help-0

[strange fractal]

so E is very long?

or is

CD very storng

long

[valid smelt]

yeah you basically just extend it until it meets the circle

and then you call that point E

[strange fractal]

i am so compused

nvm got it

wait

so wat i do now

[valid smelt]

well start labeling angles and distances and applying the geometric rules you learned in class

[strange fractal]

can i do this

i forgot to write a ab

b

i havw this rn

not sure how to prove equal sides

[valid smelt]

uhhh thats definitely not the graphic they were aiming for you to make LOL

[strange fractal]

oh.

BUT IT WORKS RIGHT

[valid smelt]

or is it..

let me check

[strange fractal]

idk i was just playing around

[valid smelt]

should look something liek this

yeah

ok

i think i got confused for some reason, yours is fine

[strange fractal]

ok i have dis so far

opposite angles in a parallelogram equal

and opposite sides in a cyclic quadrilateral add up to 180

so i have the 180-x

OH

DO I HAVE TO PROVE ITS AN ISOCELES TRIANGLE?

[valid smelt]

you can try that i honestly don't really know where to proceed from here because im kinda busy atm but yeah you just gotta play around until you find a proof that works

[strange fractal]

OMG

WAIT

so angles in a straight line add up to 180 right

so 180 = 180-x

wait what

i have 0 degrees

[strange fractal]

omg 😔😔

no one can help me MY MATHEMATICS IS TOO COMPLEX

[valid smelt]

does she know....

[strange fractal]

WHAT

angle y = 180-(180-x)

OMG

ANGLE Y=180-180+x

=X

[strange fractal]

IS THIS RIGHT

SO THEREFORE SINCE THE 2 ANGLES ARE THE SAME IT IS ISOCELES AND AE = AD

[valid smelt]

if you didn't mess up and proved that angle ade = dea then that does prove isoceles and ae=ad

make sure to double check

[strange fractal]

oops

did i mess up

WHAT WHERE DID I MESS UP

[valid smelt]

i have no clue!

i dont know if you messed up btw

you could be right

lol

[strange fractal]

bro i hate proofs

my textbook give no answere

it says

answer may vary

THEN HOW TF DO I KNOW IM RIGHT

😭😭😭😭😭😭😭

[lone heart]

@strange fractal Has your question been resolved?

[lone heart]

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[lone heart]

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[spare ocean]

how to solve ln(x) = 1?

[solemn juniper]

apply the inverse function of ln to both sides

[spare ocean]

what does that mean?

could you showcase what you mean by apply the inverse?

[solemn juniper]

what function is the inverse of ln?

[spare ocean]

ln(x)^-1?

[solemn juniper]

you can be more explicit than that

[spare ocean]

oh wait

e^x?

[solemn juniper]

better

[spare ocean]

so u get ln(x)^ex?

[solemn juniper]

no

you get e^(lnx) = e^1

[spare ocean]

okay, so we have our original equation as exponents now

[solemn juniper]

what is e^(lnx)?

[spare ocean]

e

[solemn juniper]

no

[spare ocean]

x?

[solemn juniper]

in general

x

[spare ocean]

x = e^1 so x is e

this feels strange to me

[solemn juniper]

this is true for all logarithms

log_a(a) = 1

[spare ocean]

a^ log_a(a) = a ^ 1?

[high cargo]

If you look at the definition of ln/log in general it might no longer look wrong

[solemn juniper]

sure

[spare ocean]

and then, a = 1?

[solemn juniper]

from your equation you get a=a

[spare ocean]

ah ye

[solemn juniper]

which is always true

[solemn juniper]

showing that this is always true

[spare ocean]

10log(x) = 0
10^10log(x) = 10^0?

so, = 1 since ^0

[solemn juniper]

another property of the logarithm

[spare ocean]

so this only counts when theyre the same

i dont get it, sorry..

[solemn juniper]

what don't you get?

[lone heart]

@spare ocean Has your question been resolved?

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[alpine sable]

This unit is about dot products and 3d vectors. It also covers projections on dot products. I just don't understand why the projection of P0P1 to n-> is the distance, not the actual projection. I am also confused on the transformation (equal sign) of the projection to that weird formula on the right; plus, why is |n| not a redundant? (It's 1 since it's a unit vector)

[tacit arch]

Too blurry to read. Where does it say n is a unit normal

[alpine sable]

It says it's a normal vector.

Wait is normal vector not a unit vector?

[empty moth]

okay okay

let me start you off

[tacit arch]

No. Not all normal vectors are unit length

[empty moth]

what's the normal vector in terms of a and b

the equation you are given is ax + by + c =0

the denominator gives you a hint too

cuz the magnitude of n is a^2 + b^2

[alpine sable]

(a, b)?

[empty moth]

great now

let $P_1 = (x_1, y_1)$ be a point and $P_0 = (x_0, y_0)$ be a point

[ocean seal]

nosqldb

[alpine sable]

ok

[empty moth]

@alpine sable give me

in terms of a and b

[alpine sable]

is it not (x1-x0, y1-y0)?

[empty moth]

let's call that v and now we want to compute n * v = <a,b> * <x_1 - x_0, y_1 - y_0>

can you do that for me?

[alpine sable]

ok

a(x1 - x0) + b(y1 - y0)

[empty moth]

okay okay

let's rewrite this

$ax_1 + by_1 - (ax_0 + by_0)$

[ocean seal]

nosqldb

[alpine sable]

ok

[empty moth]

but x_0, y_0 is a point on the line

[alpine sable]

so we'll call that c?

[empty moth]

so $ax_0 + by_0 + c =0$, so $ax_0 + by_0 = -c$

[ocean seal]

nosqldb

[empty moth]

exactly!

we use the equation of the line we were given

putting it all together

[alpine sable]

oh ok

[empty moth]

we get the numerator as $ax_1 + by_1 +c$

[ocean seal]

nosqldb

[alpine sable]

yes

[empty moth]

but we take the absolute value

now if $n = <a,b>$, what's |n|

[ocean seal]

nosqldb

[empty moth]

once you figure this out

ur done

[alpine sable]

sqrt(a^2 + b^2)

[empty moth]

put it all together

[alpine sable]

that's d

[empty moth]

and you get ur expression on the RHS

amazing!

any questions?

[alpine sable]

but how does this all relate to how it's perpendicular distance from P1 to the equation?

[empty moth]

so ur asking how does it relate to this?

[alpine sable]

well, I see that that d is the projection of P0P1 to n not the perpendicular distance.

[empty moth]

well what what direction is the normal in

[alpine sable]

oh

perpendicular to the line

OH

I kept thinking it was parrellel to the line

[empty moth]

yes

bc ur so amazing, I will share an diagram

[alpine sable]

yes! the perfect answer.

[empty moth]

have a good night man and close this channel when you can

[alpine sable]

thank you

.close

[lone heart]

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[primal stream]

guys i have cambride soon

can anybody help

chat me

i cant chat cuz my phone will br hidr

the topic is volume algebra and fraction

if anybody help it would help

[lone heart]

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[left scarab]

could someone explain this part of the question

[zealous lichen]

Nul A is the null space of A containing all vectors v such that Av=0

Col A is the column space of A being the span of column vectors of A

[left scarab]

ohh

[left scarab]

so any of the columns in A would work

[zealous lichen]

yes

[left scarab]

thank you

i have another question now

the next question

R^p and R^q denotes the dimension of a vector right

so i just need to figure out the dimension of a vector in nul A and col a

oh yeah that was easy

i just had to get past the wording

.solved

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[thorny copper]

!status

[lone heart]

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

[austere sable]

Hello! I was researching induction proof of the statement that an alternating sum of binomial coefficients equals to zero. I came across a method using Pascal’s identity on MathExchange. But there they use this transformation. First of all on wikipedia it is stated that Pascal’s identity is for positive n and k. And the first term of this sum is clearly has k = 0. Is this legal? Shouldn’t we first move “n choose 0” out of the sum and then apply Pascal’s identity rule? Second, is the statement here that “n choose k” = 0 if k > n or k < 0 is valid? Because if we try to calculate the value of this kind of “n choose k” where k > n or k < 0 we will get negative factorial. And this can’t be right…

[thorny copper]

yes you're correct, it's a notationally sloppy error

[austere sable]

So, am I correct for both issues?

.close

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[chilly mountain]

For c(iii), c(iv) and c(v) I keep getting the answer in terms of 'd' is there any way to solve for d?

[elder crow]

All i need is an explanation as to why the answer is C

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@chilly mountain Has your question been resolved?

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[chilly mountain]

no

[chilly mountain]

.

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[twin nimbus]

@chilly mountain these questions are related to Vieta's formulas, it's perfectly reasonable to get results that depend on d.

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[real reef]

I need help on these two problems. I missed the entire last week of my geometry class, and I do not know how to solve these problems/what theorems to use. If someone could please explain to me how to find the answers, that would be greatly appreciated.

[waxen flame]

Intersecting Secant Theorem for #6.

[real reef]

What is that? I can google it if that would be easier for you.

[waxen flame]

Basically says the measure of the arc is twice the angle of the intersecting secant lines.

[real reef]

So the answer would be 118?

[waxen flame]

👍

[real reef]

Nice, thank you so much!

[waxen flame]

Oh wait no. I'm mistaken.

You should look it up. Its something like (A-B)/2 is the angle but I'm not exactly sure.

[real reef]

Alright, thank you. I will.

[waxen flame]

Where A and B are the arc measures.

[real reef]

I got 139 by doing 59 = 1/2(X - 21)

X being the unknown arc.

[waxen flame]

,calc 59*2 + 21

[ocean seal]

Result:

139

[waxen flame]

Looks good.

[real reef]

Thank you!

[waxen flame]

The other problem uses the Inscribed Angle Theorem.

[real reef]

Alright, I’ll look that one up as well.

I’m getting some mixed results. I searched for Inscribed Angle Theorem and did not see any images that looked like the problem, so I searched for “Tangent Secant Inscribed Angle Theorem” and found the formula <x = 1/2(arcACB)

When I plugged that in, I ended up with 30x = 83 which does not seem right.

Do you know what I’m doing wrong?

I just redid it and got 5, and I’m fairly confident that’s the right answer. Thank you for your help, have a great rest of your day!

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[alpine sable]

im stuck on what cos and tan could be

Sin is 754 square

[low gyro]

do you know SOH CAH TOA?

[alpine sable]

No

[formal kite]

Try these formulea 👍

the symbol $\theta$ is just an angle

[ocean seal]

marcus

[alpine sable]

I still cant find it with that

[low gyro]

sin B = CD/BC=23/sqrt754

[alpine sable]

im sorry I think ill have to skip this question on this practice

sin(B) = 0.2,
cos(B) = 0.98,
tan(B) = 0.2

is that right

?

<@&286206848099549185>

yo

still need help @alpine sable

Yeah man

does those look right?

let me seee

so sin is opposite of hypotenuse

opposite = 23
adjacent = 15
hypotenuse = sqrt754

right

sin = 23/sqrt754

so it'd be 0.8?

@alpine sable

So sin 8.
cos 0.98
tan 0.2?

wait

do u understand why?

[alpine sable]

Yeah you divided 23 with square 75 4

yes but do u know why i did

No

here, i'll send a pic of my work

[alpine sable]

,rotate

[alpine sable]

@alpine sable

tell me f u dont get it

sin(B) = 0.8
cos(B) = 0.5,
tan(B) = 1.5?

@alpine sable

let me check rq

yes

do u get why tho?

[lone heart]

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[crisp stone]

how do i get desmos to graph these equations? https://www.desmos.com/calculator/aemabyc5gx

[crisp stone]

i want it to show combinations of values from 0 to 10 in both variables

question is where do i input them? i dont work alot with desmos so i dont know how to get it to apply to my function

[ancient temple]

Actually, im not so sure what you mean by combinations of values from 0 to 10 in both variables

[crisp stone]

like theres variables R and I in the link i sent

[ancient temple]

Yes

[crisp stone]

i want for example to be able to see 7 R and 3 I or 3 R and 7 I

[ancient temple]

That will be a region of points

Also your function is multivariable here, so a 2D graph will not do you justice in "seeing" those points

[crisp stone]

cant you see two variables when given info about them in a 2d graph?

its like the value when they intersect i think

[ancient temple]

No, in a 2d graph, your domain is just R. One of the axes is used for the input value(singular, not plural), and the other is used to plot what the value of the function is at that point

When you have 2 variables, your domain is R^2, and in order to "see" it, you'd need a 3d graph

Since you have two input values, you'll need two axes, one for each input value

And a third axis for the value of the function at that point

[crisp stone]

well anyways lets just forget graphs for now, is there any software that can help me solve this equation based on a range of values i give it? without the graphing

any site or software

[ancient temple]

What equation?

[crisp stone]

since someone gave me this link to work with i have a hard time understanding the equations written in desmos but what i need to do is just give all the answers of whole values from like 0 to 10 for both r and i

well actually 5 for both

it was 10 in total

[ancient temple]

There is no equation to solve there, you just plug in the r and i values and the function will give you the value at that point

As an example, if R=2 and I=3, f(R,I)=f(2,3)=(3•2^(3))^(10+5(2)) ≈ 4.01*10^27

[crisp stone]

yeah i just need something that does that, im just trying to get help on something someone else asked me to do so im really lost here

lost*

[ancient temple]

Im not aware of any site that will automate that process of plugging in numbers into the function and giving you the output

[crisp stone]

well ill just try to find something out

[ancient temple]

Microsoft excel does a good job at this

[crisp stone]

well ill try to figure it out, how do i free this channel?

[ancient temple]

Type .close

[crisp stone]

.close

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[trim tangle]

find area of this shape

i mean the part not covered by the triangel

[woven plaza]

just subtract them

[storm ridge]

We need lengths of sides

[woven plaza]

area of circle - area of triangle

[trim tangle]

ikl

no measurments doe

thats the thing

[woven plaza]

[storm ridge]

Whaa

[woven plaza]

do they want it in terms of radius or something

[trim tangle]

probably

[storm ridge]

How many variables are you allowed to assume?

[trim tangle]

no info on that

[storm ridge]

Damn

[trim tangle]

π r²-bxh/2

[frigid mirage]

do you know anything about the triangle

[trim tangle]

nope

thats all

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[unborn shell]

can someone give me feedback on this one?

[unborn shell]

given the null and alternative and the significance levesl we are asked to find n and c

I used CLT for this but got that n = 9 which is incredibly small for CLT

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[astral sequoia]

My question is about the value of jacobian at saddle points. At saddle point, is the jacobian zero, if so why?

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[astral sequoia]

.reopen

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✅

@astral sequoia Has your question been resolved?

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@astral sequoia Has your question been resolved?

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@astral sequoia Has your question been resolved?

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@astral sequoia Has your question been resolved?

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[halcyon lake]

i need help with piecewise functions

[halcyon lake]

do i just wait or

uhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[naive valley]

just ask your question

[torpid dirge]

you send a specific question

[halcyon lake]

theres honestly nothing specific, i need help understanding the concept of piecewise functions-- how to graph and evaluate them

and what they are

[small fog]

Piecewise functions are like an “if” statement kinda

If the value satifies this condition, do this. Otherwise if it satifies this condition, do that. etc

[ocean whale]

There are plenty of videos that explain this

[halcyon lake]

i dont get them dawg

i need stupid dumb ppl terms

[ocean whale]

What part don't you understand specifically?

[halcyon lake]

unless u have good recs

[ocean whale]

https://www.youtube.com/watch?v=OYOXMyFKotc
https://www.youtube.com/watch?v=Uzw9tsGq2Pw

[halcyon lake]

cool cool ill check those out

could i get some help with these

my teacher used like a table method and i dont really understand it

[small fog]

Ok

Whats the graph for x^2-4

[halcyon lake]

uh

how am i supposed to know that

[small fog]

Idfk go on desmos or just draw a parabola

[halcyon lake]

ok look im trying ok

ok i did it

[small fog]

So draw that graph for x < 3

Bc thats what the piecewise says to do

[halcyon lake]

so itll stop at 5

or like (3, 5) but why

why 5

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[autumn fable]

help

[lone heart]

@autumn fable Has your question been resolved?

[autumn fable]

nope

[dusty obsidian]

what working have you done so far for this

try putting it into the standerd linier form of y=mx+b

what is b? (the shift up or down)

and what is m? (at what rate does y change when x does (rise/run))

[torpid tide]

loser

[torpid tide]

bro how long has this forum been going on for

[lone heart]

@autumn fable Has your question been resolved?

[lone heart]

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[dusty obsidian]

i got in 6

by putting the whole thing to In()

to make 2x-x=in(6)

x=in(6)

but answers say in3

[summer dirge]

it doesnt work like that

there's no rule for ln(a+b+c)

notably, ln(a+b+c) is not equal to ln(a) + ln(b) + ln(c)

[dusty obsidian]

i see

[summer dirge]

here's my hint

rewrite the equation as (e^x)^2 - e^x - 6 = 0, and let u = e^x

[dusty obsidian]

woh its a quadratic

[summer dirge]

now you can solve it for u, and then sub e^x back in later!

and solve for x

[dusty obsidian]

and the -2 disappears because you cant put negatives into a log so its just 3!

thanks :)

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[dusty obsidian]

my working:

the ^-1 expands to

1+e^x

multiply in ten

10+10e^x

move the ten

10e^x=-7

[tacit arch]

No it doesn't

[dusty obsidian]

i was thinking that might be the issue

[tacit arch]

(1+2)^(-1) does not equal 1+1/2

[dusty obsidian]

oh

that does make sense

is it 1/3 then?

[inner haven]

yes

[dusty obsidian]

ok im going to try this problem again

[tacit arch]

Use (1/a) * a = 1

[dusty obsidian]

10/(1+e^-x)

then e goes to the top

10e^x=3

[inner haven]

you cant rewrite it like that

Unless its just 10/e^-x

10/1 + e^-x = 3

[dusty obsidian]

oh because you add them

i can do (1+e^-x)/10=1/3 right

then *10

(1+e^-x)=10/3

-1

e^-x=10/3 -1

1/e^x = 10/3 - 1

e^x = 3/10 - 10/10

e^x = -7/10

now i cant log it though because its negative

[inner haven]

I would simplify the right side first

so 10/3 - 3/3

[dusty obsidian]

hang on did i make a mistake

how can simplifying first give a different answer

[inner haven]

Cause if you simplify the right first

youll get 7/3

1/e^x = 7/3 --> reciprocal

e^x = 3/7

[dusty obsidian]

i understand how it gives a different answer in the literal sense

but mathematically simplifying at a different point should not change the answer should it

also the answers say that there are two answers

in(3/7) and 7

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[celest spire]

.

[celest spire]

This one is a heck of a question

Guys do u know how to solve this one

[serene ridge]

huh

whats F

[celest spire]

Inmo it's centre of square

In my opinion

[serene ridge]

also i cant visualise 3D geometry

this is hard

[celest spire]

Yeah man agreed

Btw

Arcsin(0.2√30)

Can't be it's value because its not possible

So this option is wrong

[celest spire]

What do u think

[lone heart]

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[rugged crow]

how do i factor something like x^2 - 9x + 14 over 4x

[rugged crow]

not exactly that because i want to solve on my own

but like

similar problem to that

[wet river]

factor top

then see if anything cancels

[rugged crow]

ok so im gonna try solving this

x^2 - 6x + 8 /(over) 3x

[rugged crow]

can u uh nudge me in the right direction

i do not remember this stuff

[wet river]

stop pinging me

im busy

15 min rule

[rugged crow]

im dead lmao

i pinged u once

[wet river]

!15min

[lone heart]

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

[rugged crow]

.close

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[mellow frigate]

how do i approach this question

[lone heart]

@mellow frigate Has your question been resolved?

[placid zinc]

Let's let G be the inverse of A.
What happens when you take Ax = b, and left-multiply both sides by G?

[lone heart]

@mellow frigate Has your question been resolved?

[mellow frigate]

IX = BG ?

[placid zinc]

Remember we're left-multiplying

Ix = Gb

Try not to swap capital letters - we typically use lowercase for vectors, uppercase for matricies

[mellow frigate]

aaah

so

Ix = Gb

[placid zinc]

Ix is just x, of course

So x = Gb
And we've solved the system of equations

Just get the inverse of A, and multiply that with b. This will solve any system where an inverse to A exists

[mellow frigate]

but how do I prove Axi = bi where b is identtiy

[placid zinc]

Take a second to identify what each object here is. What's A? What's x? What's b? Why does Ax = b represent a system of equations?

[mellow frigate]

so this is what I understand. A is basically a matrix like
A11 A12 A13...A1N
.
.
AN1 ...............ANN

x is basically the bariable like xyz so if i have a system of equation like
A1 x + A2y + a3z = 3
A4x + A5y + a6 z = 4
A7x + A8y + A9z = 5
so i can write it as
A1 A2. A3
A4. A5. A6 = A
A7. A8 A9
x y z = X
3
4 = b
5

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[mellow frigate]

.close

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[urban mist]

i need help getting the answer ( i know how to do it i just cant get the right answer )

[urban mist]

please

[quick tapir]

womp womp

[limpid turret]

Post nothing if you have nothing helpful to say

[quick tapir]

mb

[limpid turret]

!show

[lone heart]

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

[urban mist]

So the answer i got is Rory and 0.42 and idk how i got it 🥲

u dont need nothing for c as u can say u have written an explaination

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@urban mist Has your question been resolved?

[urban mist]

yes

[lone heart]

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[hot eagle]

Have to solve for side x to 1 dp

I know how to complete this somehow I just need to make sure I'm labelling these sides correctly

[alpine sable]

have you heard of cosine rule

[hot eagle]

This is trig

I have

[alpine sable]

what is c, a, b?

[woven plaza]

[alpine sable]

and theta

[hot eagle]

Would it be 9 = opp, 20 = ajd and x = hyp?

Or am I labelling this incorrectly

Opp and adj are correct or need to be swapped

[lusty glade]

you dont do opp adj hyp for non right triangles

[alpine sable]

it isn't a right angled triangle

[lusty glade]

use law of cosines

[hot eagle]

Oh my god

Thats the wrong image

[lusty glade]

ah

[hot eagle]

I'm sorry hold on 😭😭

Thats the one I just solved one moment

Soryruroriyou omg

[lusty glade]

x is opp

[hot eagle]

no label = hyp

8 = adj

[lusty glade]

8 is adjacent (non hyp next to angle)

yeap

[hot eagle]

So I'm using tan

[lusty glade]

yes

[hot eagle]

Ookok tan(36) = x/8

My notes are at school I completely forget how to write this out properly one moment, don't tell me

tan(36) = x/8

x/8 = tan(36)
x8 x8

x = 8tan(36)
x = 5.8

I feel like I'm missing a step

[lusty glade]

that looks good to me

extra step when the variable is the denominator

[hot eagle]

Oh damn I actually did it properly

Most success I've made all week ☠️

I've missed the entire week of school due to illness so I'm studying for an uncoming test

I just struggle pretty bad with trig

[lusty glade]

you're not alone, hope you feel better and good luck on the test!

[hot eagle]

Thank you!

.close

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[dawn quail]

not sure what I did wrong here

I checked online and according to this it's correct https://www.symbolab.com/solver/matrix-minors-cofactors-calculator/cofactors \begin{pmatrix}-8%263%26-1\\ 3%26-2%262\\ 9%269%26-5\end{pmatrix}?or=input

[alpine sable]

it seems correct to me

[dawn quail]

weird

[midnight rock]

look

[midnight rock]

3rd row 3rd column

[dawn quail]

it's saying all entries are incorrect

[midnight rock]

the signs

are changed

while they're not supposed to

[dawn quail]

oh ty

and i also had 45 for the bottom left

[midnight rock]

and

[dawn quail]

should be 4

[midnight rock]

45

yes

[dawn quail]

ty

.close

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[distant cove]

Hey, need some help in probability:

given A,B,C events in the same probabilty space, such that A,B are non independent and P(A intersect C) > 0 and P(B intersect C)>0

Prove or disprove:
P(A intersect B | C)=P(A|C)P(B|C)

[distant cove]

i know it's disprove

but i can't think of an example

[royal epoch]

Hey let me try to help you out

Quick heads-up, I'm not entirely certain this will work

But let's just try to set something up using a simple 6-sided die

We could have A be the event that we roll an even number and B be the event that we roll 4 or higher

[distant cove]

ok

[blazing sable]

We can't comment.

[distant cove]

Wdym?

[royal epoch]

Can you think of some C that would disprove the statement?

[blazing sable]

That fact that two events are independent or not has nothing to do with their conditional independence.

[blazing sable]

This statement should be enough to disprove the statement.

[distant cove]

i think about

given 2 dies

event A landed on 1

event B landed on 1

C - the sum is 2

so

p(A|C)=1/5

so Pb,c

so -
P(A|C)P(B|C) =1/5*1/5=1/25

but

P(A intersect B | C)=1/5

is it enough?

[royal epoch]

Good idea, unfortunately P(A|C) now reads as the probability that the dice A landed on 1 given that the sum of the two dice is 2, which is actually 100%

[distant cove]

aah

so

change 3,3,6

[royal epoch]

Sounds good!

[distant cove]

can u continue what u started? im intrested haha

what's ur c even

[royal epoch]

I realised I put it slightly wrong before, but decided not to correct it as you were already going a different path ahha

[distant cove]

i would be happy to hear more

[royal epoch]

What I tried was:
A: rolling even
B: rolling a 6
C: rolling >3

A and B are cleary dependent.
P(A|C) = 2/3
P(B|C) = 1/3
P(A intersect B|C) = 1/3 =/= 1/3 * 2/3

If I didn't make any errors just now ^^'

[distant cove]

but a,b needs to be independent in my question

[royal epoch]

They need to be non-independent

[distant cove]

holy

s***

i wrote it bad

[royal epoch]

Very poor sentence structure in the question haha

[distant cove]

im so sorry

the correct is -
independent

[royal epoch]

Don't worry about it

Well that slightly changes the question haha

[distant cove]

well shit, idk how to use that properlly

i hope it's clear

Given events A, B, and C in the same probability space, such that A and B are independent, and $$( P(A \cap C) > 0 ) and ( P(B \cap C) > 0 ).$$

Prove or disprove the statement:

$$
[ P(A \cap B | C) = P(A|C) \cdot P(B|C) ]

$$

[ocean seal]

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

[distant cove]

yea

sorry

now its good hahaha

[royal epoch]

Okay I think I have another example

We're rolling 2 dice.
A: dice 1 is 4 or higher
B: dice 2 is 4 or higher
C: sum of both dice is 6 or smaller
then:
P(A|C) > 0
P(B|C) > 0
P(A and B|C) = 0

[distant cove]

sounds good

A: Rolling an odd number on a fair six-sided die.
B: Rolling an even number on the same fair six-sided die.
C: Rolling a number greater than 1 on the same fair six-sided die.

so
P(A)=P(B)=1/2
P(C)=5/6

P(A and C)=1/3
P(B and C)=2/3

[royal epoch]

Here A and B are not independent though unfortunately

[distant cove]

aah

u're correct

i hate probabilty theory haha

but i think i got it

[royal epoch]

Yeah, can imagine ahha

[distant cove]

here, what's the value of P(A|C)

and PB|C

1/2, right?

[royal epoch]

uhm, I'm not sure, seems a bit annoying to calculate since you'd have to check for A being 4 and A being 5

I suppose we could make it a bunch easier by just removing the 'or higher' and 'or smaller' everywhere

Then we'd get
P(A|C) = P(B|C) = 1/6
P(A and B|C) = 0

[distant cove]

yeaaa

amazing

omg

thank you so much

i appreciate that a lot

[royal epoch]

You're very welcome haha, it was a journey for me as well ^^'

[distant cove]

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[alpine sable]

what is derivative of y=sinx

[alpine sable]

alright, what about y=sin2x

and how do you solve it?

[blazing sable]

2cos2x

[alpine sable]

why the 2 outside?

[blazing sable]

Because you are differentiating you 2x. Not x

[alpine sable]

I don't understand

[wheat isle]

Do you know the chain rule

[wheat isle]

This is just the chain rule

[alpine sable]

ah

so

is it like this: sin2x = dx/dy(2x) * dx/dy(sinx)

I am confused by the f'(g(x)) bit

so you replace the inside of the function with just x?

isn't it supposed to be just 2x?

yes, that was a typo

so you take the inside out, and then replace it with x? this the chain rule?

.solved

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[alpine sable]

thanks

.reopen

another one

[lone heart]

✅

[alpine sable]

y=sin^3(5x)

this is a two level chain rule?

[blazing sable]

Kinda

[surreal rivet]

yes you would apply chain rule twice to find the derivative

[alpine sable]

so its like this:

5cos(5x)2sin^2(5x)

then you simplify to
10cos(5x)sin^2(5x)

right?

no its not right... hmmm

[surreal rivet]

just check the first step of your chain rule ( where you differentiated the cubic in $(sin(5x))^3$ first)

[ocean seal]

coolioV2

[alpine sable]

3 in front

ahh

so its 15

thank you!

.solved

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[proper dragon]

Hi, I'm studying absolute value, can someone please give me some hint to solve this exercise :( I don't know where to start :(

[lone heart]

@proper dragon Has your question been resolved?

[foggy oxide]

look at the terms that can be added up

[proper dragon]

I don't see any terms :( a,b and c are differents

and a^2 + 1/a^2 = a^4 + 1 / a^2, I don't see how that can helps me :(

[surreal meadow]

@proper dragon para que clase es esto

[proper dragon]

Es para fundamentos de analisis

Estamos definiendo el conjunto de los números reales de forma axiomatica

Hasta ahora hemos visto los axiomas de campo y orden

Además de inducción, sin embargo, ese ejercicio no hace parte de los ejercicios propuestos en el capitulo que trata inducción, sino que esta propuesto en el capitulo de valor absoluto

[surreal meadow]

te mostraron la AM-GM inequality?

la inegualdad del medio aritmetico y el medio geometrico?

[proper dragon]

Como tal en clase no, pero, las aprendí haciendo ejercicios del bartle

es esta verdad?: sqrt(ab) < 1/2 (a+b)

[surreal meadow]

si, <=

te diria que consideres esa como una pista

si no es suficiente avisame

[proper dragon]

Okay, hare lo mejor que pueda

Muchisimas gracias

[proper dragon]

Ya conseguí probarlo! :D

Usando la desigualdad AM-GM probe que 2 <= a^2 + 1/a^2, analogamente para b^2 y c^2, sumando se tiene lo que se quiere probar!

Muchas gracias

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[sick meadow]

How can I use elimination for this question

[left isle]

form a basic linear function to represent f(x)

like f(x)=ax+b

from there find f(f(x))

that should get you started

[sick meadow]

Like this ?

[left isle]

the values you are given are for f(f(3)) and f(f(2))

not f(3) and f(2)

[sick meadow]

Would I make the Fs represent what’s in the equal sign ?

[left isle]

have you found f(f(x)) yet

[sick meadow]

No

[left isle]

do you know how?

[sick meadow]

In the other case I did, it was a bit more simpler since it was just substitution but now I have to turn f(fx)) into a linear eq

And I’m assuming I gotta do that for both eqs given

But idk how to

[left isle]

well f(x) is linear

you don't have to make it linear. it already is

say f(x)=ax+b

then f(f(x))=a(ax+b)+b

[sick meadow]

Ok

[arctic peak]

this isn't strictly a linear function. It might work out for this particular problem, but it's poor phrasing on the side of the textbook.

[sick meadow]

That’s like the 5th time that’s been said for each quesrion I have asked

Like both practice test question and text book has poor phrasing

[sick meadow]

Would I be able to eliminate like this?

[left isle]

uh. well you need equations to use elimination. otherwise you'll just end up with f(f(3))-f(f(2))=1

which doesnt seem very helpful

[sick meadow]

Yeah

[left isle]

if f(x) isn't linear then the problem becomes much harder

[sick meadow]

Cuz when I did it, it = 0

Cause it becomes 1=1

Apparently the answers -1 but I’m not sure how they got to that

So I gotta know what f(x) is

[left isle]

tbh i'm not sure either. i've gotten as far as having equations to work with but....

you're gonna have to assume a form of an equation for f(x) if you want hope of solving it

[sick meadow]

How do I assume it ?

F=f3-2?

[arctic peak]

f(x) isn't linear, it's just of the form ax +b which is a linear polynomial. b should be 0 for it to be a linear function.
But to be fair, I'm just being picky with notation.

[sick meadow]

I tried using ChatGPT to explain it but it says this

[left isle]

Please do not use that forsaken chatbot for math

[sick meadow]

When I’m confused I use it

[left isle]

It is very convincing but doesn't know math

[sick meadow]

Since it sometimes explains it

And it doesn’t make sense here

[arctic peak]

if you assume f(x) = ax+b, then f(f(x)) = a(ax+b) + b which is a^2x + ab + b
we can call: a^2 = d, ab+b = e
so now f(f(x)) = dx + e

from here it should be a system of 2 equations with 2 unkowns

[sick meadow]

Ok I get that

Still a bit lost

[lone heart]

@sick meadow Has your question been resolved?

[sick meadow]

If someone can solve this and just show me the steps I think that would help

Or at the least show it in linear form

that’s a bit more clear

[sick meadow]

.close

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[ashen mortar]

i got D E F right but cant get A B C

[jagged cobalt]

whats the problemo

[ashen mortar]

find the limit

i solved D E and F but i cant get A B C

[jagged cobalt]

i mean whats your problem with it, i can indeed read

is the notation confusing or something

[ashen mortar]

yes

notation confusing

[jagged cobalt]

the - and + represent left limits and right limit respectively

left limits you look what the function approaches coming from the left of the point, right you look what it approaches coming from the right

[ashen mortar]

oh ok

so x -> 2- means approaching -2? or approching 2 from the left

[jagged cobalt]

approaching 2 from the left

[ashen mortar]

and when you mean left, is it literally from the left of the graph

[jagged cobalt]

literally, yes

[ashen mortar]

so is the answer to a) 3?

[jagged cobalt]

it is

[ashen mortar]

oh right i understand now

thank you

[jagged cobalt]

np

[ashen mortar]

.close

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[ashen mortar]

.reopen

[lone heart]

✅

[ashen mortar]

sorry wait 1 more thing

[jagged cobalt]

sure

[ashen mortar]

is 2- means approach from left, and 2+ means from right, then was does x -> 2 mean

[jagged cobalt]

the limit exists if the left and right limits are the same

that would then be just->2

[ashen mortar]

but in this case the limits arent the same right? because x -> 2+ does not exist

[jagged cobalt]

2+ does exist

2 itself doesnt

remember youre not looking at the value of the function exactly at 2, just what its tending to

[ashen mortar]

oh so x -> 2+ is also 3

[jagged cobalt]

nope

look at the graph

[ashen mortar]

1?

[jagged cobalt]

yes

[ashen mortar]

oh ok and since 2- and 2+ are different, therefore x -> 2 does not exist

[jagged cobalt]

correct

[ashen mortar]

alright cool that makes sense

thanks

[jagged cobalt]

nw

[ashen mortar]

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[alpine sable]

I have this question in a textbook and I would like to know which one is correct and why

Which of the following statements is true when differences are found (res-
tas) of two matrices?
a) The matrices must be the same size.
b) The matrices must be square.
c) The matrices must be both row vectors or column vectors.
d) One matrix must be a row vector and the other a column vector.

[alpine sable]

<@&286206848099549185>

.close

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[peak cloak]

a) is right

[alpine sable]

thx

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[high wolf]

Can anyone help me with this problem?

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[lethal belfry]

so I started by finding $U \cdot a$

[rose sigil]

hello @slender gull

HAHAHA

[lethal belfry]

where $*$ is the dot product

[edgy night]

i think u wanna use \cdot :)

[ocean seal]

Why am. I here

Why am. I here

[lethal belfry]

thanks

which is $|a|^2-|a \cdot b|^2$

[ocean seal]

Why am. I here

[lethal belfry]

which is essentially $1-cos^2(\theta)$

[ocean seal]

Why am. I here

[lethal belfry]

where theta is the angle between the vectors

so is it just $|a \cross b|$?

[ocean seal]

Why am. I here

[edgy night]

yup

[lethal belfry]

thanks!

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[tawny crater]

Hi

[tawny crater]

How do I do this

Do I use small angle approximation

[ancient temple]

Easiest way I can think of is by using geometric series expansion

[tawny crater]

huh

u mean a GP

[ancient temple]

Yes

Expand 1/(1+x)

[tawny crater]

but this is a maclaurin

tutorial

[ancient temple]

Ah okay, so use the mclaurin series formula then

[tawny crater]

but how do i apply the " x is so small"

[ancient temple]

You just ignore everything x^3 and onward

They have mentioned "x is so small that x^3 and higher order terms can be neglected"

So do exactly as they say

[tawny crater]

so i just expand to xcube

then the rest i ignore

[ancient temple]

Only upto x^2, since they said even x^3 can be ignored

"all the terms in x^3 AND higher powers can be ignored"

[tawny crater]

oh okay

thank u

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[mellow iris]

how to do c answer is supposed to be 37.1

[lone heart]

@mellow iris Has your question been resolved?

[red shuttle]

1

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@mellow iris Has your question been resolved?

[mellow iris]

?

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[alpine sable]

I need help painting a hallway with these The door is 2 metres and 75cm
And the window is 2 metres and 80cm
And i need to calculate how much paint buckets i need

[alpine sable]

Ye im gonna fail tmr thx for the help tho

[livid sage]

nonsense

[upbeat eagle]

A** for english exam

But F- for Math

[alpine sable]

Whats that math ai called

[upbeat eagle]

???
also a troll?

[alpine sable]

Im just stupid man

[upbeat eagle]

ur question isnt a question

which is solvable

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[uncut siren]

hi i need some help with a polynomial exercise that i dont understand

[uncut siren]

i need apply ruffinis theorem

i dont understand what to do if the same exponent is repeated

[4x³+x-2x³-3x²-1+(2x+1)²] : (x+1)

[mortal trellis]

well you could simplify first

[uncut siren]

yh

[mortal trellis]

or you can first divide the first part, then divide the second part and then add results

[uncut siren]

2x³+3x²+2x+0

it is right?

or 2x³+3x²+4x+0

[mortal trellis]

neither

[uncut siren]

im lost

[mortal trellis]

show your work

[lone heart]

@uncut siren Has your question been resolved?

[uncut siren]

'k

waitm e