logarithms 🙂

[limpid glen]

the value i got from calculator

[fast peak]

well you could also get that by hand. it would just take a while

[dark jetty]

you could use a taylor expansion

then any computer program that can add, subtract, multiply and divide (IE: most of them) can crunch the numbers for you much more easily

but obviously it'll be an approximation

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[silk radish]

i need some help w this, i get the base method and how to solve it but i keep getting it wrong ( this is the 6th time i got it wrong )

@minor crater maybe u can help i dont think anyone else is here

[minor crater]

why me lol

sorry, can't help

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[slate ivy]

Can someone verify if a B c and D are good

i dont know

In my book for D it says DNE so diverges

for the d) limit

i dont see how

<@&286206848099549185>

its the (-1)^n thats annoying me

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[strange cobalt]

Hello :3 I need help with a graphing problem, It's Algebra 2 in Highscool

[minor crater]

there's a horizontal line called the x-axis right

where is the curve of function f(x) just visually below the x-axis? (answer in plain english, not math)

[strange cobalt]

uhm 😓 below the x-axis

[minor crater]

okay

let me think how to explain this then..

[strange cobalt]

right

[minor crater]

i have no idea how to describe this process but if you are given a graph can you just like colour in where the graph is below the x axis?

i think you can

[strange cobalt]

yeah i can lmfao

[minor crater]

amazing!

this is the only actual step here

the rest is just writing it down

notice that it says f(x) ≤ 0

so if f(x) = 0 it's fine

[minor crater]

so the first red line goes from -infinity (because of the arrow at the end)

and up to the first zero-crossing

where does the function have its first zero crossing?

at x=__

[strange cobalt]

-7

[minor crater]

no, it just touched 0

it didn't cross it

[strange cobalt]

oHH

[strange cobalt]

-3*

[minor crater]

mhm

yep

so youe first interval is (-inf, -3]

[thin coyote]

but its says where f(x) is less than or equal to 0

[minor crater]

yes

so it's part of the same interval

[thin coyote]

so wouldnt you have to include [-7,-7]

OH

[minor crater]

no

[thin coyote]

im stupid

my bad

sry for butting in

[minor crater]

it's okay lol

[minor crater]

@strange cobalt does that make sense?

[strange cobalt]

yeah it does !

[minor crater]

amazing

now let's do it for the next interval

you do the same procedure

[strange cobalt]

sorry i know u said this but i just want to clarify, the -inf is because of the arrow on the bottom left side of the graph right?

[minor crater]

yes

that's correct

if you see arrows at the end it means the function extends to infinity (either positive or negative) beyond what they show

[strange cobalt]

(-inf, 3] ?

[minor crater]

yes

that's the first interval

but you have two more

oh you wrote 3

[strange cobalt]

yeah

[minor crater]

no, that's not right

[strange cobalt]

hm

[minor crater]

what about x=-2? f(x) is not below 0

ignore the previous interval, completely

they are disjoint

[strange cobalt]

Wdym? x=-2 is above it but i thought it'd be part of the previous interval because it's still above the x axis

[minor crater]

it's above the x axis but we are looking for values that are below it :)

[strange cobalt]

:3

[minor crater]

again, you have 3 disjoint intervals

the first one is (-inf, -3]

the other two you have to find

disjoint intervals mean that they do not intersect or interact with each other in any way, they are completely separate

[strange cobalt]

im sorry 😓 (0,0 ) and (7,0) ?

[minor crater]

well, those are zero-crossings, those will be important

but not intervals

focus on just the blue highlighted part

[strange cobalt]

um

[minor crater]

your functions goes below zero after x=0

[minor crater]

right?

sorry but i need to go now. it's 3 am here and i have classes early tomorrow (as in, today)

[strange cobalt]

have fun :-p

[minor crater]

i hope someone else will help you 🫡

[strange cobalt]

thank u

i got this !

<@&286206848099549185> :3

[tough patio]

?

[strange cobalt]

do u want me to resend?

[tough patio]

sure

idk if i could solve it

but sure

[tough patio]

is that sine?\

[strange cobalt]

no i just have to write out interval notation

[tough patio]

damn i forgot

uhhhhhh

hol up

[strange cobalt]

real 💀

[tough patio]

lemme use my 2 brain cells rq

[strange cobalt]

first one is, (-inf, -3]

[tough patio]

ic

which is secnd?

[strange cobalt]

idek mane

[tough patio]

im zoro rn

cs im lost asf

ummm

[strange cobalt]

🦮

:3

[tough patio]

🐒

[strange cobalt]

me asf

[tough patio]

🐒 💨

me asf

🐟

alr

[strange cobalt]

this test is gonna shatter me.

[tough patio]

ummm

[tough patio]

this question finna shatter me

[torn jolt]

hi

[strange cobalt]

YES JUSTIN612 !

]

[torn jolt]

[0,3] is next

then [7, inf)

[tough patio]

bro u ruined my acreer by solving that in 0.3 seconds

like i had a career to begin w/

ill leave u to it

[torn jolt]

This is basic algebra 2

[strange cobalt]

please

[strange cobalt]

real ^.^

this is my last problem

[torn jolt]

(-inf,-3]

Basically you’re finding when the graph is going up in y values

[0,2]

[7, inf)

[strange cobalt]

why 2? i thought it'd be 0?

tysm btw ^.^

[torn jolt]

Starting at x = 0 and going towards 2, the graph is increasing in y values

[strange cobalt]

oh i see

ty :3

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[shadow silo]

Isn't there a possiblity that 3x^3 can be a integrating factor?

[shadow silo]

Bcoz e^integral p dx it will give a constant too

I.F. = cx^3?

Answer is given only b,c

[velvet sedge]

what is p(x)?

you turn the above differential equation into a linear one. It isn't linear in the state its in right now

do you see why it isn't linear the way its written? Do you see how you can make it linear ODE?

@shadow silo

[shadow silo]

coefficent of y i.e. 3/x

[velvet sedge]

yes

and you integrate that and get

3ln(|x|)

[shadow silo]

+c?

[velvet sedge]

yea +c

[shadow silo]

So a is correct?

[velvet sedge]

c|x|^3 not equal to cx^3

@shadow silo

[shadow silo]

Won't c take care of sign

either + or -

[velvet sedge]

c(|x|)^3

this is an even function

cx^3 is an odd function

[shadow silo]

Whoa

But wait

if x is negative, c can be negative?

[velvet sedge]

you can't change c

while you change x

c is fixed

[shadow silo]

Right

So if it was cx^2

then it might've been the ans

[velvet sedge]

yea

since squaring makes negative values positive, and abs doesn't effect it, so if it was 2/x

it would've been an answer

[shadow silo]

Great, thanks a lot!!

[velvet sedge]

np :)

you can discuss this more with your professor/TA/someone else as well just to see if this is what they meant by the way also :)

but i'm fairly certain this is why the answer purely wasn't cx^3

[shadow silo]

Yeah But classes are not for few days and this was an olympiad question so i was eager

[shadow silo]

👍

[velvet sedge]

ah, ic, if it's olympiad then there are probably answers with explanations out there already

[shadow silo]

Yeah but that too isn't till mid december

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[frank beacon]

how?

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@frank beacon Has your question been resolved?

[shadow silo]

By hit and trial it is seen that -1 is a root

,w solve -x4+x2+2x+2

[glossy valve]

Results provided by WolframAlpha

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

Requested by mr.perfect7

[shadow silo]

Others are cetainly bad

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[torn jolt]

I need help getting started on finding the equation to differentiate later

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@torn jolt Has your question been resolved?

[gritty rose]

They give you C in terms of V

[torn jolt]

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[untold vessel]

need someone to look at my proof

hehe

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[quaint mortar]

hi i dun understand this question.

[worn matrix]

what part are you having prob with

[quaint mortar]

i dont understand how it is a right angle

[worn matrix]

can you see 2 isosceles triangles?

[quaint mortar]

yes

[near mirage]

Start by putting alpha and beta to their side angles

Then check what alpha plus beta is

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[tiny drum]

an object is place 7 cm from a concave mirror whose radius of curvature is 10 cm the image formed  will be

[worn matrix]

the image formed will be like what?

what we have to tell?

like iverted or not

[tiny drum]

yes

[worn matrix]

oh

so its placed in between the radius of curvature and focus of the concave mirror

[tiny drum]

yep

[worn matrix]

where do you think it forms

[tiny drum]

idk

that’s why i asked

the rays prob meet

because concave is

converging ken’s

lens

sooo

it’s real

anddd

upright

?

[worn matrix]

first of all its not a lens

its a mirror

[tiny drum]

oh yeah i mean mirror

[worn matrix]

this might help

also you have to learn these cases

or you have to use all the formula's to find its length, and all to determine if its inverted or not

[tiny drum]

ok so what is it

[worn matrix]

so learning these cases would be better

[tiny drum]

real and inverted.

?

[worn matrix]

it the d one in this case

[tiny drum]

😭oml i hate geometrical optics wth is this

isn’t it placed beyond f

u said it’s between f and radius of curvature

[worn matrix]

yea it is

because f is R/2

[tiny drum]

💀oh yeah

[worn matrix]

that means its focus lies on 5 cm from the mirror

and yea its real and inverted

[tiny drum]

thanks 😭😭😭

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[hoary linden]

How do I solve part c?

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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

[robust slate]

Being answered in #help-4

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[proud spear]

Hi.
Kate has some 10% vinegar. How much water does she need to add in order to get 3 kg of 6% vinegar?(options shown in picture 1)
I am doing this type of question for the first time, so I would appreciate my work getting checked.
Thanks in advance.

[proud spear]

<@&286206848099549185>

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[proud spear]

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[west crescent]

Guys how do i solve this?

They say find the constant term

I just need a guide on what will i do....

[compact briar]

You can use the binomial theorem

[torn jolt]

yeah binomial theorem most ideal

[knotty marlin]

yea binomial thrm

[west crescent]

<@&286206848099549185>

?

(16)(1/x³)¹⁶+...+(16)(-2x)¹⁶
0 16

My messages aren't sending :((((

[knotty marlin]

🌞

[west crescent]

Ill just sleep 😭

Hahahaha

Thank you and goonight guys hahahaha

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[old zinc]

need help with cauchy criterion

[old zinc]

I want to be able to prove that a series is convergent, using the cauchy criterion

and I'd like to use a hard example e.g. sigma 1/n as my series since it's simple.

[compact briar]

sigma 1/n is not convergent

[old zinc]

it's not ??

but 1/n is a null sequence

I thought the sigma of null sequences are convergent

[compact briar]

No

It has to be a null sequence, but thats not enough

[forest widget]

now use cauchy criterion to show it doesn’t converge

[old zinc]

wait or was it the other way around

if a series is convergent, then it would mean that the sequence within it is a null sequence?

is that right?

[compact briar]

Yes

[old zinc]

ok ok

i mixed that up

now how do apply the cauchy criterion to a series

,rotate

[old zinc]

this is what I'm working on

I have the cauchy criterion written there on top

it's for a series with the seqeunce 1/k^2

where I have z.z. is what I have to show

and I saw a video on youtube where the guy set n=m+1 so that the thing would simplify to a_n < ε

and then he did something with that

but I didn't really get it

[compact briar]

Pick a random m and n, n greater or equal to m

[old zinc]

like an actual number?

[compact briar]

Imagine its an actual number

[old zinc]

ok

it would turn out like this

the line on the bottom

and that need to be smaller than ε

wait

i think i need to get rid of 1/m^2 on the left

that would be cancelled out

@compact briar

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[glossy blaze]

I need help

[lavish sequoia]

Shut up you took my channel

[glossy blaze]

Am I wrong?

<@&286206848099549185>

[full hawk]

I can help you

First you do -5 x 27

Which is -135

Then you do - 3^2 which is -9

Then another minus 9

Then +6 = 147

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[torn jolt]

i want help with a question

[torn jolt]

can anyone help me?

[torn jolt]

ok

the number of ordered pair(s) (x,y) of real numbers satisfying the equation 1+x^2+2xsin(arccos(y))=0 is

the answer which i got is 0 but the given answer is 1

is anyone there

[velvet flint]

hello

can you send the question in written

so it is easier to understand

@torn jolt the equation comes out to be this

and as you can see the graph or by just putting the value x=0,y=0

you get only 1 ordered pair

[torn jolt]

try something with this using domin and range of arcsin

thanks

[torn jolt]

why did u do y+yx instead of x^2+1

[velvet flint]

just solve using basics of inverse trigo

like take theta = cos^-1 y

then make a triangle and find the sin of that same angle

and make the implicit function of y.x

then just put x=0,y=0

its the only root

to prove it i showed you on desmos

[torn jolt]

lol

i did sin(cos^-1 y)=+-(1-y^2)^1/2

is that wrong

[velvet flint]

ruk bhai bhejta hu solution

[torn jolt]

ok so like

get the inver trig function on one side

and other aljebraic on one side

then try to graph both

and yes use sin(pi/2 - y ) = cosy

and use principle range domain thing

lol

[velvet flint]

according to me i get 2 ordered pairs

i got the initial solving wrong

my bad

[torn jolt]

i got it

i was also making a mistake

[velvet flint]

did you understand now?

just put in values rn

[torn jolt]

i forgot the 2 that is coefficient of x

[velvet flint]

alright

gl now ig

[torn jolt]

its done

[velvet flint]

the answer is 2 right?

[torn jolt]

no

[velvet flint]

bruh

[torn jolt]

the answer is 1

[velvet flint]

how wtf

(0,0) and (0,1)

[velvet flint]

wtf did u just send

[torn jolt]

lol

unique value should satisfy according to me

[velvet flint]

your method is very good

mine is alright ig

lmao

mine is just so informal

[torn jolt]

question is dumb

[velvet flint]

tbh

fr

[torn jolt]

ahh

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[tiny drum]

why is convex mirror used in cars

[torn jolt]

u can google that tbh

[light mortar]

It's not Math

The reason for use of convex lens is it helps to spot more background than a normal mirror. So, you can see more in a convex mirror than a flat one.

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[tulip zodiac]

Help

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[weary siren]

I’m trying to learn elasticity in economics

[torn jolt]

Is this from economics

Yep

[weary siren]

What is E of i

[torn jolt]

No idea, I wanna take econ next year lmao

[weary siren]

I thought you find derivative of Q and P

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[dense edge]

am I supposed to do A - 1 first?

-4 -1 -9 4 3 0 7 -8 1 -1 4 -4 -1 -1 -1 -1

[kind jay]

Yes

[dense edge]

what is the next step?

[kind jay]

Do RREF

[dense edge]

are we treating column 4 as b?

x1, x2, x3, b

[kind jay]

No

[dense edge]

I get identity matrix when I do RREF

1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1

what is the next step?

[kind jay]

,w rref {{-4,0,-8,5},{4,0,8,-7},{2,0,4,-3},{0,0,0,-1}}

[glossy valve]

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[kind jay]

U did it wrong

[dense edge]

why did you use those entries

[kind jay]

If it was identitu the basis would have 4 elements

It's the correct entries

[dense edge]

,w rref {{-4,-1,-9,4},{3,0,7,-8},{1,-1,4,-4},{-1,-1,-1,-1}}

[glossy valve]

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[dense edge]

how did you get

[kind jay]

How did u get that

[dense edge]

from this

[kind jay]

Did u literally subtract every element with 1

[dense edge]

yeah

[kind jay]

U need to do it to the diagonals only lol

Its A-lambda*I, where I= identity matrix

[dense edge]

ohhh

OK

-4 0 -8 5 4 0 8 -7 2 0 4 -3 0 0 0 -1

,w rref {{-4,0,-8,5},{4,0,8,-7},{2,0,4,-3},{0,0,0,-1}}

[glossy valve]

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[dense edge]

so after I find RREF what is the next step?

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@dense edge Has your question been resolved?

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@dense edge Has your question been resolved?

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[torn jolt]

Idk if this is right

[torn jolt]

Just checking

If not, a push in the right direction would be solid

[muted matrix]

Yeah that's right

the only thing left to do is solve for x by divding by 4 and taking the square root

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[torn jolt]

Thank you

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[ashen depot]

How would I factor the denominator?

[slate zenith]

look at the bottom, any way to factor out something to make it a quadratic?

[ashen depot]

x?

[slate zenith]

great!

[ashen depot]

oh

I see

so

So the factored form would be x(x^2 - 2x + 1) -2?

[slate zenith]

yup!

[wide sundial]

What

[ashen depot]

?

[ashen depot]

Thank you

[wide sundial]

That’s not “factored”

[slate zenith]

oh wait

keep going

[wide sundial]

You want to group x³ with x and x² with the constant

[slate zenith]

ah, yes that would be better

sorry

[ashen depot]

How would I group x^3?

[slate zenith]

whats a common factor between x^3 and x

[ashen depot]

1?

[slate zenith]

no, we can just factor out x

like before, but instead lets factor it only out of x^3 and x

[shell laurel]

just x

[shell laurel]

a common factor is just why you can divide both of these terms by

[ashen depot]

hmmm

I am thinking and rationalizing it in my mind rn

SO

x^2(x-2)+x-2?

[shell laurel]

uh

provide me the original problem

[ashen depot]

okay

gimma a sex

sec*

lord have mercy

[shell laurel]

NA

thats crazy

[ashen depot]

NAH

I DIDN'T mean to type that

[shell laurel]

sure buddy

anyways

just schow me the problme

[ashen depot]

okay

[shell laurel]

ok so

i assume yk powers rules

right?

like when your multipying to powers with the same base you add the exponents

2*

I would advice factor the top then bottom

if i did it right the bottom should be x(x^2 - 2x + 1) -2

[ashen depot]

okay

That's what I had though originally. Thank you!

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[torn jolt]

How can i do partial fraction decomp on this

[umbral dome]

partial fraction decomposition only works when the denominator has a higher degree than the numerator

so before you can perform partial fraction decomposition, you should start by performing polynomial long division

once you do that, you should end up with:
(quotient) + (remainder)/(denominator)

the quotient is just a polynomial. perform partial fraction decomposition on the (remainder)/(denominator) as normal

[torn jolt]

ok

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[torn jolt]

.reopen

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✅

[torn jolt]

ok so

i am supposed to get this from it

from

i do not know how

[umbral dome]

those are not equivalent expressions. can you show the entire problem?

[torn jolt]

wants me to use laplace transform to solve this

this is the solution

as u can see it says they are equivalent

i think

im so silly

i forgot the s^2 + 1

[umbral dome]

yeah

it's easy to miss

[torn jolt]

shouldnt it be s^3 + 10s still on top though?

why is that reduced

by 1 degree

[umbral dome]

i'm honestly not sure. the strange thing is that the partial fraction decomposition doesn't seem to be correct, either

[torn jolt]

ok i think i will skip this problem lol (solution sheet can be wrong a lot)

thanks for sticking it out with me though, i appreciate the help of excellent citizens like you 🫡

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[dark egret]

old channel closed due to inactivity

but heres what ive got

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[fickle finch]

Could someone help with this q on probability. STudying for test atm.

three cards drawn from a deck without replacement. Find the probability the last card is a spade given that exactly one of the first two cards are a diamomond

[unkempt thorn]

!status

[full forum]

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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

[fickle finch]

1. ive done questions like it b4 but this is more complicated

[unkempt thorn]

sorry thats not the best way to do this

lets start with the definition of conditional probability

whats the formula for P(A|B)

[fickle finch]

P(A and B)/P(B)

[unkempt thorn]

right

so what's P(B)

(B in this case is exactly one of the first cards being a diamond)

[fickle finch]

would it be (39/52)(13/51)+(13/52)(39/51)

[unkempt thorn]

should be yeah

now what is P(A and B)

[fickle finch]

(39/52)(38/51)(13/50)

?

[unkempt thorn]

is there any reason one of the first two cards cant be a spade?

[fickle finch]

(39/52)(38/51)(13/50) + (13/52)(39/51)(38/51) + (39/52)(13/51)(12/50)

spade 1 and 3, spade 2 and 3, no spaced?

[unkempt thorn]

remember, we also have to account for the fact that one of the first cards is a diamond

[fickle finch]

2 (13/52)(13/51)(13/50) + (13/52)(26/51)(13/50) + (26/52)(13/51)(13/50)

[unkempt thorn]

ok so the third term is (non spade and non diamond)(diamond)(spade), the second one is (diamond)(non spade non diamond)(spade)

but im not sure i understand the first term

[fickle finch]

diamond diamond spade?

but now that i think abt it idk y i put times 2

[unkempt thorn]

exactly one of the first two is a diamond

[fickle finch]

nvm

im trolling

so just the second and third term?

[unkempt thorn]

well

[fickle finch]

wait no theres 2 more terms?

[unkempt thorn]

why cant we have (spade)(diamond)(spade)

or (diamond)(spade)(spade)

[fickle finch]

u right

imma move on (tight schedule) but thanks for the help. I know what i gotta do know

[unkempt thorn]

ye

[fickle finch]

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[tawdry horizon]

im not sure how to do the recursive relation part

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[daring eagle]

Please help #4

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[cursive dock]

1/5th of her money was used to buy 12 files and 5 pens. each pen cost as much as 3 files.
fill in the blanks:
1 file costs as much as ??? ____ pens
so 12 files costs as much as ???___ pens
and 1/5th of her money bought her the equivalent of ???___ pens
so each pen cost her ????____ of her money

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@ruby stag Has your question been resolved?

[ruby stag]

3 pens
5 pens
Idk

[cursive dock]

not quite, not quite

1 pen costs as much as 3 files

which means each file costs what fraction of a pen

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[jade fulcrum]

How do i do this

Will it be

5(-2y) - y = 6

?

[hot herald]

no

[jade fulcrum]

Oh then how

[hot herald]

firstly address what you just typed,
it seems that you attemtped to replace x with (-2y)
but there's no justification for that as you don't have anything saying that those two things are equal

have you been introduced to solving these types of questions with elimination or substitution?

[jade fulcrum]

Yes

i substitute

How do i do this then

i am supposed to replace x

Right

[tired birch]

You can replace either

I like to replace the one that looks cool

[jade fulcrum]

Oh

How do i know which one is cool

[tired birch]

In this case it's $x=2y-6$

[glossy valve]

diaas_(yt)

[jade fulcrum]

Replace that?

[tired birch]

Why not? It Looks cool

[jade fulcrum]

Why does it say x = 2y -6

i thought it’s x - 2y = -6

[tired birch]

Because x-2y=-6, add 2y to both sides bro

[jade fulcrum]

How do yoy flip it

Oh

[tired birch]

So put the value of x in the first equation

$5x-y=6$

[glossy valve]

diaas_(yt)

[jade fulcrum]

?

[tired birch]

$5(2y-6)-y=6$

[glossy valve]

diaas_(yt)

[tired birch]

There

I did a sillly

silly billy i am

[jade fulcrum]

10y - 30 -y = 6

[tired birch]

9y-30=6

y=4

[jade fulcrum]

it is 9y = 36

right

y is 4

Now what

5(4) - y = 6?

But what is y

or x

[hot herald]

you found y = 4
you should be subbing y with 4

and solve the resultant equation to determine x

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[orchid breach]

@jade fulcrum u still need help?

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[dense edge]

how should I proceed with this question?

[velvet sedge]

use definition of eigenvector

:)

[sonic quiver]

and distribute matrix multiplication over vector addition

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@dense edge Has your question been resolved?

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[river mulch]

a question is asking me to find the region between z = x^2 + y^2 and the a 4x4 square of (2,2) (-2,2) (-2, -2) (2,-2) on the xy plane

[river mulch]

i figure i have to employ green's thereom somehow

but such that i only have one function for P(x,y) = x^2 + y^2 and some idea of the double integral, how do i find the Q 👀

[spiral vigil]

wait what does it want? the volume of that space?

,w plot z = x^2 + y+2, -2 < x < 2, -2 < y < 2

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[river mulch]

yes

here's what i got so far

not sure if im missing the mark though

[spiral vigil]

if it's just the volume that you want that's just the double integral over that domain

no need for Green's theorem

[river mulch]

wait so like

double integral of z

over the square's domain

[spiral vigil]

yea

[river mulch]

REALLY?

[spiral vigil]

yeah... it's just like the area below a curve

for single variable calc

[river mulch]

oh so like

cus it uses both x and y

we emply the double integral

[spiral vigil]

ya

[river mulch]

ahhh that makes sense. my intuition is just broken xD

thanks boss

.close

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[torn jolt]

!heko

!Help

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[torn jolt]

my teacher only taught how to derive equation but iv never done this with numbers

[analog tundra]

I'll help

[torn jolt]

thanks

[analog tundra]

I've done similar things

I'll try

String constraint?

[torn jolt]

never heard of that

this is a pretty new topic for us

[analog tundra]

Length of string is constant

[torn jolt]

right

only x and y change

[analog tundra]

Is the string always taut

[torn jolt]

straight

yes i beleive so

[analog tundra]

How tho

[torn jolt]

the rope?

[analog tundra]

Yes

[torn jolt]

im assuming it stays straight as my teacher has done question like that

still there?

!report

[analog tundra]

But if there's motion how can it be possible

[analog tundra]

Is it like a tape

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@novel flower Has your question been resolved?

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[lucid sierra]

hello i need help

[glacial pasture]

are those superscripts the bases?

[lucid sierra]

wht are superscripts

[glacial pasture]

the numbers hanging above stuff

[lucid sierra]

oh yeah

my teacher write it like that

idk xd

[lucid sierra]

i think im stuck here

[minor crater]

tetration 💀💀💀