help-27

[sand dove]

.

[lost laurel]

this is the OG problem

$V_1 $ has some elements that it doesn't share with $V_2$

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

ok

take one

name it x

[lost laurel]

okay

[sand dove]

V2 is not a subset of V1, this means

[lost laurel]

x belongs to $V_1$ but not to $V_2$

[sand dove]

um

don't take the same one

THERE EXISTS some guy that belongs to V2 but not to V1

who said it was x

and it's not even possible

since x doesn't belong to V2

[sand dove]

.

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

that's what we got from "V1 is not a subset of V2"

[lost laurel]

yes

[sand dove]

now what does it mean for V2 not to be a subset of V1

[lost laurel]

$y \in V_2 , y \notin V_1$

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

Yes

So, what happens with x+y?

(Recall we're trying to prove V1 U V2 is not a subspace)

[lost laurel]

so $x+y \notin V_1 \cup V_2$

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

a counter example would be let $V_1$ be the x-axis, let $V_2$ the y axis

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

their union isn't a subspace of $R^2$

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

Hence $V_1 \cup V_2$ isn't necessarily a subspace

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

No

We didn't want that

We don't want to prove "V1 U V2 isn't necessarily a subspace"

We want to prove "V1 U V2 is not a subspace AT ALL"

It's like I asked you to prove (1+b)² != b² + 2b and you said "Well taking, b=1, we do have 4!=3 so we don't necessarily have (1+b)² = b² + 2b"

[sand dove]

So not this

Stay general

And I want you to show that x+y is NEVER in V1 U V2

[lost laurel]

this?

not too sure tbh

of how to go about proving it

[sand dove]

By contradiction

[lost laurel]

but $x+y \in V_1 \cup V_2$

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

intuitively

[sand dove]

No, you're getting mixed up

$x+y\in V_1+V_2$

[woven radish]

rafilou2003

[sand dove]

That's always true

[lost laurel]

oh, right

[sand dove]

But V1+V2 is not V1 U V2

[lost laurel]

so how do I prove it by contradiction

[sand dove]

Suppose by contradiction that x+y is in V1 U V2

Reason by cases

[lost laurel]

so there are 3 cases

right

[sand dove]

No

V1

Or V2

2

[lost laurel]

I don't follow

[stone stump]

what 3 cases were you thinking of

[lost laurel]

sorry 2

1. Their intersection is only {0}

2. they have a finite intersection

3. One set subset of another

[stone stump]

subspaces nearly never have a finite intersection (only if you are working over finite fields). maybe you are thinking of finite dimensional but that would also be false

well we excluded case 3 from the start

but no, these arent the cases rafilou meant

x+y is in V1 u V2

what does that mean

[lost laurel]

the element x+y belongs to the union of the spaces V_1 and V_2

[stone stump]

ok and what does that mean

[lost laurel]

the element x+y is either in V_1 or V_2 or both

[stone stump]

ok

so lets say it is in V_1

so x is in V_1, y in V_2 and x+y in V_1

[lost laurel]

yes

[sand dove]

So

Can you reach a conclusion on x or y?

Knowing x+y is in V1, and x is also in V1?

[lost laurel]

it belongs to $V_1 \cup V_2$ then

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

No we're getting further from the point

x+y is in V1

x is in V1

Can we deduce something about y maybe?

[lost laurel]

$y$ is in $V_1$

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

Yes, because...

[lost laurel]

their sum is in $V_1$

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

You still didn't tell me what you used

Their sum is in V1, so what?

Why can't y be outside?

[lost laurel]

if it were outside, the sum wouldn't be in $V_1$

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

Why?

[lost laurel]

don't know how to reason it out formally

it just makes intutive sense

[sand dove]

Well this doesn't suffice

x is in V1

x+y is in V1

How do we get y from these values?

[lost laurel]

by subtracting them ?

I don't think we can subtract spaces though

only add them

[sand dove]

Really?

Never heard of additive inverse?

[stone stump]

why are you trying to subtract spaces

x, y, x+y are all vectors

[lost laurel]

I have

[lost laurel]

$x+y-x=y$

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

Ok

And -x is in V1 because...

[lost laurel]

V_1 is closed under addition

[sand dove]

No, that's not how you get -x

How do you get -x from x?

[lost laurel]

by taking its additive inverse

[sand dove]

Yes but we don't have "closure by additive inverse" as a property

So maybe another way?

[lost laurel]

0 belongs to it

to obtain 0 from x,-x has to be an element too

[sand dove]

Again.... you're running in circles

Why couldn't -x be outside?

[sand dove]

Same problem as here

Use a given property of subspaces

Try to write -x differently

[lost laurel]

closure under scalar multiplication?

[sand dove]

Yes!

-x = (-1)x

[lost laurel]

how does that help us though

[sand dove]

-x is in v1 now

Because of closure under scalar mult

[sand dove]

So y is in V1

[lost laurel]

yeah

okay

[sand dove]

So...

[lost laurel]

y belongs in $V_1$?

[woven radish]

ƒ(Why am. I here)=I don't know

[sand dove]

y is in V1, and yet y is not in V1

So

[lost laurel]

we have arrived at a contradiction

thus the assumption is wrong

[sand dove]

Which assumption?

We made multiple

[stone stump]

uhm subspaces are closed under inverses. so no need to do that scalar mult stuff. we dont have to reinvent the wheel everytime

[sand dove]

@lost laurel stated that they did, in fact, needed to reinvent the wheel

They are at the beginning of linear algebra

So while additive inverse closure is true

[stone stump]

even then. by definition a subspace is a vector space and therefore closed under inverses

the subspace criterion is not the definition of what a subspace is

[sand dove]

I'm not exactly sure that's how they defined subspaces but yes

Bc some define it with the closure conditions

[lost laurel]

this assumption was wrong

[sand dove]

x+y in V1 was wrong

[lost laurel]

right

let me think about this a bit

thanks for the help

.close

[devout snow]

Channel closed

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[lost laurel]

.reopen

[sand dove]

btw @lost laurel what is your def of subspaces?

[devout snow]

✅

[sand dove]

And have you seen if subspaces <=> subset that is a vector space <=> closure conditions?

[sand dove]

Ok then nvm

[devout snow]

@lost laurel Has your question been resolved?

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[slender spade]

can you give a solution to this without using the knowledge that 2^(3^8) can fit in there?

from a yt vid btw

[slender spade]

cause that's sort of like knowing the answer already, and applying the answer

[lost laurel]

not really

I mean re-write the RHS as $(2^3)^8$

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

and you're done?

[loud belfry]

There is one more but like its not practical

which is use the lambert W function

[lost laurel]

yeah

or series expansion

,w series expansion of x^x

[woven radish]

Results provided by WolframAlpha

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Requested by physicsrocks

[loud belfry]

its jsut rewriet as e^(xln(x))

and series e^z

[lost laurel]

or that

[loud belfry]

also i dont know if this works at all

but i was trying continued fractions?

[lost laurel]

here?

[loud belfry]

yeah

dont know why

[devout snow]

@slender spade Has your question been resolved?

[devout snow]

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[devout snow]

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[smoky gyro]

Just making sure for part ii), am I supposed to plug in 4.5 for x?

[smoky gyro]

Or do I put 4.5 for y

Oh nvm

.close

[devout snow]

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[lilac widget]

yes guys

give me anything to solve

[smoky gyro]

Blud what are you doing

[lilac widget]

what

im trying to help

[smoky gyro]

No

[lilac widget]

ok

[smoky gyro]

Just go other help channels and help them

This is closed

[lilac widget]

i love tou'bye'

[smoky gyro]

no

[lilac widget]

ok bye

sorry

[devout snow]

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[restive river]

If 2x-3y=3 find the ratio 4^x/8^y

[sonic smelt]

!status

[devout snow]

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

[restive river]

I found y

Insert it i got 3=3

Bc equation is true.

And after that i found x

Put x on a system with y

Got mad

Erased it all.

Moved to another exercise.

[eternal tapir]

hint 4^x = 2 ^(2x) and 8^y = 2^(3y)

[restive river]

Oh brb

[devout snow]

@restive river Has your question been resolved?

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[devout snow]

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[bright juniper]

Log_10(x+1.5)<-log_10(x)

[bright juniper]

Need help solving this

[bright juniper]

Well i moved it to the right and the factored it so now i got

(x+1.5)x<1

[empty flame]

how did you get that

[empty flame]

can you write the steps and send them

[bright juniper]

Ok so

Ill write log_10 as lg bwt

[empty flame]

ok

[bright juniper]

so the original equation is lg(x+1.5)<-lg(x)

[empty flame]

wait

[bright juniper]

lg(x+1.5)+lgx<0

[empty flame]

i wasnt focusing but yes i see what you did

lg(x(x+1.5))<0 so x(x+1.5)<1

[bright juniper]

Yes

[empty flame]

yes that works for sure

so from here what can you do

[bright juniper]

Well we know it can’t be negative so

x(x+1.5)>0

I got those two

But that where im stuck

[empty flame]

work from here

[empty flame]

what can you from here

[bright juniper]

x^2+1.5x-1<0

[empty flame]

ok

[bright juniper]

But the answer from that are terrible

[empty flame]

how did you know

i mean they arent terrible at all

[bright juniper]

What are they?

[empty flame]

,w solve x^2+1.5x-1=0

[woven radish]

Results provided by WolframAlpha

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[empty flame]

these are the roots

[bright juniper]

I see

You are correct i was messing up

Well solved it thanks

[empty flame]

what did you get

[bright juniper]

(0,0.5)

Thanks

I got another one mind helping?

sin^2x=3sinx+a need to solve this if one of its answers is pi/2

[empty flame]

you need to find a ?

[bright juniper]

No x

Well i think ik what to do

First i replace x with pi/2

And find a

[empty flame]

you have one of its solutions which is pi/2 and you need the other solution ?

[bright juniper]

Nvm i think i can do it on my own thanks a lot for the help tho ❤️

[empty flame]

np

but one note

dont forget that you will get infinitely many solutions

[bright juniper]

Yea ik that

[empty flame]

because of the periodicity of sinx

[bright juniper]

Ok

Thanks

[empty flame]

ok then you are good to go

[empty flame]

you did it all i didnt do anything

[bright juniper]

Nah without you i was stuck

[empty flame]

if you are done please type .close

[bright juniper]

.close

[devout snow]

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[devout snow]

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[tidal basin]

what function can I use to show two such curvatures side by side instead of one? (please ping to respond)

[devout snow]

@tidal basin Has your question been resolved?

[nova glen]

$a(x,y)=-e^{-x^2-y^2}-e^{-(x-4)^2-(y-4)^2}$ should work

[woven radish]

esca (@ with reply)

[tidal basin]

it worked

thanks

[nova glen]

np

[tidal basin]

i expected it to combine two functions if i just added them

.close

[devout snow]

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[wispy seal]

anyone knows which forces are being applied?

Im confused and dont know any

[soft umbra]

Friction, gravity

[wispy seal]

can u pls point it on the drawing

[acoustic leaf]

friction to the right, gravity down and left, normal force up

[wispy seal]

oh so both of those arrows pointing are gravity

[soft umbra]

Purple forces are both derived from gravity

[wispy seal]

I see

tysm

[echo sandal]

Wair

I might be mistakened

Gravity is only a downwards acting force

You can’t have gravity horizontally

That’s not how it works

@wispy seal @soft umbra

[quaint citrus]

There are vertical and horizontal components to the gravitational force vector

If that’s what you’re referring to

[devout snow]

@wispy seal Has your question been resolved?

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[lucid crypt]

if i have to expand a natural log like ln(abc)^1/2, is the answer 1/2lna + 1/2lnb + 1/2lnc?

[lusty sapphire]

yes

[lucid crypt]

sounds good

[devout snow]

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[remote sparrow]

I have a random game. I already know the odds of winning, and would like to know if playing this game will give profit on average. Theres a 20% chance to get a 1, a 20% chance to get a 2, a 5% chance to get a 3, a 2.5% chance to get a 4, and a 1% chance to get a 10. Whichever number you roll is the multiplier for how much you win, so if you play for 5 dollars and roll a 4 you get 20 dollars, but theres only a 2.5% chance to do so.

[eternal tapir]

and if none of those happen you lose the game?

[runic prawn]

what happens if you don't get 1,2,3,4, or 10

[remote sparrow]

Yes and the money you bet is taken away

[remote sparrow]

^

[runic prawn]

let's say you bet N

find the expected profit from playing the game

[remote sparrow]

I dont know any formulas to apply

But wouldnt expected profit be 0 since youre more likely to lose the game than win?

Or would that formula be on average

[runic prawn]

negative profit means loss

so there's a -N because you're betting that

[remote sparrow]

Oh ok yea fair

[eternal tapir]

each outcome has an expected portion that contributes to the overall profit/loss

[remote sparrow]

Right

[runic prawn]

then +(0.2)(1*N)

[eternal tapir]

you have 48.5% chance of winning and a 51.5% chance of losing

[runic prawn]

etc

[eternal tapir]

so 51.5% of the time you lose your initial bet

[runic prawn]

if u want to be rigorous use law of total probability or smth

[eternal tapir]

20% of the time you break even,
20% of the time you double your intial bet
etc

[remote sparrow]

But the money is paid before you ever roll, so if you get a 1 you arent profiting youre only getting the money you paid back

Oh you understand

Awesome mb

[eternal tapir]

yes those values sum up to the 48.5

[remote sparrow]

Yes

[remote sparrow]

This is to calculate profit?

[runic prawn]

expected profit yes

[eternal tapir]

i would look up a video or something on expected value

[runic prawn]

profit can be treated as a random variable

[remote sparrow]

So if i bet $5 each time the formula would be (.2)(1x5)+(.2)(2x5)+(.05)(3x5)+(.025)(4x5)+(.01)(10x5)

And if the answer is more than 5 its profitable and if its less its not?

[eternal tapir]

it would be

(.2)(1x5 - 5)+(.2)(2x5 - 5)+(.05)(3x5 -5 )+(.025)(4x5 -5 )+(.01)(10x5 -5) + (.515)(-5)

[runic prawn]

yes

[eternal tapir]

,calc (.2)(1x5 - 5)+(.2)(2x5 - 5)+(.05)(3x5 -5 )+(.025)(4x5 -5 )+(.01)(10x5 -5) + (.515)(-5)

[woven radish]

The following error occured while calculating:
Error: Undefined symbol x5

[eternal tapir]

cba fixing

[remote sparrow]

Wait so is my formula not right?

Yours has X*N-N

[eternal tapir]

it's the same idea

[runic prawn]

your formula gives the expected return from playing the game

not including the bet amount

[eternal tapir]

mine is just relative to 0

[remote sparrow]

Oh i see

So should i go put it in google

[runic prawn]

basically their formula is yours minus N

[runic prawn]

if you want? just calculate it

[remote sparrow]

Yeah so that the answer will come up relative to 0 instead of 5 since i made N = 5

Google says the answer is -0.25

49.75% of people would profit right?

[eternal tapir]

no it means

on average you lose .25$ every 5$ you bet

[remote sparrow]

Which is 5% of your money

So 45% of the population would profit

??

[runic prawn]

for that you want to find P(X>0)

not E(X)

@remote sparrow

[remote sparrow]

I dont have a graphing calculator

And the internets graph looks like crap

So i got a graph where y = p(x>0)

Is that what you mean when you say find p(x>0)

.close

[devout snow]

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[bright juniper]

In a road 2 guys are walking. One of the guys steps are 10% shorter then the others but also 10% more frequent the the other guys. Which one is faster

[barren prawn]

its the same

thats my first assumption

didnt even think about it

[proud perch]

Guy 1 takes 10 steps in 10 seconds and moves 10 meters. His speed is 1 m/s.
Guy 2 takes 11 steps in 10 seconds, but each of his steps are only 90% of Guy 1. 0.9 * 11 = 9.9 meters. His speed should be 0.99 m/s if I'm not mistaken

[devout snow]

@bright juniper Has your question been resolved?

[devout snow]

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[devout snow]

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[devout snow]

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[dull forum]

I have a general question about gauss jordan elimination and augmented matrices

[dull forum]

When simplifying a matrix using gauss jordan how do you determine the best number to apply to rows when multiplying and dividing. Is there a way to determine the best number or is just just have a knowledge of basic arithmetic

[devout snow]

@dull forum Has your question been resolved?

[dull forum]

no

<@&286206848099549185>

[outer yarrow]

If I get 1 more @ I’m going to cry

[dull forum]

lmao sorry man i need help finite killin me

[outer yarrow]

it’s alr man

I’m trying to sleep lol

[dull forum]

my bad can i not ping someone else?

[barren prawn]

no

there is no best number

[dull forum]

i always seem to use the wrong numbers

I dont know how

I also struggle knowing when the best times to add subtract multiply and divide is

[barren prawn]

there is no wrong numbers also

with practise you get better

[barren prawn]

as long as you are getting to the reduced form ofc.

[dull forum]

My point is i always end up with ridiculous solutions which end up being wrong in the constant matrix or I multiply or divide by a number which changes one of my 1s or 0s to another real number like 3,4 you know what i mean?

I get the reduced form but the answer is still wrong

[barren prawn]

idk man you gotta show me some of your work to see what u are doing

[dull forum]

okay sorry give me like a minute i can get something up

Work starts in the top left and moves down

The math seems to check out in that I ended up with reduced form and homework said it was wrong

I didnt label it either but my first step was switching the first and third row

[barren prawn]

you checked for calculation errors?

[dull forum]

Not thoroughly but quick glance it seemed good

[barren prawn]

i see one

in third matrix drawn you have -18 + (-13 * -3 ) = 17

but -18 + (-13 * -3 ) = 22

otherwise your method is good

[dull forum]

It helps to know im doing it right but i still struggle on everyone that I do in the same way I cant get the math right to produce the right answer so I wasnt sure if there was a way to check the numbers other than going through them over and over until its right

[barren prawn]

maybe just do the calculations slower, you have the method figured out and its better do go once slow then do it wrong and then check again three times and still not being sure

[dull forum]

okay gothca

appreciate the help

.solved

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[elfin widget]

Can someone walk me through please?

[viscid ivy]

Hm

sin^-1(-x) = -sin^-1(x) you know this ?

So it will be - sin^-1(√2/2)

[elfin widget]

.solved

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[echo sandal]

ABCD is a trapezium

EAC IS A straight line

write an expression in terms of X for cos <BAE

Don’t understand how you get to -7/x

[radiant anvil]

cos(180-x)=-cos x, for 0<x<90

[devout snow]

@echo sandal Has your question been resolved?

[echo sandal]

Got it thx

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[lost crag]

should here the "=" sign be replaced by "≡"?

[lost crag]

≡ is equivalence as i know it right?

= is just equal value,

so when we have a lot of possibilties like with groups ≡ seems more suitable?

[sand dove]

you can have equality between sets

[lost crag]

whats the difference

i said as much myself

but im also asking if ≡ is more suiting here

[sand dove]

the sentence "when we have a lot of possibilities" doesn't make sense to me

[lost crag]

with Venn diagrams i can illustrate principly totally different sets

can i put ≡ instead of =?

for all of them

[sand dove]

no, rather =

since

A set contains given elements

[opal cloak]

I dont know nothing about sets, but i'd say equivalence by definition, makes more sense

[sand dove]

so no matter how I write {2,3,4} as {n in N, 2<= n < 5} or something else

[lost crag]

yeah exactly

[sand dove]

We can say statements are equivalent because when we're in a certain configuration, they're either both true or false

so yes no matter the "lot of possibilities" for configuration A and B will say either both "true" or "false"

here for sets

[lost crag]

those are indetical, therefore ≡

[sand dove]

we only have one possibility

[lost crag]

not just equal

but identical

[sand dove]

what does equal mean to you?

[opal cloak]

i think u got it mixed up

[sand dove]

equal means identical in ALL aspects

[opal cloak]

2+2 is equivalent to 4, but 4 equals 4

[sand dove]

whereas equivalent means identical in a CERTAIN aspect

[sand dove]

nono you're getting more mixed up

2+2 does equal 4

[lost crag]

5=5, as in objects, specific stuff

but for theorems like this one we say sin2x+cos2x ≡ 1, because we know it's always true

[sand dove]

by the definition of addition, 2+2 is exactly the same thing as 4

[lost crag]

as in, identical sides

[sand dove]

but, the "=" sign specifically says both sides are identical

and yet you see that 2+2 = 4

that means that 2+2 and 4 are identical

[lost crag]

oh

i thought equal < equivalance, in terms of levels of equality

[sand dove]

no, the opposite

here's one way to see it

[lost crag]

then what's the difference between = and ≡ in meaning?

[sand dove]

take statements P : "1 < x < 5" and Q: "2 <= x <= 4"

and I investigate those statements when x is an integer

well

P ≡ Q

[lost crag]

theyre not equal

only rarely

[sand dove]

.

[lost crag]

thats bs, only sometimes P≡Q

[sand dove]

P ≡ Q over the integers, because no matter the integer x I pick, P and Q will either be both true or both false

[lost crag]

you cant generalize like that, you need to give the range of x

[sand dove]

Ahem, x in an INTEGER

[sand dove]

I wrote this countless times

[opal cloak]

i think u wrote that once

[sand dove]

and here

[lost crag]

oh i see you're right

[sand dove]

so

P ≡ Q over the integers, because no matter the integer x I pick, P and Q will either be both true or both false

[opal cloak]

i mean those statements are quite different

[lost crag]

so they're equivalent

but why not equal?

[sand dove]

they're the same on the PERSPECTIVE that x is an integer

what if x is not necessarily an integer?

[lost crag]

i mean the value of the statements P and Q is identical

[lost crag]

oh

but listen

[sand dove]

so we had to RELY on the perspective that x is an integer to assert that P and Q mean the same thing
so they're only equivalent

for 2 objects to be EQUAL, they need to mean the same thing NO MATTER the perspective

so x+x for example

and 2x

No matter where x lies
if the addition and multiplication by 2 are defined the way we do

then x+x = 2x

[lost crag]

what if i take
$(P is \forany x \in N. 1<x<5) \rightarrow (Q is \forany x \in N. 2<=x<=4)$

[sand dove]

this is even more flagrant with sets

[woven radish]

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

[lost crag]

what if i take
$(P is \forany x \in N. 1<x<5) \rightarrow (Q is \forany x \in N. 2<=x<=4)$

[woven radish]

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

[sand dove]

so P is the statement $"x\in \bN \wedge \ 1 < x < 5"$ and Q is the statement $"x\in \bN \wedge 2 \leq x \leq 4"$

[lost crag]

i see

[sand dove]

oops wrong side

[woven radish]

rafilou2003

[sand dove]

well here since for P and Q to be both true, you added the condition of x in N, this SETS the framework

meaning you're forcing a UNIQUE perspective

[lost crag]

so why is
cos^2(x) + sin^2(x) ≡ 1?

it should be equal as i see it

[sand dove]

where did you see that?

[lost crag]

let me check if it was wikipedia

[lost crag]

yes it wasa video i think i have saved then

[sand dove]

in any case

equivalence is weaker than equal

so if P = Q, then P ≡ Q

[lost crag]

so ≡ is "identical" under certain circumstances

and = is "always, under all perspectives and possibilities, identical"

?

[sand dove]

yes

[lost crag]

i see

[sand dove]

so when you create sets

two sets can be equal

since their definition will not depend on the circumstance

[lost crag]

but if i defined x to be a natural numbers in the previous P, Q example you gave me, does that make it "="?

[sand dove]

like

$A = {x\in \bN, 1 < x < 5}$ and $B = {x\in \bN, 2 \leq x \leq 4}$

[woven radish]

rafilou2003

[sand dove]

what is A ?

it's just {2,3,4}

same for B

[lost crag]

same group

[sand dove]

so A = B

[lost crag]

so =?

i see

[sand dove]

they always have the same elements

[lost crag]

and what if i have 2 statemnts that are equivalnt sometimes? as in sometimes
P=True, Q=True, for a "certain" x, does that means they're sometimes equivalnet?

as in let's say i do in a proof a case where x=0 and i get both P=Q, can i say they're =, or ≡?

[sand dove]

well again it depends on where you study those statements

and how they're defined

ofc some things are tautologies in disguise

[lost crag]

yes, in those cases it's "="

[sand dove]

but if you have a doubt for STATEMENTS

it's never too cautious to use "equivalent"

[lost crag]

but let's say for 2 statements i know for a fact aren't always identical for every x value

[lost crag]

but if i divide let's say x to cases in a proof (x=0, x>0, x<0), do i say when x=0 for example when P and Q are identical (both either True or False) that theyre "≡", or "="?

in this case i think it's safe to say "="

since this is a subcase that is very clearly constrained

[sand dove]

you can say P(x) = Q(x)

since True = True

[lost crag]

great, then i think i get it

yeah exactly

ty mate for explaining everything

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[arctic cloak]

3d geom

[devout snow]

@arctic cloak Has your question been resolved?

[radiant anvil]

it has to also pass through some (a,0,0), (0,b,0), (0,0,c) where a+b+c=0 and a is not 0

[devout snow]

@arctic cloak Has your question been resolved?

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[raven swift]

help?

[raw knoll]

hi

U can start by just multiplying everything by the bottom parts ie

7-2(2x)/(x+1)=x/2

and then change the x onto the top which should be a lot easier to solve

@raven swift

[raven swift]

huh?

[restive river]

cross multiply everything

multiply it all by 2x, x+1, and 4

eliminate the denominators

[raven swift]

each fraction?

[raw knoll]

yup

just make them integers basically

[raven swift]

so 7-2=.5?

[raven swift]

i have been trying but i cant find a common denominator

[raw knoll]

what

[raw knoll]

you don't need to

[raw knoll]

^

that's the first fraction

then just repeat the other two

(x+1)7-2(2x)=x(x+1)/2 etc

[raven swift]

i do that and get -3x-7=-2x^2-2

all under 4

[raw knoll]

then u can just multiply both sides by 4

then solve

[raven swift]

why? they are all under 4 so i can just equal them, no?

[raw knoll]

also I don't think that's right...?

[raw knoll]

yeah that's the same thing lol

[raven swift]

look im not understanding this at all

[raw knoll]

ok wait

[raw knoll]

^do u see what I'm doing here

[raven swift]

not like that

you have to use that $ form

or smth like that

[raw knoll]

huh

you want me to use latex?

[raven swift]

yea

[raw knoll]

$7-\frac{2(2x)}{x+1}=\frac{x}{2}$

[woven radish]

Kai The Catgirl

[raw knoll]

I mean like generally it doesn't matter if you get it...

[raven swift]

yea i got that

[raw knoll]

do u see what I did

[raw knoll]

then just continue multiplying everything by the denominators

basically

what I mean is like 1/4 *4 to make it an integer

[raven swift]

i got 4x-7x-7=(-2x^2+2)/(4)

[raw knoll]

alr lemme check brb

[raven swift]

alr

[raw knoll]

I seem to get x²-5x-14=0

[raven swift]

how?

[raw knoll]

I'll show the working one sec

[raven swift]

alr

[raven swift]

yea and multiply left side by 4

[raw knoll]

wait this is almost right you just removed an x

[raw knoll]

yeah

[raven swift]

and what do you get?

[raw knoll]

Yeah exactly what I said-

[raw knoll]

anyway this is almost right but when u multiplied 2x by -1 you got -2 instead of -2x

[raven swift]

what? if u multiply left side by 4 u get 12x+28

[raw knoll]

I divided by 2 first

[raven swift]

why?

[raw knoll]

there's a common factor

[raven swift]

?

[raw knoll]

might as well simplify to deal with smaller sums

the top of the right side has a multiple of 2

So just cancel that with the bottom

[raven swift]

can i show u how im doing it?

[raw knoll]

go ahead

[raven swift]

give me a sec

[raw knoll]

your method is right but you just forgot an x

4x-7x-7=(-2x^2+2x)/(4) would be right

(it's all the same, yours is just multiplied by -2)

[raven swift]

since its a -4/x+1 do i multiply by -(x+1)?

[raw knoll]

uhm u can just multiply by (x+1)

[raven swift]

ok

@raw knoll

[raw knoll]

when u multiplied 2x by x+1 you got 2x+2

just change it to 2x²+2x and it'll be right, then u can just solve

[raven swift]

yea i did do that

but im not getting what ur getting

[raw knoll]

it's the same

since you get a multiple of mine=0 it's the same answer

the only difference is that mine is simplified

[raven swift]

?

[raw knoll]

okay wait make yours integers first then equate it to 0

so like ax²+bx+c=0

[raven swift]

give me a min kai

@raw knoll u here?

[raw knoll]

yeah?

[raven swift]

thats wgat i did

[raw knoll]

I don't see it...

[raven swift]

wdym

wanna see?

[raw knoll]

yeah

[raven swift]

@raw knoll

[raw knoll]

Your 2x-12x isnt simplified

Also you get 0=2(x²-5x-14)

[raven swift]

2x^2

[raw knoll]

that 2 is supposed to be 2x, I reminded you like 4 times

[raven swift]

yea 2x^2-12x-26=0

[raven swift]

wait how?

[raw knoll]

it's -10x

[raw knoll]

^

[raven swift]

OHHHH

[raven swift]

so 2x^2-10x-26?

[raw knoll]

brother are you not reading my messages 😭

[raw knoll]

yeah then u can do this

0=2(x²-5x-14)

[raven swift]

sorry

[raw knoll]

-28

[raven swift]

and divide by 2

[raw knoll]

^

[raven swift]

and get x^2-5x-14?

,calc 26/2

[woven radish]

Result:

13

[raw knoll]

exactly my answer

[raven swift]

no -13

not -14

[raw knoll]

no it's 28

[raven swift]

how? -26-2?

[raw knoll]

you took away 2 for the 2x

[raven swift]

oh

yea

@raw knoll one more quick question

can i ask?

[raw knoll]

okay then solve using quadratic eqn

[raw knoll]

go ahead

[raven swift]

so im going to take my finals for alg 2 and this is like a sample, can u check and tell me how it feels?

https://www.nysedregents.org/algebratwo/618/algtwo62018-exam.pdf

[raw knoll]

I don't know what year alg 2 is

but uhm it feels decently easy?

u should be fine

just be careful, you seem to be very careless

[raven swift]

but im stuck with probabilty and statistics a liitle

like u see question 34?

[raw knoll]

yeah what about it

[raven swift]

for the first one u would do 23-18?

[raw knoll]

it's just the interquartile range calculations but for 95% instead of 50%

[raw knoll]

probably

that's what it's asking for, right?

[raven swift]

and if the answer lies in the 95% interval it is not significant?

[raven swift]

ur the smarter one

lol

[raw knoll]

I don't know phrasing there lol

[raven swift]

wdym?

[raw knoll]

phrasing everywhere is different I can't be sure that's what they're asking for lol

[raw knoll]

yeah

[raven swift]

so do 23-18 tho, right?

[raw knoll]

I think so???

[raven swift]

alr then

i think i got it

take it easy

kai

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[raw knoll]

gl

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[last egret]

aye we had the same number in mind

I am having problem with this

what should i do it is suppose to find a continous function

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@last egret Has your question been resolved?

[trail eagle]

I'm not sure I get what the question is. Could you restate it?

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[frosty turret]

Two questions

The first red highlighted line:
One is allowed to choose a point pn1 such that d(p, pn1) < 1 because p is a limit point of E, right?

[candid timber]

yes

[frosty turret]

For the second highlighted line:
There are infinitely many point in the neighborhood of p and so there must exist a ni > ni-1 such that d (p , pni) < 1/i
Does it matter if the ni process continues infinitely?
Like what if I found a ni+1 > ni
And also, don’t I have to get d( p, pni) < epsilon?
Why do they just leave it as 1/i?

[candid timber]

You continue on this process so that for each i you have an n_i such that d(p,p_ni)<1/i. Then for any epsilon>0, there is a natural N such that 1/N< epsilon (aka the Archimedean property of the reals).

And for all n_i >= N, you have d(p,p_ni)<epsilon

[frosty turret]

So before I write pni converges to p I should write. There exits an N such that 1/N < epsilon. Thus,
d( p, pni) < 1/i <= 1/N < epsilon for all ni >= N. Therefore, pn on converges to p.

[candid timber]

I mean yeah, you could do that. Although the construction of such a sequence already implies the convergence. Yeah, it's implied

[frosty turret]

would I be able to put down d(p, pni) < 2/i and by just getting this i can say pn conv to p
Asking to see what I can and can’t do

[candid timber]

Yeah you can do that. The point being is you want to construct a sequence of numbers such that their distance from d is decreasing to 0. You could even take d(p,p_ni)<10000000000000000000/i, or d(p,p_ni)<1/2^n

[frosty turret]

Got it
Thank you for all the help

🙂

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[heavy current]

[heavy current]

I'm a bit stuck with this

if x is an irrational number in [0,1], how do I ensure that it is in A?

clearly, I need to use the fact that A is closed, but how exactly eludes me at the moment

I'd appreciate any hints

[candid timber]

Do contrapositive

Show R\A is a subset of R\[0,1]

[heavy current]

hmm

[wicked turtle]

do you know the result that the rationals are dense in [0,1]?

[heavy current]

I do

[wicked turtle]

you can use that

[heavy current]

[candid timber]

R\A is open and the rationals are dense

[wicked turtle]

take an irrational x in [0,1]
since the rationals are dense in [0,1] you can find a sequence (x_n) of rationals in [0,1] that converges to x
now how can you use the given facts about A?

[candid timber]

Oh wow this is simpler than what I had in mind

[heavy current]

thinking...

[heavy current]

since A is closed, it contains all of its limit points - in particular, it contains x?

I think that would work?

[arctic field]

[heavy current]

ig my only issue is that

since A is closed, it contains all of its limit points

is not something Spivak has said

but it's smth that I know from elsewhere

[arctic field]

what has spivak said about closed sets

[wicked turtle]

oh i see, what's his definition of closed?

complement is open?

[heavy current]

yes

[candid timber]

Then my hint is another way to go then

[arctic field]

show that the complement is empty then

[heavy current]

hmm

[wicked turtle]

(complement relative to [0,1])

[arctic field]

i would think you're just working in [0,1]

a way to characterise dense sets is to show that every ball intersects it nontrivally

[autumn fjord]

yar

[heavy current]

oh hi smay

[autumn fjord]

helo

[heavy current]

I will not lie: I kind of want to just move on from this problem via this argument

[arctic field]

[heavy current]

I'm starting to care less whether Spivak has said smth or not 😭

[arctic field]

it's a fine argument

[autumn fjord]

what is the definition of closed in your book

[heavy current]

the complement is open

[heavy current]

but... I know this fact from elsewhere so...

can I just use it and move on

[autumn fjord]

you can just prove that a set is closed iff it contains its limit points

[arctic field]

do you know how to prove it

[heavy current]

I can try to prove it

[autumn fjord]

you can also take some irrational in A^c

and argue via the fact that A^c is open that x isn’t in [0,1]

[arctic field]

this feels a lot like exactly the same as proving that a set is closed if it contains limit points

[autumn fjord]

okay, what is the exercise though

[arctic field]

maybe only if direction

[autumn fjord]

[heavy current]

maybe I will try this...

[arctic field]

the proofs will verbally be the same i reckon

[autumn fjord]

just use something similar to show that A is closed iff A contains its limit points tbh, this exercise is useless

[heavy current]

okay, I shall try

[lusty sapphire]

$A=\overline{[0, 1]\cap\bQ}$

[woven radish]

SWR

[devout snow]

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[honest thorn]

Integrate 3x^2 * sin(x^3)

[honest thorn]

i first differentiated sin^2(x^3)

which gave me 6x^2 * cos(x^3) * sin(x^3)

so i need to multiply by a factor of 1/2cos*x^3)

i did it wrong though

[lost laurel]

$\int 3x^2 sin(x^3)dx$?

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

well, can you see a nice u-sub?

[honest thorn]

i havent learnt that yet

i only learnt to like add a exponent and just test differentiation to find answer

[lost laurel]

then this is going to be pretty hard imo

like you can't let $x^3=u$ and solve it?

[woven radish]

ƒ(Why am. I here)=I don't know

[lost laurel]

haven't learnt that yet?

[honest thorn]

ok ill learnt tit now hten

but the ttext tbook im using is giving me this uestion before the part where they teach the.u sub

[lost laurel]

hmm

then try differentiating -cos(x^3)

[honest thorn]

im learninig iit rn

thanks

.close

[devout snow]

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[unborn atlas]

I know this is physics but the physics server is asleep and this is just trigonometry basically, so i determined that s has a magnitude of 6.4, and that it has a direction of 51.3 degrees from the +x axis, beyond that I don’t know what to do

[unborn atlas]

It’s question 10.4

[lost laurel]

,rotate

[light pecan]

thanks mate

[unborn atlas]

Any thoughts

[lost laurel]

part a or b

[unborn atlas]

Both

[lost laurel]

well, can you define the dot product for me

[unborn atlas]

sfcos(phi)

s is 6.4

I have no idea what f is

And I have no idea how to get the angle between them

[alpine geode]

well if you want to do it the hard way, f = sqrt(f_x^2 + f_y^2), where you know f_x and f_y is the unknown. cos(phi) can also be expressed in terms of f_x and f_y, but it's long and convoluted.
There is another way to define the dot product for 2 vectors tho, try looking at that one

[unborn atlas]

Yeah and I get f_y =23.1

But that’s wrong

Is F not 26?

[unborn atlas]

Also wdym hard way what other way is there

[alpine geode]

this one's better

going back to the hard way, cos(phi) = cos(arctan(5/4) - arctan(f_y / f_x))

[unborn atlas]

14.8?

[alpine geode]

correct

[unborn atlas]

Yeah but why don’t I get 14.8 the other way

And how am I able to directly add Newtons to Joules

I’m brand new to physics

[alpine geode]

where are you adding newtons and joules ?

[unborn atlas]

26J= 4m* -12N+5m* xN

[alpine geode]

one newtonmeter is one joule

[unborn atlas]

O

Wait so like I said I’m brand new to physics here so when you say joule and newton I am clueless, I thought they are both units of energy or something

So what is a joule in simple terms

[alpine geode]

a joule is a unit of energy

[unborn atlas]

Force per meter?

[alpine geode]

force * meter

[unborn atlas]

Then what is a newton

[alpine geode]

a unit of force

force isn't energy

[unborn atlas]

Why does multiplying a force by a distance make energy

[alpine geode]

that's just how it's defined but you can maybe think of it this way. When you have a box, you can think of force as the force with which you push on it. The further you want to push the box, the longer you need to apply that force. In the end, the force * the distance gives you the energy you needed to move that box

[unborn atlas]

So I can say it takes 25 joules to move a box 5 meters with 5 newtons of force

Okay well anyways

Why don’t I get 14.8 the other way

[alpine geode]

yeah

show how you got there

[unborn atlas]

$$26N = \sqrt{-12^2+F_y^2}$$

[woven radish]

Nathan

[unborn atlas]

Square both sides

$$676= 144+x^2$$

[woven radish]

Nathan

[unborn atlas]

$$532=x^2$$

[woven radish]

Nathan

[alpine geode]

nonono

the magnitude of the force isn't 26N

you don't know the magnitude

[unborn atlas]

Then what is 26N

[alpine geode]

you only know that the work done by the force over s is 26 Nm

[unborn atlas]

Wdym

26N * 1m

What is the 26 N

[alpine geode]

26 Nm

that is a unit of energy

it's the work done

you can't equate newtons to Nm

[unborn atlas]

But it’s over 1 meter

Wait is it saying 26N times 1m or 26N dot m

[alpine geode]

same thing

it's just that it's a unit of energy

[unborn atlas]

How is it the same if n dot m is nmcos(phi)

[alpine geode]

cosine has no unit

[unborn atlas]

?

What do you mean

[alpine geode]

give me a min

[unborn atlas]

You there?

[alpine geode]

one sec. doing the long way

[unborn atlas]

Ok

[alpine geode]

so this is how you do the long way, correctly

now, back to your question. In physics, when you multiply two quantities that have units, you get a quantity with a new unit, that is the product of the two

for example, multiplying Force by distance, gives energy (that's how it's defined)

when you calculate the work done by F over s, you multiply a force by a distance, giving newton meters

[devout snow]

@unborn atlas Has your question been resolved?

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[unborn atlas]

.reopen

[devout snow]

✅

[unborn atlas]

Misclick

[unborn atlas]

What’s the short way then

[alpine geode]

what you did earlier

-12 * 4 + 5 * y = 26

you can see that it ends up being the same, but one is just faster

well anyways it's probably better if you meditate on it, I'm tired so I'll go to sleep if you really have trouble with this send me a dm i'll answer when i wake up

[devout snow]

@unborn atlas Has your question been resolved?

[devout snow]

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[devout snow]

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[visual jewel]

Stuck at how to find ALPHAt

[devout snow]

@visual jewel Has your question been resolved?

[devout snow]

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[lost laurel]

$3^x+3^y=234$?