help-27

[radiant drift]

cause like if ur given a question like find the GDF/LCM of 3a+5 and a+8, traditionally ways wont work

ig euclidean alg only gets u GCF

[devout snow]

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[past crystal]

How would you prove two triangles are congruent if they all of their angles are congruent?

[plush knot]

I don't think you can, those are similar triangles
To be congruent, they also have to have an identical side

[prime egret]

Similarity doesn’t imply congruency

[past crystal]

Thank you guys

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[fluid pagoda]

can someone help me with this

should i start by squaring both sides?

[tepid dust]

Yes

I think -rooting both sides

Then changing the greater than to less than. After that it’s important algebra I think 🤔

[hollow pollen]

?!

[plush knot]

You can't with inequalities, unless you consider each case (sqrt(x-4) is positive, sqrt(x-4) is negative, -sqrt(x+2) is positive, -sqrt(x+2) is negative)

Here's why:
-9 < 5, but if you square them:
81 < 25, which is clearly wrong

[tepid dust]

Welp looks like I’m dumb

[plush knot]

I'd suggest the method I've been taught, but I've seen other people explaining easier ways to solve inequalities like that, so I won't

[bright burrow]

Squaring should work if we take intersections with their domain tho?

[fluid pagoda]

i think this ineq does not have solutions

[bright burrow]

The lhs is non negative always, and the rhs is never positive

[plush knot]

Nice

[fluid pagoda]

but in this case left side is always bigger than right

right?

[bright burrow]

Yea

[fluid pagoda]

so we basically just solve the ineq

[bright burrow]

So there is a solution, not sure abt the squaring method tho, ive not used it in a long time

[fluid pagoda]

we don't have to consider both cases

[bright burrow]

Theres nothing to solve

[fluid pagoda]

but why

[bright burrow]

Dont we already have the answer?

🤔

[fluid pagoda]

yeah but i don't understand

[plush knot]

sqrt(x - 4) is always positive, -sqrt(x + 2) is always negative

(something positive) > (something negative)
When is this true?

[tepid dust]

Your supposed to flip the inequality

[bright burrow]

Sqrt (x-4) non negative is better wording

imo

[fluid pagoda]

always

[tepid dust]

When u devide both sides by -square root it changes the inequality

[fluid pagoda]

which is still the same

@plush knot is right i understand it

[bright burrow]

Not exactly

[fluid pagoda]

but from now on idk what to do

[bright burrow]

I mean this is always true, however in this case you have to consider something else too

[plush knot]

there's one last thing to notice, those square roots may not be always defined

[fluid pagoda]

so this is not true?

i'm getting confused

[bright burrow]

Follow this

[slim roost]

Move all to left then see

[fluid pagoda]

if i move them to the left what do i doo after

[bright burrow]

What no😭

[bright burrow]

Consider the domain of the square roots

[slim roost]

Let me check

[plush knot]

Here's what going on
You know that that inequality is true, when the left and right parts are defined
You can't have undefined < something (or undefined > something of course lol)

And square roots are not defined when you take the square root of..??

[slim roost]

Ah not working

[fluid pagoda]

dom right side is x>=4 left side is x>=-2

it's the opposite sorry

[plush knot]

yes nice

So now, when are those both defined at the same time?

[fluid pagoda]

x>=4

[plush knot]

Yup

So that's the final answer

[fluid pagoda]

what

but photomath said there is not solution

[plush knot]

it does

I mean, there is a solution

We just found it

[fluid pagoda]

yeah you are right]

sorry

[bright burrow]

Photomath dumb

[fluid pagoda]

no no i was wrong

thanks a lot guys

[plush knot]

[fluid pagoda]

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[pulsar ermine]

I don’t understand how they came up with what v hat is?

Is it essentially x=cos(theta)u and y=sin(theta)u’ where both x and y are vectors

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[chrome nymph]

Can someone explain to me why when have a numbered list with entries under numbers from 10 to 20 inclusively we have 11 entries in total? I need some explanation which explains it more logically than "just add 1"

[stone stump]

count them on your hands?

[chrome nymph]

LOGICALLY

[stone stump]

I don't see how this is weird. if you have entries 1 to 10 (inclusive), how many entries are that in total

[chrome nymph]

What if I have to count thousands?

And not from 1?

[stone stump]

well imagine you start labelling the list new again

the 10th item is now number 1, the 11th item is now number 2 and so on until the 20th item is now number 11

which you can see cause you have to subtract 9 from 10 to get to 1, 9 from 11 to get to 2 and so on

you subtract 9 from every number and now the list starts at 1 and ends at 11

so it contains 11 elements

[chrome nymph]

Not wanna be rude, but I asked for a logical explanation why it is so, not for a way to bring it to counting from 1

I don't have problems with calculating the correct answer, I have problem with lack of logical explanation on why I can't simply substract

[stone stump]

I'm sorry but I can't give you a logical explanation why a list of 10 items does not have 9 items

[chrome nymph]

😦

[untold oak]

this pretty much sums it lmao
a list of 10 items must have 10 items

[topaz axle]

suppose you want to count item 2 to 4

[untold oak]

if a list of 10 items has 9 items, itd be a list of 9 items, not 10. logically.

[topaz axle]

4 is the number of items behind the 4
you subtract 2, the number of items behind 2

the 4 itself is special, it's not counted in either

logically

i guess we can define it to include the item new picture then

now 2 gets overexluded

[lone ravine]

Let's subtract 10 from each number in your list. Now we have the list of numbers from 0 to 10. Obviously, if you count the amount of numbers from 1 to 10, that's just 10 (because you're counting like 1, 2, 3, ... to 10 up). But now you need to add 1 because of the 0

[topaz axle]

so it's like, you want to subtract the number before the 10, 20 − 9 = 11

[chrome nymph]

👀

Seems more like what I would like

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[restive river]

Please explain to me, which ODE technique I have to use here:

[wild creek]

this is a 1st order linear equation

so use the thoery on that

[restive river]

Would you please tell me which?

[wild creek]

well there should be something in your course about 1st order linear equations

or look up online

solving 1st order linear differential equations

[restive river]

I am sorry, but may I know which ODE technique I need to apply?

[wild creek]

it doesnt really have a specific name

https://brilliant.org/wiki/first-order-linear-differential-equations/

[restive river]

Ahh! Yes 🙂

separable differential equation.

[wild creek]

oh yeah you can do that to prove it

but for this kind of equations you can right away give the answer usually

since its always the same

[restive river]

That's also correct.

Thank you, mate 🙂

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[bright gulch]

Did i get my derivative right?

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@bright gulch Has your question been resolved?

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[tepid orbit]

can some1 explann

what this means

[midnight dirge]

well

if weve some positive number

its cube is postive

[tepid orbit]

yes

but

[midnight dirge]

and well

[tepid orbit]

where does -x^3

[midnight dirge]

• of any positive number is negative

idk

randomly

[tepid orbit]

ok

so

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[scarlet sequoia]

sin(A+B) formula

[thorny bear]

hmm

so i need to convert the arcsin to sin?

[scarlet sequoia]

imagine arcsines are angles

in fact they're

[thorny bear]

ahh i see now

thanks 🙂

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[quartz oriole]

I need help on the b) part. There should be marginal pdf of Y for cases 0 < y < 1 & y>1 but I do not understand how to find the latter.

[devout snow]

@quartz oriole Has your question been resolved?

[quartz oriole]

<@&286206848099549185> anyone who knows about conditional distributions?

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[toxic tendon]

Hello

I am having issues drawing this picture in mathematica

These are my parameterizations of the lines

However, mathematica is only showing the first one

And the line doesnt start at the origin but instead of the end point of the first line at (1,1)

Why is that? how do I get it to display all the lines

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[muted berry]

hello there

[muted berry]

how do you do this

[turbid marlin]

what is induction?

i forgor

[muted berry]

base and induction case

[turbid marlin]

well isn't the denominator just n!n!?

[muted berry]

yeah

how do you simplify

after inductive step i got

(2k+2)(2k+1)(2k)!
/
(k+1)(k!)(k+1)(k!)

[turbid marlin]

idk i suck at combinatorics

[lone ravine]

Note that you want to get to 2n C n

[muted berry]

wdym by that

yeah

oh yea

i do

i kinda see it

because theres a k!k! in denominator

[lone ravine]

Yes

[muted berry]

idk how to simplify the other stuff

can you help with that please

[lone ravine]

Move the (2k+2)(2k+1)/(k+1)^2 out

[muted berry]

so we have (2k)! / k!k! * (2k+2)(2k+1)/(k+1)^2

?

[lone ravine]

Yes

[muted berry]

then we get (2k+1)/ 2(k+1)

on the right side of the multiplication

[lone ravine]

Yes

[muted berry]

ohh

then we get

wait idk how to simplify

the rest

oh wait 2k+2 > 2k+1

[lone ravine]

Wait. Wouldn't it be 2(2k+1)/(k+1)?

[muted berry]

yeah but i brought the 2 to the denominator

[lone ravine]

The left part is just 2k C k

[muted berry]

and since the denominator > numerator that whole expression will be less than the base case

yeah

the right part of the 2k C k

is

2k+1 / (2k+2)

[lone ravine]

wdym? numerator is obviously bigger

[muted berry]

bring the 2 down

wait

we cant do that

my bad

so how do we simplify

from here

[lone ravine]

If we apply the induction hypothesis we know that 2k C k < 3^k

[muted berry]

yeah

we needa prove RHS for 3^(k+1)

[lone ravine]

So we have to show now 3^k * 2(2k+1)/(k+1) < 3^(k+1)

[muted berry]

so we know that right

[muted berry]

what we do from here

[lone ravine]

There seems to be something wrong 🤔

Devide by 3^k

[muted berry]

and how do we conclude that 2k C k is 36K

3^k

what happens after we divide

[lone ravine]

2(2k+1)/(k+1) < 3

[muted berry]

yea

and then how do we know thats true

[lone ravine]

That is the issue. It seems to be wrong for large k

[muted berry]

yea

wheres the error we made

[muted berry]

do u think the statement is wrong

idk if thats possible tho

[lone ravine]

No, I think I made a mistake somewhere

[muted berry]

let me see too

(2k+2)! = (2k+2)(2k+1)(2k)! right

did u find the error is

hmm

[muted berry]

did we mess up simplification?

[lone ravine]

Wait

[lone ravine]

Isn't it > ?

10 Cr 5 > 3^5

[muted berry]

wdym

that doesnt work

try 0

[lone ravine]

I found the same question with > 3^x

[muted berry]

can u send link'

[lone ravine]

For large enough x

https://math.stackexchange.com/questions/2228320/show-that-for-any-integer-n-ge5-the-inequality-2n-choose-n3n-holds

[muted berry]

idk if that works

hm

[woven radish]

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

[muted berry]

nah it was on the stackexchange

[lone ravine]

Yes, he got to this using the same method

[muted berry]

i honestly dont know because for a smaller n

dont work

[lone ravine]

Try n=5,6,...

[muted berry]

yeah

but

it doesnt work for 4321

wait lwk it might be >

[lone ravine]

His question says show for n > 5

[muted berry]

yea

that may be the case here too

[lone ravine]

Can you show the original question?

[muted berry]

no matter what tho

it simplifies to 3 + k-1/k+1

right

[lone ravine]

Yes

[muted berry]

only works for n>=1

right

[muted berry]

can u check my reasonin rq

also how do i conclude the proof

[lone ravine]

Yes, the factor of increase is bigger than 3 for n >= 1. But it takes until n = 5 for 2k C k to overtake 3^k

[lone ravine]

Note that for n = 5, 2k C k > 3^k

[muted berry]

oh ok

[lone ravine]

Than do what we did

[muted berry]

would i calculate base case from n=5?

[lone ravine]

Yes

[muted berry]

ok thanks

[lone ravine]

And show If 2k C k > 3^k then this is true for k+1, too

[muted berry]

this is the question

[lone ravine]

Also, you can show the cases 1 to 4 per hand

[muted berry]

yeah

theres no n value

i think its greater

[lone ravine]

It's greater for n >= 5

[muted berry]

yeah

[lone ravine]

Anyways, I got to sleep now

[muted berry]

kk thanks

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[umbral roost]

.reopen

[restive river]

Where do I go from here?

[wicked turtle]

,rotate

[wooden wraith]

When you came to 5x+b = 22(x+d) - 17(x+c)

You assumed that x^2+3bx+80 = (x+c)(x+d) in order to do that

[restive river]

Yeah

[wooden wraith]

Well, if that's true, that's a lot more information to work with

You have three variables b, c, and d. If you can make a system of three equations relating them, you can solve for them

You already have one.

[wooden wraith]

The other two come from here

[restive river]

How? I can only formulate 1 = 22d - 17c

[wooden wraith]

x^2+3bx+80 = (x+c)(x+d)

Expand the right hand side

[restive river]

$x^2 + 3bx + 80 = x^2 + dx + cx + cd$

[woven radish]

Jared

[wooden wraith]

Yeah

So combine the like terms

And do the same thing you did to find 1=22d-17c

Equate the coefficients

[restive river]

Like this?

[wooden wraith]

Yes. What does that tell you?

[restive river]

3b = d + c

and 80 = c

[wooden wraith]

80 = cd

[restive river]

cd*

[wooden wraith]

Yep

Real quick

Your first equation 1 = 22d-17c was not quite right

It should be b = 22d - 17c

[restive river]

ohh

alright

[wooden wraith]

But now with three equations and three variables

You should be able to solve the system for all three

[restive river]

How is 80 = cd used in the system?

Since it's c * d

[wooden wraith]

Well, you could do d = 80/c and substitute that for d

Or c = 80/d

[restive river]

Where do I go from here?

[wooden wraith]

That's right

But maybe kind of a dead end

I'd eliminate b instead, so that your new equation is in terms of c and d

Because the remaining unused equation is only in terms of c and d

[restive river]

Ohh alright

I'll try eliminating b instead then

$-51c = -67d$

[woven radish]

Jared

[restive river]

c = 67/51d

[wooden wraith]

That's not exactly what I got, can I see your work?

[wooden wraith]

So looks like you're multiplying by -3 on top and then adding the equations right?

[restive river]

Yeah

[wooden wraith]

Double check. Should be -52c = -65d

[restive river]

wait one sec

[wooden wraith]

-3(17c) -1c = -52c
-3(22d) +1d = -65d

[restive river]

Oh wait

you're right

alright so now c = 65/52d

[wooden wraith]

Yeah

And cd=80

[restive river]

c = 80/d

[wooden wraith]

c=80/d

[restive river]

yeah

Where do I go from here?

[wooden wraith]

Substitute that for c

[restive river]

Wdym

[wooden wraith]

$\frac{80}{d} = \frac{65d}{52}$

[restive river]

Oh

Right k

[woven radish]

tatpoj

[wooden wraith]

Sorry 52 not 62 lol

[restive river]

d could be either -8 or 8

[wooden wraith]

Yep.

There are two solutions to the entire problem

[restive river]

Do I try for both?

[wooden wraith]

Your goal is to find b. And the problem specifies b < 0

So yeah, try for both. See which one gives negative b

[restive river]

alright

c = -10 or 10

[wooden wraith]

yep

[restive river]

oh

b is -6

[wooden wraith]

Yep 👍

[restive river]

oh!

I see, tysm for the help :)

[wooden wraith]

No problem 👍

[restive river]

.close

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[pale stag]

the question wants me to prove the height of the fountain is 34m.

ive found the radius and i have the angle of elevation, how do i find the slope?

[bleak sun]

you mean the height of the triangle you made?

just use trig

tan(theta) = opp / adj

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[tiny rain]

hi can someone give me tips for this qn pls

[elfin hill]

By contradiction

If det(A) ≠ 0 then A is invertible

[tiny rain]

Ok

Lemme try

[remote meteor]

I

Wokay

Is it proved or proven

My English

[tiny rain]

wait im gonna

try first

before looking lol

[tiny rain]

yo

if det(A) = 0

that means the matrix is invertible?

not invertible i mean

[remote meteor]

There’s another way of doing it I’ll show it

[tiny rain]

hold on

i wanna attempt it first haha

[tiny rain]

is this true?

@remote meteor

[elfin hill]

Yes

You can also use det(AB) = det(A)det(B)

And chosing an infertile matrix for B gives you the result

[tiny rain]

hmm

okay

[tiny rain]

breh i cant even solve the first part

sad

what can i even get from this lol

hmm

@remote meteor yo

ill have to prove both ways right?

like, this is only one way

[remote meteor]

Oof ig

[tiny rain]

hmm

should i do it by contrapositive

[tiny rain]

or contradiction

[remote meteor]

Contradiction is ok

[tiny rain]

what do i do from here lol

[elfin hill]

A is invertible

[tiny rain]

wait wat

but det(A) = 0

[elfin hill]

Ooooh sorry

Misread

[tiny rain]

lol

hah

i dont quite see where to go

from here

[elfin hill]

Then A is inactive

Injective

[tiny rain]

whats thaat

[elfin hill]

And thus invertible

[tiny rain]

err

could i say instead that

[elfin hill]

Ker(A) =0

[tiny rain]

havent learnt that haha

[mighty minnow]

hi

can someone help me out

with algebra 2

[tiny rain]

#❓how-to-get-help

[mighty minnow]

im kinda stupid

oh

[unique sequoia]

Is calculus hard?

[tiny rain]

bruh

#❓how-to-get-help

[devout snow]

@tiny rain Has your question been resolved?

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[restive river]

Obtain the differential equation that describes the following family of curves. All straight line with x-intercept at x=2.

[topaz beacon]

lines are of the form y=mx+b

we know that they go through the point (2,0), so plug that in

[restive river]

like this y=m(2)+b?

[topaz beacon]

0=m(2)+b

[restive river]

then the arbitrary constants are m and b?

[topaz beacon]

yes, but you can find the value of one in terms of the other

[restive river]

wait i canot differentiate it if i dont have y

it should be dy/dx

[topaz beacon]

we do not have any slopes

so no need to differentiate

[distant harbor]

that's not quite right

let's start with the eqn:

y = m(x-2)

solve for m

[restive river]

is this point slope formula

[distant harbor]

yep

[restive river]

and m is the only AC?

[distant harbor]

yep

solve for m

[restive river]

i can isolate m so it would be like m=y/x-2, then differentiate?

[distant harbor]

yes

[restive river]

for the denominator (x-2)^2, can the equation be multiplied to the denominator for simplification?

[distant harbor]

you should get something in the form "=0", the denominator should not matter at all

[restive river]

so it would be [y'(x-2)-y]/(x-2)^2

would there will be no simplification then?

[distant harbor]

= 0

[y'(x-2)-y]/(x-2)^2 = 0

so just get rid of the denominator, lol

[restive river]

okay so, y'(x-2)-y=0

[distant harbor]

yep

now solve for y'

[restive river]

then is that the final answer? it'll be y'=y/(x-2)

[distant harbor]

yes

notice that in our original equation we had:
y = m(x-2)
y/(x-2) = m
so that means:
y' = y/(x-2) = m

makes sense because it's a line doesn't it?

[restive river]

yeah, looks like the intercept form

[distant harbor]

now that i think about it there's an easier way to do this

we can start with:
y = m(x-2)
take the derivative:
y' = m
plug in y/(x-2) = m:
y' = y/(x-2)

[restive river]

ah yes, using the substitution method

btw, thanks a lot, i always have a hard time choosing the formula

.close

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[strong hazel]

Obtain the differential equation by eliminating the arbitrary constants

y=Ae^(-3x)-Be^(2x)+Cx^3

I tried getting up to the 3rd derivative, and was thinking to use matrix but its 4 equations and, i think matrix can only be used for 3 equations

[devout snow]

@strong hazel Has your question been resolved?

[analog trellis]

Might just want to try using the same idea with y, y’ and y’’.

Use Cramer’s rule to solve for A, B and C.

[strong hazel]

So there will be three matrix tables?

[analog trellis]

I don’t know what a matrix table is.

[strong hazel]

Okok, I'll try to solve it using Cramer's rule

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[tiny rain]

hi, can someone help me with this question

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[tiny rain]

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[short wigeon]

In machine learning, we use sigmoid function to setup the logistic regression.

Why we can define P(y=1|x;theta) and P(y=0|x;theta) with a pair of totally different function?

Thank you,

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@short wigeon Has your question been resolved?

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[thick reef]

Missed some notes for this and I’m confused on how to approach and solve it can someone help me any help is appreciated!

[devout snow]

@thick reef Has your question been resolved?

[tall knoll]

Seems like you want to use bayes theorem for this

[thick reef]

Yeah that’s the thing I’m confused on how to approach it

Or like apply it-

There we go

[tall knoll]

Start by defining some events, that way you can rewrite things using probability notation

[thick reef]

like into percents?

Like rewriting the probabilities?

[tall knoll]

So define an event where someone is a smoker, and another where someone dies

Then yeah, translate the probabilities into something more useable

[thick reef]

Someone is a smoker

Probability of dying ~ 6%

Probability of not dying ~ 94%

Nonsmoker

Probability of dying ~ 1.1%

Probability of not dying ~ 98.9%

[tall knoll]

Technically correct but not exactly what I'm asking for

What is bayes theorem?

[thick reef]

Like the probability of an event based off previous information we knows related to the event?

[tall knoll]

That's a reasonable physical interpretation

For the purpose of solving the question, do you know its formula?

[thick reef]

No but I remember my professor using a table of some sort for a problem related to using bayes theorem

[tall knoll]

That's strange

I'd like to see how your prof solves this type of problem

[restive river]

10%

[tall knoll]

[thick reef]

.close

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[restive river]

How many ways to put 13 indistinguishable balls in 4 indistinguishable boxes?

[potent moon]

https://math.stackexchange.com/questions/373202/distributing-identical-objects-to-identical-boxes

This might help

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[reef prism]

Hello everyone! I am so confused about limits at infinity when the highest degree of numerator is greater than the highest degree of the denominator. My professor told us to always divide the variable with the highest degree to the numerator and denominator. The attached photo is the sample he kind of explained to us.
My question is, isn’t 1/0 considered undefined? How come the limit became positive infinity? Is there a theorem that mentions the limit of undefined is equal to a (or the value it approaches)?

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[maiden zinc]

$f_a = \frac{x-a}{|x-a|+a}$

[woven radish]

Max.cc

[maiden zinc]

i have to find that when we variate a all the curves pass through a point

ie $\exists x\in \mathbb{R}, \forall a \in \mathbb{R}, f_a(x) = constant$

[woven radish]

Max.cc

[stuck axle]

If I understood, it’s one common point f_a(x) for all a ?

[maiden zinc]

ye

[stuck axle]

Okay so you can just take two values of a, and ask for an equality

[maiden zinc]

when we evaluate in 0 it doesn't work, same for a multiple of a

[maiden zinc]

good idea

[stuck axle]

Like $f_{a_1}(x)=f_{a_2}(x)$

[woven radish]

DonutsenPLS

[stuck axle]

I guess it would work

And some values seems more efficient than others

[maiden zinc]

i don't know how do solve that

[stuck axle]

Have you never done polynomial equations with absolute value ?

[maiden zinc]

never

[stuck axle]

Oh i see

So let’s try learning the basics then

Let y=|x|

[maiden zinc]

y = -x or x

[stuck axle]

Indeed

But when ?

[maiden zinc]

-x when x <= 0 and x when x >= 0

[stuck axle]

Yes

I was gonna say don’t forget the >=

But you did it first

[maiden zinc]

ye i thought about it x)

[stuck axle]

So now, how about y=|x-a| ?

[maiden zinc]

if x-a >= 0 y = x-a
if x-a <= 0 y = a-x

[stuck axle]

Mhm

You can pass the a on the other side it’ll be clearer

[maiden zinc]

ye

if x >= a, y = x-a
if x <= a y = a-x

[stuck axle]

Great

Now let’s consider a and b

a > b

|x-a|=|x-b|

You can make a little value table if you think it’ll help

[maiden zinc]

here we have 3 case, right?

[stuck axle]

Exactly

[maiden zinc]

hum

[stuck axle]

Now is the most important part

Once you tell me what you think I’ll give you an easy way on how to represent that

[maiden zinc]

if x >= a, x-a = x-b <=> a = b
if a < x < b, a-x = x-b <=> x = a + b / 2
if x <= b, a-x = b-x <=> a=b

[stuck axle]

Be aware that a is bigger than b

Thé second would be then b < x < a

But otherwise it’s correct

Look :D

Now that this is understood

Let’s get back to the original problem

[stuck axle]

Which values do you think would be the best for a_1 and a_2 ?

[maiden zinc]

we have : $\frac{x-a_1}{|x-a_1|+a_1} = \frac{x-a_2}{|x-a_2|+a_2}$

[woven radish]

Max.cc

[stuck axle]

Exactly

[maiden zinc]

0 and 1

[stuck axle]

Right

[maiden zinc]

but it doesn't prove for all value

[stuck axle]

You said it’s common to all a ?

[maiden zinc]

i means we can have a function that works only ofr 0 and 1

[stuck axle]

^

[maiden zinc]

yeah

but proving with 0 and 1 would prove it just for 0 and 1

not for all a

[stuck axle]

Then there is one point that passed by f_a regardless of a ?

[maiden zinc]

or we find a value with 0 and 1 and then we prove that it works for all a

then i'm ok

[stuck axle]

Or you can take 0 and a ?

I don’t know

[maiden zinc]

since a is not define it works ye

but how solve it

[stuck axle]

I’m going to try it first brb

[maiden zinc]

What do you think about it

it seems weird

cause i took the value of a for x >= a into x <= a to find x

[stuck axle]

It’s not x/x that you have

[maiden zinc]

if a = 0 what would we have then

[stuck axle]

Ohhh yeah

Well I have tested with a1 and a2

And made them equal to one and the other

And there is one solution

[maiden zinc]

how do you found it

[stuck axle]

By making f_a1(x)=f_a2(x)

Sorry for the long delays

[maiden zinc]

Yeah but how

it looks pretty complicated

[stuck axle]

Well let a1>a2

And do lien we’ve done before

[maiden zinc]

but we don't know if a1 > a2 or a2 > a1

[stuck axle]

It’ll be the same result

And we have to let one value bigger than the other

Also it doesn’t work if a is negative

[maiden zinc]

Did you have something like that to solve

[stuck axle]

Nor if a = 0

[stuck axle]

Yeah

But it’s useless to do it

[maiden zinc]

hum why

[stuck axle]

Cause you’ll have some quadratic equations

[maiden zinc]

and how do you solve it then

[stuck axle]

Eh i gtg, if I have the time I’ll explain it later

[maiden zinc]

ok np thx anyway

[devout snow]

@maiden zinc Has your question been resolved?

[devout snow]

@maiden zinc Has your question been resolved?

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@maiden zinc Has your question been resolved?

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[neon folio]

hi

if integers are consecute

then

they are just 1 more than the last one

so our numbers are: a,a+1,a+2,a+3

[rich summit]

1/2 * n * (n + 1)

[neon folio]

and we know the sum of the first three is 206

[rich summit]

Sum of n first integers

[neon folio]

now we find a

oh

ok

ok

[rich summit]

2 odd, one even integer

3 even isn't possible

[neon folio]

a+a+2+a+3 works

[rich summit]

Maybe you can use this to approximate the solution

[neon folio]

find a form here

since a is an integer

we need to try all possible combinations

until a is an integer

[neon folio]

and it works with this

with the other combinations a wont be an integer anymore

yes

yes

yes

you can check too if you want

np

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[neon folio]

[devout snow]

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[high chasm]

dont understand how f1(0) = 2, f1(1) = 4 etc...

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[restive river]

R is a subset of C. x is a member of R = (x, 0) is a member of C. we know that with the extended real line, x - infinite = negative infinite which should equal (x, 0) - infinite but isnt considered the case in this image. why is that?

[stone stump]

well "positive" or "negative" wouldn't make sense in C anyway

essentially we just call all of them the same thing, just infinity. it doesn't matter in which direction you go

also as a sidenote, if x is in R, then x is also in C

if you want you can write it as x+i0 or something I guess but you don't need any tuples (x,0) here

[restive river]

.close

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[shut whale]

it is possible that A still does not contain 1 2 3 and 4

so while it is possible that A=N, there are also cases where A \neq N

[outer stirrup]

So answer c and answer d are eliminated. We know for sure that 5 is an element of A, so answer b is still valid. Is there anything in the question to imply that 1, 2, 3, 4 are or are not elements of A?

[shut whale]

nope

there is not

[restive river]

Hey ik this isn't an answer but just a Lil curious if a is in the set then if a = 1, 2 will also be there as a + 1 is also in the set, and more, unless there is something else, cuz ik little about sets

So won't it be d?

[shut whale]

but here they have said nothing about 1 being in the set

[outer stirrup]

So I guess if A is out of the question the only possible answer is b. So was the whole 'If a is an element of A...' thing just a red herring?

[shut whale]

sure, if 1 was in theset, then it would be d, but here 1 may not necessarily be in the set

we only know for a fact that 5 is in the set

[restive river]

ok i guess idk anything about sets, we only have that to study in our school from next year (Indian grade 10 student here btw)

[outer stirrup]

Yeah I was trying to figure out why the a + 1 part is relevant but I guess it's just not

But I was almost convinced the answer was a because of that

[shut whale]

pretty sure you'd have had some of it in grade 8 or 9

[shut whale]

it is

it shows option (a) is correct

[restive river]

Idk if u are Indian but in our education system sets are only starting from grade 11

[shut whale]

it basially shows A contains all natural numbers greater than or equal to 5

[shut whale]

ok, different boards then

or I've just forgotten

[outer stirrup]

But what if a = 1? Then 1 + 1 is 2 which is also an element of A

[shut whale]

yeah, but the question says nothing about A

the question wants you to say what you know for certain

for certain from the question, we know A contains 5, as it says so

and we know that all integers greater than 5 are in A by the a+1 rule

[outer stirrup]

Ohhh I see. We know for sure that 5 is in A, and if a + 1 is in A then every integer greater than 5 is also in A. We know nothing about integers <5. So b can't be true, neither can c or d. So the answer is a 🙂

Thanks a lot mate! Hope our boy Sharl wins the WDC next year! 😄

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[cosmic trail]

what's the square root of 3 in exponent form

also damn mekmek on the grind today

[wheat vale]

I have an exam the day after tomorrow, this card is worth 1.5 in the final grade and I've been doing this shit since 8 in the morning

[left robin]

sqrt(3)=3^(0.5)

therefore

1/sqrt(3)=3^(-0.5)

now divide by this

wait

nvm

that is one step to much

just multiply by sqrt(3)

and then write sqrt(3) as 3^(0.5)

[wheat vale]

ok

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[wheat vale]

i think i got it

[cosmic trail]

damn, good luck bro

don't overdo it though, get godo rest

[wheat vale]

thnks

ok

.close

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[high chasm]

How does this show that the Cardinality of N = Cardinality of Q

but what ive shown so far

how does that show that card N = card Q

[prime egret]

Q+ at least, you can order the rationals

This means you can put them in order 1 2 3 …

[high chasm]

sry bit lost

tbh

i dont acc understand

whats going on in the pic i sent

[prime egret]

You’ve just put the rationals in a one-to-one correspondence with natural numbers

Hence showing their caardinality is the same

1:1
2:2
3:1/2
4:1/3
…

[high chasm]

is the red column the reals?

and the blue row the rationals?

[prime egret]

All of these are rationals

That’s the point of this construction

[high chasm]

basically im confused on how this table is used to show that the card of N = card of Q

[prime egret]

How do you show card N = card Z?

[high chasm]

show that f is onto and 1-1

[prime egret]

You can also show that you can arrange elements of Z in some order in a certain pattern.

0,-1,1,-2,2,…

You can count these and if i ask you what the 43th term is you can tell me which means there’s a one-to-one correspondence with the natural numbers.

[high chasm]

can we go over the proof of #N = #Q from scratch together cuz im really confused icl

[prime egret]

Do you get this

Z can be listed hence the same cardinality as N

It’s the same with the table you have. If you follow the direction of the arrow, eventually you’ll list all the rational numbers.

[high chasm]

idk why im not understanding this

im just kinda clueless to whats going on

[prime egret]

Do you get this

[high chasm]

why have u ordered the Z in that way

[prime egret]

We want to be able to list Z in a particular pattern so that we can guarantee every element of Z will appear on this list

This by listing we can call 0 the first term, -1 the second term, 1 the third term. This means there’s a one-to-one correspondence between N and Z.

[high chasm]

but how does this show theres a one to one correspondence

[prime egret]

There’s a function f that maps 1 to 0, 2 to -1, 3 to 1, 4 to 2 …

[high chasm]

sry im just still confused why youve ordered z in that way

[prime egret]

An infinite set is countable iff it’s possible to list all the elements in a sequence. This is because a one-to-one correspondence f from N to our infinite set can be expressed i terms of our sequence a1,a2,…,an where a1 = f(1), a2 = f(2),…an=f(n)

For example bijection from N to Z can be worked out quite easily by looking at our list

$f(n)=\begin{cases}
\hspace{1cm} \frac{n}{2} \hspace{0.89cm}\text{if n even}\

• \ \frac{n-1}{2} \ \ \ \text{if n odd}\
  \end{cases}$

[woven radish]

Pure

[high chasm]

ok isee

imma watch a few videos and come back to u

[devout snow]

@high chasm Has your question been resolved?

[devout snow]

@high chasm Has your question been resolved?

[high chasm]

think i might have got it now

https://www.youtube.com/watch?v=Lfw96n0m1js

this video helped

[prime egret]

[high chasm]

thanks for the help

[prime egret]

Np

[high chasm]

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[pseudo stratus]

how would i find g(x)

[void rain]

You cant

[pseudo stratus]

so how would i go about this q

[wild creek]

sandwich theorem

[pseudo stratus]

yea

[void rain]

Do you know what that is?

[pseudo stratus]

yep

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[jaunty gazelle]

Never faced a problem like this one. Would I multiply by the conjugate of 1/square root of x -5, or would I multiply the whole thing by the conjugate?

[hollow pollen]

multiply 1/square root of x -5 by the conjugate of square root of x -5 over itself

[jaunty gazelle]

are you able to visualize it here, I don't understand what you mean by "over itself"

[hollow pollen]

multiply the left side by $\frac{\sqrt{x} + 5}{\sqrt{x} + 5}$

[woven radish]

Gijs

[hollow pollen]

i think that should get you further along

[jaunty gazelle]

Left but not right, right?

[hollow pollen]

multiply this by that fraction

[jaunty gazelle]

Yeah, I get sqrt(x)+5/x-25

Anything afterwards?

continue to multiply by conjugate

so multiply numerator and denominator by sqrt(x)+5?

[hollow pollen]

okay so you get then $\lim_{x \to 25} \frac{\sqrt{x} - 5}{x - 25}$

[woven radish]

Gijs

[hollow pollen]

this is what you have right

[jaunty gazelle]

No I get sqrt(x)+5..

on numerator, not -5

not sure what I did wrong

[hollow pollen]

yes but you bring in the other fraction

i did that step for you

they have the same denominator now so you subtract 10

5 - 10 = -5

[jaunty gazelle]

Oh.

I didn't think we could do that

[jaunty gazelle]

So wait, you merged both fractions?

[hollow pollen]

yes

[jaunty gazelle]

And since both fractions have a common denominator of x-25, you just kept one of those denominators and attached the numerator we just computed on top?

[hollow pollen]

yes i think you understand

perhaps it would be good to write the steps out on paper though

[jaunty gazelle]

Do you then substitute 25 in or multiply by conjugate again?

I GOT it.

1/10

[hollow pollen]

how did you do it

[jaunty gazelle]

wait

[jaunty gazelle]

multiplied by the conjugate,got x-25 on numerator, canceled out x-25 on denominator and numerator, then got 1 on numerator and the square root of x +5.

I then substituted in 25 to find the limit, and simplified to 1/10.

[hollow pollen]

uh

1 secx

oh nice

yes thats right

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@jaunty gazelle Has your question been resolved?

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[deep fable]

Hello, I was wondering if the answer I got for this problem is correct, Not really asking for the answer outright just want to know if I am correct, My answers for this is x = -1, x = -1/2, and x = 1. I would enter the answer myself but after many attempts I am now on my last try before the question locks itself from future attempts 😅

[wooden wraith]

Well, by inspection we can see that x=1 is not a zero

2+3-1 =/= 0

So something's not right

[deep fable]

true, i got it by quadratic formula and i was confused myself

guess ill recheck my notes

oh i got it right sick

welp thanks

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[short iron]

I am confused on what this means

[topaz beacon]

a reference angle is the smallest angle made by the ray and the x axis

[short iron]

Right I get that

The trig values wouldn't be the same because they're just different angles?

[topaz beacon]

yes

ref angle values are always positive

[short iron]

Regardless of what quadrant they're in?

[topaz beacon]

yes, the quadrant tells you which values to make positive or negative for the actual angle

[short iron]

Shit I need to go back and look at this

I'm really not getting it

Thanks

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[silent sonnet]

Hey, I need some guidance on how to solve this problem. I don`t if I have to use the conjugate method or do something else to get a and b.

[winter patrol]

use conjugates

[silent sonnet]

Do i have to move the square root of 3 to the left?

[winter patrol]

no

[silent sonnet]

Im stuck at a+bx-3/x(square root of a+bx+squareroot of 3)? Nothing is canceling out. Do i factor, distribute, or something else?

<@&286206848099549185>

[devout snow]

@silent sonnet Has your question been resolved?

[silent sonnet]

no it hasn`t

[devout snow]

@silent sonnet Has your question been resolved?

[silent sonnet]

no

.close

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[fluid rampart]

Hi, just a general question, if the vecteurs u, v, and w are linearly independent, would u+v, v+w and u + w are also linearly independent?

[left robin]

my intuition says yes they are still independent

but im not sure how to prove

maybe i can come up with something

lets check:
a(u+v)+b(v+w)+c(u+w)=0
au+av+bv+bw+cu+cw=0
u(a+c)+v(a+b)+w(b+c)=0

if this is equal to 0, then we have a,b,c pretty restricted
if we however say that u,v,w are linearly independent, then this should not work for any combinations

the wording at the end could be better

but i think this proves it

by contradiction

different wording:
let a+c=n, a+b=m, b+c=k
then we get:
nu+mv+kw=0
which is impossible since:
au+bv+cw=/=0 for all a,b,c

[fluid rampart]

wait what do you mean by if "u,v,w are linearly independent, then this should not work for any combinations"

[left robin]

linearly independent means
au+bv+cw=/=0 for all possible a,b,c

where =/= means does not equal

im setting it equal to 0 to look for a contradiction