help-10

[slate zephyr]

what about it?

[gusty pelican]

is there a name for this process

[slate zephyr]

I would call it finding a counter example by consturcting a model.

[gusty pelican]

ah okay thank you

[slate zephyr]

however

have you noticed anything

[gusty pelican]

okay wait sooo the first statement was true

and the second statement was also true

so they are logically equivalent

[slate zephyr]

😧

yes

but

only for this P and Q

We chose bad P and Q.

[gusty pelican]

wottttt

[slate zephyr]

I will let you alter one of them in a way to make them different.

[gusty pelican]

wdym?

[slate zephyr]

that we did not find a counte rexample where they disagreed.

[gusty pelican]

how? is x not supposed to be {a, b}

[slate zephyr]

So from our computation we might even start suspecting that they are same.

[slate zephyr]

still that

but different P or Q

[gusty pelican]

what do you mean by different P or Q?

change P is true only when x=a

to P is true only when x=b?

[slate zephyr]

We could define P always true or P also true when b

[slate zephyr]

yes

[gusty pelican]

so that type of change

hm

[slate zephyr]

youz will need to try different variants

[gusty pelican]

I need to change it until I find a counterexample?

[slate zephyr]

yes

[gusty pelican]

if I don't find a counterexample

[slate zephyr]

OR you would need to proof that they are same.

[gusty pelican]

I can say that they are logically equivalent

[slate zephyr]

well, you need to proof no counter example exists.

Probably eaisest to argue in some way why they are same.

I assume you larnt some transofmrations where they stay identical?

De Morgans rule etc

[gusty pelican]

yes

but I'm not really adept at them

[slate zephyr]

[gusty pelican]

why

[slate zephyr]

(you probably already noticed but this is one of my favorite symbols to put)

[gusty pelican]

I know P -> Q is ~P v Q

[slate zephyr]

Why "why"

[gusty pelican]

well I am only 8 days into this course

so I don't know many things

¯_(ツ)_/¯

[slate zephyr]

one step at a time

[slate zephyr]

^ let's see why they are different first

Here we do not need to transform

it suffices to find some situatiosn where they disagree

[gusty pelican]

x from {a,b}
P(x) = {true if x=b}
Q(x) = {true if x=b}

should I try with this

I'm short on time tho

😬

[slate zephyr]

how so

[gusty pelican]

I have to go to school tomorrow

[slate zephyr]

generally rushing is bad for learning.

:/

You could try figure it out tomorrow

[gusty pelican]

it's okay

I think I've learned a lot today

[slate zephyr]

let's see

[gusty pelican]

thanks for the help

[slate zephyr]

no that won't do

[gusty pelican]

better than my professor

[slate zephyr]

still both true

[slate zephyr]

[gusty pelican]

change Q?

Q true when a,b

P true when a

[slate zephyr]

It should be a good thing but you make it sound like a bad thing by putting "at least"

[gusty pelican]

my bad

I'll remove it

[slate zephyr]

nah don't worry

[gusty pelican]

:B

[slate zephyr]

hm

yeah

[gusty pelican]

professor doesn't go over these hard examples in class

[slate zephyr]

now it will disagree

I think

[slate zephyr]

^

Q=true, P only true for a.

[gusty pelican]

∃x(P(x) → Q(x))

P(a) -> Q(a)
true

P(b) -> Q(b)
true

so there exists is true

∀xP(x) → ∃xQ(x)

for all x we have P(x)
but P(a) is true and P(b) is false

and then....

[slate zephyr]

yes

SHIT

they still agree

[gusty pelican]

ah

[slate zephyr]

perhaps you should put the "at least" back

[gusty pelican]

are you sure there is a counterexample

do you just know that bc you are good at this

[slate zephyr]

Yes, as you can see i definitely never make mistakes.

[gusty pelican]

lol

[slate zephyr]

Still I think they are different.

Okay

But at this point I need to fidn them...

[gusty pelican]

how many combinations do we need to try

[slate zephyr]

16

[gusty pelican]

P only true when b
Q only true when a
is the same thing as before right?

[slate zephyr]

4 for P and 4 for Q

Some are very similar and hence redundant to try.

[gusty pelican]

P only true when b
Q true for a, b

[slate zephyr]

P = false does not work, becausew both sides true.

Wait okay

so we can see

the statement on the left implies the statement on the right.

we can only hope for the right statement true and the left statement false.

[gusty pelican]

so the combination for that would be

wait

how do we make that false

isn't there no way we can make that false

[slate zephyr]

P = true also des not work.

I am actually confused myself now.

I think there is no reason why they should be the same but I also can't see the counterexample yet.

wait

they are the same.

Proof

Assume the left statement is true.

This either means that one x have P(x) is false

then the right is trivially also true

otherwise the left statement implies that there exists a Q(x).

Hence the right is also true.

in the other direction.

Assume the right statement is true.

then if for all x P(x) is true

we can find an x such that Q(x) is true

This x will also do on the left

Otherwise there exists an x where P(x) false

[gusty pelican]

I see....

[slate zephyr]

hence the left side is true for that x because of viscious truth.

So they are the same.

This really comes as a surprise to me.

This goes to show how implications in the middle of stastements with discious truth can throth a wernch into our intiotion.

[slate zephyr]

I wonder if this equivlance has a name.

[gusty pelican]

O.o

so you haven't found anything that makes it not equivalent

[slate zephyr]

I believe i have actusally proven them to be equivalent.

If you can show A => B

and B => A

then that means they are equivalent

[gusty pelican]

ah

I gotta go D:

[slate zephyr]

I that case if one is true

the other is true too

So they cannot differ.

[slate zephyr]

Alright

[gusty pelican]

thanks so much for your help

[slate zephyr]

I think we finished anyway

[gusty pelican]

have a good day

[slate zephyr]

I need to digest my defeat also...

[gusty pelican]

loooool

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[radiant spindle]

how i can prove that a < f(x) < b if x is constrained by c < x < d? like the concept in general

[wooden cipher]

This isnt always true

[radiant spindle]

i meant more like

for a specific function f(x) = ... , if it asks me to prove that f(x) < b for x < a specific number, how should i go about proving it algebraically?

[tacit scarab]

find the maximum

[radiant spindle]

so lets say
if 2 < x < 4 for f(x) = x^2, and it asks me to show that f(x) < 17
how should i go about finding the maximum without using limits? (this is a proofs course)

like i can't just sub in 4 into f(x) to show this because x is defined as having to be less than 4?

[tacit scarab]

If you can't use differentiation you can try to prove that f is increasing

[radiant spindle]

can you elaborate on that

[tacit scarab]

Prove that f is increasing at the interval 2<x<4

Then plug in f(4)

Which equals 16

Then f(x) < 16 because f is increasing

[radiant spindle]

ohh so you would still plug it in

[tacit scarab]

Yes

[radiant spindle]

thank you

[tacit scarab]

No problem

[radiant spindle]

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[radiant spindle]

.reopen

[obtuse pebble]

✅

[radiant spindle]

sorry for reopening this but can you explain why you need to prove that f is increasing? (and how you would do so without differentiation?)

would u still need to plug in values for this? (aside from the final part where u plug in f(4)

[tacit scarab]

An increasing function f have this property:
If a<b then f(a)<f(b)

So if x<4 f(x)<f(4)=16

To prove f is increasing, compare f(x) and f(x+a), where a is positive

[radiant spindle]

so you would still plug in a number for a then to prove that its increasing?

[tacit scarab]

You need to prove it for all a

So don't substitute a number to it

Let's do f(x) = x as an example

Let a be positive
f(x)=x and f(x+a) = x+a

Obviously x<x+a

thus f(x)<f(x+a)

So f(x) = x is increasing

[radiant spindle]

ohh

[radiant spindle]

so as f(x) is increasing, it doesn't reach f(4) before x = 4.

[tacit scarab]

Yep

[radiant spindle]

thus u have proved that f(x) < 4 for x < 4?

[tacit scarab]

Exactly

[radiant spindle]

thank you so much

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[timid silo]

When do you not need to use absolute value for ln|x|?

[timid silo]

My professor told me today that this is for particular solutions for differential equations (where you just need 1 solution)

[sage geode]

Usually the absolute value is put there so that the domain is not so restricted

[timid silo]

hmmm, in the integral 1/x dx she said it wasn't necessary since it's positive.

see, no absolute value there. is that train of thinking wrong?

[sage geode]

Well yeah because |x| is the same as x for positive x

So it doesn't really matter which one you write I suppose

[timid silo]

what about here? why is there an absolute value here?

[sage geode]

Well is it given that y(x) > 0 for all x? (or y(x) < 0)

[timid silo]

oh i see

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[zinc tundra]

Suppose we know that the number of items produced in a factory during a week is a RV with
mean 500. Find an upper bound on the probability that this week’s production will be at least
1000? If the variance of a week’s production is known to equal to 100, then give a lower bound
on the probability that this week’s production will be between 400 and 600? (Hint: Moment
Inequalities)

[zinc tundra]

should we use markovs inequality for the first part??

[polar fossil]

the chebyshev ineq requires that the variable be nonnegative

[zinc tundra]

and?

<@&286206848099549185>

[polar fossil]

oh yeah i guess you can't produce negative units yeah that will work then

[zinc tundra]

markovs?

what does it mean by upper bound ..at least thousand?

P(X>=1000)<=?

this?

[polar fossil]

yeah

and yeah Markov and Chebyshev are two names for the same inequality

but Chebyshev sounds cooler lol

[serene fossil]

technically they arent the same

because markov is a special case of chebyshev im pretty sure

actually thats not even true

they are just different theorems

in fact markovs is used to prove chebyshev

[zinc tundra]

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[vast glade]

finding derivative using implicit differentiation

what’d i do wrong here?

[polar fossil]

the chain rule wasn't done right

derivative of sin(πx) is πcos(πx)

[zinc tundra]

differentiation of sin(pi x)

yeah

[obtuse pebble]

@vast glade Has your question been resolved?

[obtuse pebble]

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[twilit onyx]

how to do d ?

[twilit onyx]

rest is p easy but i am not able to eqaute x

I get a eqaution in form of ax^4 - bx^3 +cx^2-dx

[obtuse pebble]

@twilit onyx Has your question been resolved?

[formal wigeon]

you need to find the points of the graph that satsify the question

ie they intersect

at least thats what i think the question asks

[twilit onyx]

no I need to fond the definite value for x

in which f(x)=inverse f(x)

[formal wigeon]

yeah

[twilit onyx]

so is that just intercept

?

[formal wigeon]

i imagine so

wherever they intersect take the x value there

but its asking for exact value

[twilit onyx]

cause they intercept at (1,1)

and one more point

[formal wigeon]

are you sure they intersect at 1,1?

[twilit onyx]

the solution sheet says says x=(3+underoot 5)/2

oh wait they dont

[formal wigeon]

so you would just set

1 + underoot x

equal to the inverse

[twilit onyx]

they interest at some point beyond 2 before 3

on both

[formal wigeon]

yeah its ~2.68

but i cant remember how to get that value

[twilit onyx]

hmm btw this is a non gdc question

[formal wigeon]

wdym gdc

[twilit onyx]

calculator

or digital graph

[obtuse pebble]

@twilit onyx Has your question been resolved?

[obtuse pebble]

@twilit onyx Has your question been resolved?

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[gusty wraith]

Hi. I have a question:
"Let A, B, and C be three points that are not in line. Draw a point D which is equidistant from all three points!"
I asked chatgpt and the answer is on the picture. is that correct?

[latent walrus]

dont use chatgpt for maths

[gusty wraith]

any idea or keyword to find the answer?

[kind hawk]

surprisingly it is correct as far as I see. but yes in general dont use chatgpt

[gusty wraith]

Aa I see... thank's for confirming!

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[timid silo]

In how many different ways can eleven people be seated at a circular table with eight seats, if three are left waiting?

[timid silo]

I think it could be
11!/(11-(8-1))! = 11!/4!
or
(11-1)!/(11-8)! = 10!/3!
or
(11-1)!/(11-(8-1))! = 10!/4!

or maybe (11!/3!) x 7!

Which one is correct and why? Or why they are all incorrect?

I think it should be 10!/3!
Because in circular permutation you eliminate 1 for n, and n in this case is 11

<@&286206848099549185>

[wide rose]

C 8 from 11?

11x10x9x8x7x6x5x4?

wait

is it circular for rotational symmetry?

then 11!/3!/8?

[timid silo]

mm well, I know that you would have to eliminate 1

to account for one fixed place

[timid silo]

yep

mm no

P

Is permutation right?

The order matters

[wide rose]

isn't 11!/3!/8 right

because there are 8 sets which cancel to only 1 for a circle

[timid silo]

Wdym

[timid silo]

I don't get this

[wide rose]

is 1 2 3 4 5 6 7 8 = 2 3 4 5 6 7 8 1

there

[timid silo]

No

[wide rose]

why?

i mean as people

[timid silo]

I mean yes
I thought you put 1 2 3 4 5 6 7 8 = 2 1 3 4 5 6 7 8

Which matters is the order

so it's not combinations

[wide rose]

so

[timid silo]

is Variations

[wide rose]

so

there are always 8 (7 other) variants

for 1 ordered

[timid silo]

Also you need to take into account they are in groups of 8 and there's onw fixed place

[timid silo]

yes, but there are 11 people

[wide rose]

-_-

[timid silo]

so those are just the places available

[timid silo]

what's that

?

[wide rose]

imagine

a1 - a8 are selected people

let's work with that

[timid silo]

Yes I know

[wide rose]

what is the ratio of a1-a8 to a1-a8 in a circle?

[distant flax]

can someone help me find the 3 quartiles for 5,2,1,12,6,9,9,10?

[wide rose]

#❓how-to-get-help

[distant flax]

o

[timid silo]

wdym by that

[wide rose]

ways of alligning people(a1-a8) in a line

and in a circle

spoiler

it is 8

[timid silo]

in a line
If you have only 8 people, there's 8!

[wide rose]

i mean

[timid silo]

It's a variation wothout repetition

[wide rose]

in distinct order

[timid silo]

Yes
variations are for different orders

Because it matters the order

so you would have 8! different variations having 8 people in a line

[wide rose]

that's right

and for circle?

[timid silo]

that is permutations

and here

you need to eliminate 1

[wide rose]

so we...

[timid silo]

Having only 8 people

no 11

here the difference is that there are groups

[wide rose]

just do it for 8 people

[timid silo]

a group of 8

[wide rose]

ok

[timid silo]

if I do it, I need an explanation to discarding the fact that I could choose from 11 people first

[wide rose]

lets make it super ez

[timid silo]

actually 10! to take into account the fact that one place is fixed in a circle to make it relative to that place

[wide rose]

i want to show that, u can know the ways you put people in a circle, knowing how the would sit in a line

[wide rose]

lets use 2 people and 2 sit

[timid silo]

Ok

[wide rose]

how many ways for line and circle?

[timid silo]

this is permutations without groups

[wide rose]

i don't get it

[timid silo]

for a line there are 2! or just 2

[wide rose]

yes

and

[timid silo]

For variations you have groups ot you pick

for permutations you don't have groups

[wide rose]

i think i get it

but

[wide rose]

selecting 8 people

and arranging them

are 2 steps

not a whole

i mean, you multiply

[timid silo]

In both the order matters, but for variations n and k are different
in permutations n and k are equal

[wide rose]

now I'm confused

[timid silo]

mm but this is prior knowledge to solve this

[timid silo]

why

I want to understand your point of view

I want to learn

[wide rose]

ok

my point is

we know all of the variants of putting 8 people in a line

(forget about 11 rn)

[timid silo]

Yes is 8!

for a line

[wide rose]

then you know in how many ways you put em in a circle

[timid silo]

and 7! for a circle

yes

this is circular permutations

[wide rose]

oh

ok

then multiply by the ways u select 8 people from 11

[timid silo]

why would you multiply 11!/3!

[wide rose]

because

[timid silo]

why (11!/3! )x 7!

how do I know that's correct

[wide rose]

check for simpler examples?

[timid silo]

ok

[wide rose]

oh wait

[timid silo]

n = 3

k = 2

[wide rose]

nvm

good luck

[timid silo]

in a circle of two spots

[timid silo]

why

[wide rose]

why not

[timid silo]

because luck is not a factor to resolve a problem

[wide rose]

I'm just being kind

or life doesn't matter

[timid silo]

why

[wide rose]

what why

[timid silo]

life doesn't matter

[wide rose]

O_O

cause you are asking why

wishing luck is just a sing of kindness

and gives motivation

[timid silo]

okay, good luck

I 'll post it again

.close

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[sly gorge]

hlo

[sly gorge]

i want to ask simply does this mean P(A/B) or P(B/A) i think theres some misunderstandin

or way to interpret

?

?

ik this is dumb q but still help me

[tardy epoch]

do you know the definition of independence?

[sly gorge]

independence means when p(A intersection B)= P(A) * P(b)

@tardy epoch i answered , sry for ping.

@tardy epoch

[tardy epoch]

yea so calculate that for your A and B

and the intersection

[sly gorge]

logically it should be like :- P(B intersection A)= p(b) *p(a/b) ?

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[undone sentinel]

i got 4.6

is that right

my textbook doesnt have a solution for this example so idk if im doing it right

[vast willow]

,calc sqrt(29)

[warm shale]

Result:

5.3851648071345

[vast willow]

how do you calculate distance between 2 points ?

[undone sentinel]

oops i subtracted 4

the distance formula

[vast willow]

$\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}$ right

[warm shale]

calculus is fun

[undone sentinel]

yea

[vast willow]

ok just sub

,calc sqrt(21)

[warm shale]

Result:

4.5825756949558

[undone sentinel]

okay so i did it right

[vast willow]

did you round ?

[undone sentinel]

ye

[vast willow]

ok wrong

[vast willow]

this is the right answer

you have to add not subtract

[undone sentinel]

oh

[vast willow]

there is a + not a - between the paranthesese

[undone sentinel]

i accidentally subtracted

[vast willow]

.

yea

you should've added instead

thats all

[undone sentinel]

okay thanks

these problems are easy my eBook isnt displaying shit right so maybe im doing it wrong

[vast willow]

ohhh

[undone sentinel]

am i tweaking or is there not a single natural number here?

[karmic heron]

Is sqr of 36 not a natural number?

[undone sentinel]

is square root anything irrational?

[karmic heron]

All good

[timid silo]

shat is the square root of 36

[undone sentinel]

6

[timid silo]

so

[undone sentinel]

but its in square root form so doesnt that make it irrational

[timid silo]

how?

no

[karmic heron]

No

[undone sentinel]

okay

i understand

what about 0

[karmic heron]

1 to infinity

[obtuse pebble]

@undone sentinel Has your question been resolved?

[obtuse pebble]

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[frail depot]

Are these calculations right?

Mostly the $e^{(qx)^2}$ part

[warm shale]

SimonWin

[frail depot]

I forgot to put parenthesis on the picture but they are now

or shouldnt it be the whole thing actually

${e^{qx}}^2$

[warm shale]

SimonWin

[frail depot]

Aah... yes

so it's really

$(e^{qx})^2$

[warm shale]

SimonWin

[frail depot]

So like this, and then they cancel out, and I could possible even do it before the 2qx part...

But wolfram says they're equal

So

[daring rock]

Wolfram says what is equal?

[frail depot]

when i multiply in

[daring rock]

Yeah

[frail depot]

So, that should be fine, right

[daring rock]

Yeah, that looks good now

[frail depot]

Cooool! Aweeesome!

[daring rock]

You did need the brackets around the exponential, to answer your original question lol

[frail depot]

Yes, like I did above

[daring rock]

But yeah you got it

[frail depot]

Look (23)

[daring rock]

Oh ya I see now

[frail depot]

Which makes sense because it's (e^{qx})^2 not the q^2 too

So it completely makes sense!

Verry nice!

Thank you! ❤️

[daring rock]

Np 👍

[frail depot]

🫶

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[wooden edge]

How can I evaluate the limit of the sum to infinity of (3^n - 2^n)/5^n

[wooden edge]

where n starts at 1

[reef inlet]

$\frac{3^n - 2^n}{5^n} = (3/5)^n - (2/5)^n$

[warm shale]

exophades

[royal basin]

^

what you have here is the sum of two geometric series

[wooden edge]

oh im so dumb.. sorry for wasting your time

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[wooden edge]

.reopen

[obtuse pebble]

✅

[wooden edge]

thanks though

.close

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[round wing]

Hey can someone guide me through this

[round wing]

I'm not sure what Im doing wrong

I multiplied te^t to each component and integrated

[tepid mountain]

You would be able to do it in an easier way if you considered
$$(-5i-j+5k)$$
As a constant vector that can be moved out of the integarl

[warm shale]

Sherif Player

[round wing]

so v1 * integral (te^t)?

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[round wing]

i got it

[obtuse pebble]

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[drifting merlin]

So I'm supposed to solve this system of linear equations. There are more unknowns than equations, so does this have a solution? Am I supposed to use the x = xp+xn thing here?

[drifting merlin]

Any help would be appreciated.

[obtuse pebble]

@drifting merlin Has your question been resolved?

[sage geode]

There are infinitely many solutions and in order to write them down you will need to express some amount of unknowns in terms of the other variables (in this case, you will have 3 unknowns dependent on other numbers)

E.g. x - y = 1 has infinitely many solutions for the same reason (there are not enough equations), but you can express every solution with the form (x, x - 1) once you see that y = x - 1

[obtuse pebble]

@drifting merlin Has your question been resolved?

[drifting merlin]

Ah so I need to express it in a general form.

Thank you.

[obtuse pebble]

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[mortal lava]

I just want to know whether or not ( 48m^2n) and ( 12mn^2) are like terms or not

[tight patio]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@mortal lava Has your question been resolved?

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[ashen perch]

I just need help with PT2 No. 1

[obtuse pebble]

@ashen perch Has your question been resolved?

[ashen perch]

<@&286206848099549185>

Help what do I do with number 1?

[obtuse pebble]

@ashen perch Has your question been resolved?

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[jovial fern]

how would u factor this

[tardy epoch]

,w roots x^4 -24x^2 +x+132

[warm shale]

Results provided by WolframAlpha

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

Requested by rie.mann

[tardy epoch]

I suppose you plug in small integers to check

[zenith spade]

Rational root theorem and trial and error are your best friends

[wispy wadi]

^

[tardy epoch]

Then polynomial division

[jovial fern]

thought there mightve been a faster method

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[autumn marsh]

someone teach me how to do this please 🤞🙏

[high lily]

which one?

[autumn marsh]

all three

[high lily]

do you understand function notation

[autumn marsh]

yeah

idk what to do with the table

[high lily]

the table tells you the corresponding value f(x), g(x) value at each value of x

not really that different from other tables

e.g in that first table

first column of values

[autumn marsh]

how do i know which number to choose

[high lily]

1 and a 3 below it, indicates that
f(1) = 3

[autumn marsh]

yeah

[high lily]

you choose values based on what the question asks for

you are asked to find
$$g(f(2))$$

[warm shale]

ℝam()n()v

[autumn marsh]

OH

is that right

@high lily

[high lily]

yes

[autumn marsh]

AAAAAA

how about the bottom

[autumn marsh]

scuse me

r they right

[pale scaffold]

,rotate

[pale scaffold]

Looks correct

[autumn marsh]

thanks broski

[obtuse pebble]

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[warm reef]

Can someone help me with this question

[obtuse pebble]

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[empty cypress]

yes, and all the notes are correct as well

[open forge]

Alright, thank you!

[empty cypress]

np

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[open forge]

.reopen

[obtuse pebble]

✅

[open forge]

I also want to confirm two other questions.

It is defined at 0 (0,0) so D and E are eliminated. B is simply the shape, A would be the shape of an absolute value graph so that's elminated.

I'm confused about C.

[empty cypress]

sure doesnt look like it has a vertical tangent

one of (A) or (B) is true

[open forge]

We haven't covered vertical tangents, and from what I see on Google, I think C should be eliminated.

[empty cypress]

answer: ||It has a corner because the derivative coming from the left is not equal to the derivative coming from the right||

cusp is when theres a vertical tangent

its a cusp if the two sides intersecting makes an angle of 0 degrees

the definitions are all a bit silly, but this one is clearly a corner

[open forge]

Oh, I understand it now.

[empty cypress]

i gotta go soon but if you have another quick one to check i have some time

[open forge]

Let f be the function given by f(x)=|x|. Which of the following statements about f are true?
I. f is continuous at x=0
II. f is differentiable at x=0
III. f has an absolute minimum at x=0

I have I and III since, both LH and RH limits equal f(0) and it has no transformations so abs min is 0 for the mother absolute value function.

[empty cypress]

yeah thats right

at x=0 its continuous but not differentiable

[open forge]

Alright, thank you for the help! Have a great rest of your day/night.

[empty cypress]

you too

[open forge]

Thanks!

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[proven musk]

Uhh when you horizantally stretch a function by b is it stretched ed by 1/b or b?

[karmic mural]

b tightens it

makes it go "faster"

1/b stretches out

goes slower

[proven musk]

That doesn’t answer the question

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[lean sedge]

can someone help?

[obtuse pebble]

@lean sedge Has your question been resolved?

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[uncut tree]

.help

[obtuse pebble]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[uncut tree]

I was doing this ealier

And someone told me to find the mid point

I dont know how to figure out the other end point for this

[unborn ice]

Okay

It seems pretty simple

[uncut tree]

I think it is but our teacher didn’t go into depth about this😭😭

[unborn ice]

Do you got a graph of coordinate?

[uncut tree]

No I’ll graph it rn tjo

[unborn ice]

HMMM

Seems like it already told you where the mid point is located at

[uncut tree]

Yeah

[unborn ice]

Quadrant 3rd

[uncut tree]

I need to find the other end point

[unborn ice]

Wait

I get it

You can use the distance between the mid point and coordinate K

It says its a mid point of those two coordinates

[uncut tree]

Yeah

Then

[unborn ice]

Let me graph it

[uncut tree]

-17??

Yupp it is

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[ebon tendon]

I know I need to setup an equation and solve for the coefficients that multiply to equal the U vector but cant seem to get it right. Any Help.

[obtuse pebble]

@ebon tendon Has your question been resolved?

[obtuse pebble]

@ebon tendon Has your question been resolved?

[hot sonnet]

set the equation a*s1 + b*s2 = u. Add all components on the lefts side so you get (vector in terms of a and b) = u

then solve for a and b

given that they are 2 unknowns, you just need 2 equations

find a and b, those values must make as1+bs2 = u for each component. If at least one component doesn't match, then there's no linear combination that satisfies a*s1+b*s2=u

[ebon tendon]

so take 6a+4b=6?

[hot sonnet]

yes

you need one more eq.

[ebon tendon]

-7a +6b = 25

[hot sonnet]

ok, now solve for a and b

[ebon tendon]

4b=6-6a

b=3/2-3/2a

Plugging that into the first equation

6a+6-6a = 6

0=0?

[hot sonnet]

that happens 'cause you are using the same equation

you get 0=0 which is always true. It's like saying a=2, then plug that into "first eq." and get 2=2

That's why you need 2 different eq. , so you can use the other one for plugging in

[ebon tendon]

I got it! thank you so much!

.close

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[stuck herald]

can i consider two different rectangles like this or why not?

will it be either but not both at a time?

[tacit scarab]

both cases are possible

so you need to calculate both cases

[stuck herald]

so two types of equations?

x+2y=180 or 2x+y=180

but how should i go abt this?

@tacit scarab

[tacit scarab]

the area too

[stuck herald]

yeah its same 3600 =xy

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[dire nacelle]

Hi, I am currently stuck on this part of my question. I believe it is the fact that I don't quite understand what I am trying to do in this question.

[dire nacelle]

I am not sure what "Plot V as a function of R-naught" means

[tacit scarab]

we need more context

[fathom flicker]

show both part (a) and the original question

[tacit scarab]

treat V as a function of R_0

and plot it

the function is given

V=1-1/R_0

it's the same as y=1-1/x

[dire nacelle]

.close

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[stuck herald]

50/3 how to convert this to mixed fraction

[stuck herald]

.close

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[spark wasp]

can someone please explain the span of this vector
i spent over 5 hours trying to get help

and some dude would just ignore my questions

@grizzled shore

can you please tell me what hte span is

if ur still here

[tacit scarab]

span of a set of vectors v1,v2...,vn is the set of vectors of the form a1v1+a2v2+...+anvn

[spark wasp]

yea

but i cant get a and c in terms of x y and z

so wtf do i even do in this case

[tacit scarab]

techincally this is the answer

except that they are linear dependent

so you can simplify them

[spark wasp]

wdym

what i dont get is the asnwer lol

[spark wasp]

what is?

[tacit scarab]

the span = {2a+b,a+c,2a+b}

[spark wasp]

yea

but we have to say

in R2

[tacit scarab]

the vectors (2,1,2),(1,0,1) and (0,1,0) are linearly independent

[spark wasp]

R3

etc..

[tacit scarab]

oh you just need to determine the dimension

[spark wasp]

but prove it

as well

[tacit scarab]

the dimension equals the number of linearly independent vectors

linearly independent means that if a1v1+a2v2+...+anvn=0 then a1,a2,...,an = 0

i.e. no vectors can be written as linear combinations of other vectors

[spark wasp]

like right here

i proved that you can get a and b in terms of x and y

meaning the span is R^2

[spark wasp]

i dont understand i need to be explained in dummy language

if you get what i mean

[spark wasp]

how?

they are dependent

2a+b=0

2a=-b

2a+(-2a)=0

0=0

[tacit scarab]

oh

mb

I meant to write dependent

why are you solving for A and B and C though

[spark wasp]

to see if you can get it interms of x y and z

[tacit scarab]

the span might not contain them

[spark wasp]

thats how to find the span

[spark wasp]

like i did here

[tacit scarab]

it only works if the vectors span the whole space?

I think

[spark wasp]

so what do we do when it isnt

[tacit scarab]

how are you taught？

[spark wasp]

using this method

we havent covered

matrix and stuff

this is like the second lesson

we also have to say if its a plane

or a line

[tacit scarab]

the method seems useless though

[spark wasp]

to tell if its a line or plane we see eif its a multiple

[spark wasp]

so what do i do

in this situation

do you want to see my notes?

[tacit scarab]

find the maximum linear independent set of vectors

[tacit scarab]

that'll help

you want to prove there are only trivial solution of ax+by=0

it comes down to simultaneous equations

[tacit scarab]

,rotate

[tacit scarab]

I see

[spark wasp]

,rotate

[spark wasp]

the notes have examples for kids and the homework is brutal

[tacit scarab]

lol

so you know how to prove lienar independence

[spark wasp]

yea

[tacit scarab]

ok

find a linearly independent set of vectors

the dimension equals the number of remaining vectors

[tacit scarab]

in your question the span is 2 dimensional for 3 dimensional vectors

so it's a plane in R3

[spark wasp]

what

how can i do that

when the vectors are dependent

[tacit scarab]

so you have 3 vectors x,y,z

[spark wasp]

yea

but one vector is a combination of the others

[tacit scarab]

yes

so pick any one of the vectors

and just remove it from the set

[spark wasp]

wdym

[tacit scarab]

repeat until you have a linearly independent set

[spark wasp]

could you provide an example

[tacit scarab]

let's say we have vectors [1,1] and [2,2]

[spark wasp]

ok

[tacit scarab]

obviously 2[1,1] = [2,2]

[spark wasp]

yea

so this is a line

not a plane

[tacit scarab]

let's choose [1,1] and remove it

we're left with [2,2]

[spark wasp]

ok

[tacit scarab]

so span([1,1],[2,2]) = span([2,2])

[spark wasp]

so we just solve for span of [2,2]?

x=2a y= 2B

[tacit scarab]

???

[spark wasp]

A[2,2] = [x,y]

[tacit scarab]

no

[spark wasp]

wdym no

[tacit scarab]

I don't get why you are doing tha

t

[spark wasp]

to see if we can get a in terns of x and y

x= 2a y=2a

[spark wasp]

to prove we can get a in terms of any x y combination

but with [2,2] we know that it spans the entire R2

[tacit scarab]

R1

[spark wasp]

this tells us it spans r1 then?

since we got a in terms of x

and a in terms of y

[tacit scarab]

I don't get what you're doing

[spark wasp]

Im doing what my prof is doing

[tacit scarab]

but if you get a linearly independent set, then the dimension of the span is the number of vectors in the set

[spark wasp]

a is a coeff

[tacit scarab]

what's the point of it though

you should solve a[2,2]=[0,0]

to prove its linear independence

[spark wasp]

but why

[tacit scarab]

it tells us the dimension of its span

[spark wasp]

how does knowing if its linear independence

show u its span

[tacit scarab]

the span is just ax+by+cz

but I thought you needed its dimension?

[spark wasp]

yes

[tacit scarab]

the span is just this

[spark wasp]

man why is this so confusing

[tacit scarab]

idk man

ok let's organize everything

span(v1,v2,...,vn) = {a1v1+a2v2+...+anvn}

if the set of vectors are dependent, we can remove any from the set and have the same span

if v1,v2,v3 are dependent

then span(v1,v2,v3)=span(v1,v2)=span(v1,v3)=span(v2,v3)

once you have a independent set, the dimension of span = remaining vectors

[spark wasp]

i dont understand why

how will it be the same

if depedent means the coeff is not all 0

for the vectors

[tacit scarab]

if a set is dependent, then any one of its vector can be written as linear combination of other vectors

say we have x = [1,1] and y = [2,2]

y=2x

span(x,y)={ax+by}={ax+2bx}={(a+2b)x}=span(x)

or geometrically

the vectors pointing at the same direction and thus describe the same line

[spark wasp]

ok

[tacit scarab]

even if you remove 1 vector they still represent the same line

or thinking it backwards

if you add 1 vector on an already existing line you don't get more dimension

it's still the same line

[spark wasp]

ur example doesnt make sens to me

where did u get y =2x

[tacit scarab]

[2,2]=2[1,1]

[spark wasp]

you and the guy are confusing me in #linear-algebra

[tacit scarab]

?

[spark wasp]

there is a guy in #linear-algebra talking the opposite of u

[tacit scarab]

we are literally saying the same thing

[spark wasp]

ok

so

i get this part now

now how can you tell if each other are indepdent or depdent

because i only know how to tell if the whole set is linearly independent

[spark wasp]

okay

lets say

we remove

1 vector

span(v1,v2)

now we see if this is independent?

[tacit scarab]

yes

remove until independent

[spark wasp]

ok i did span(v1,v2)

and that is also

dependent

[tacit scarab]

what's v1 and v2

[spark wasp]

[2,1,2],[1,0,1]

now we remove 1 vector

[2,1,2] is also dependent

so what do we do now

[spark wasp]

u here?

[tacit scarab]

no

[spark wasp]

this is independent

but

v1 v2 v3 how are u gonna figure if that is independent

or depdent

since u get 0=0

[tacit scarab]

?

set av1+bv2+cv3=0

[spark wasp]

2a+b=0
a+c =0
2a+b=0

[tacit scarab]

and look for non trivial solutions

[spark wasp]

2a = -b

-b+b =0

0=0

-2a=b

2a-2a=0

0=0

a+c =0

so u cant do anything

[tacit scarab]

???

why you doing -b+b and 2a-2a

b=-2a

c=-a

[spark wasp]

b=2c

[tacit scarab]

thus we have solutions av1-2av2-av3=0

[spark wasp]

uh what?

[spark wasp]

yeah i get this

[tacit scarab]

a+c=0 so c=-a