help-26

[cold crag]

Can anyone help me with 2a

I'm confused whether you suppose to do the middle part first or make the statement (p ⇒( q ∧ ¬r ))∨ ((¬p ∧ q) ⇒ r)

[ruby tree]

Writing that without parentheses is wild

Are you given any precedence rules?

[cold crag]

Yes

[ruby tree]

So are you sure it's equivalent to (p ⇒( q ∧ ¬r ))∨ ((¬p ∧ q) ⇒ r) ?

[cold crag]

Isn't it suppose to be

p ⇒( q ∧ ¬r ∨ ¬p ∧ q )⇒ r

Since you should work on disjunction and conjunction fisrt

[ruby tree]

That's what I'm asking

[cold crag]

I guess it's this one

[cold crag]

I put it on chat gpt, and it gave me this

Samething with deepseek

[ruby tree]

I don't care what LLMs gave you, the course material / textbook should have a part on operator precedence

[cold crag]

They did

Its this one right?

[ruby tree]

I guess; where is that from?

[cold crag]

Well I took screenshoot it from gpt

[ruby tree]

[cold crag]

Tho my book source has it too, it's basicly thesame

[ruby tree]

Are you sure about that

You also need operator associativity, which should also be in your book

[cold crag]

Alright, lemme check again

I translated my book source

[ruby tree]

That isn't about associativity

The problem is that (a => b) => c is not the same as a => (b => c)

Without knowing whether the question expects left-associativity or right-associativity for the => operator, you can't really answer

Well I guess you could give two answers...

[cold crag]

I see

[craggy haven]

where are you seeing a -> b -> c?

[craggy haven]

ok yeah i thought i could separate that into two different implications but nothing mskes sense

[ruby tree]

It's usually right-associative (see for example https://math.stackexchange.com/questions/12223/associativity-of-logical-connectives and http://intrologic.stanford.edu/dictionary/operator_precedence.html), but apparently some sources take it as left-associative instead (https://www.cs.utexas.edu/~dnp/frege/logical-operator-precedence.html)

And tbh I wouldn't be surprised if OP's material doesn't specify

I'm gonna have to go soon, sorry @cold crag

[cold crag]

It's okay, you've helped a lot

Thanks @ruby tree

[ruby tree]

All I can suggest is to give two answers, one for
p ⇒ (( q ∧ ¬r ∨ ¬p ∧ q ) ⇒ r)
and one for
(p ⇒ ( q ∧ ¬r ∨ ¬p ∧ q )) ⇒ r

[cold crag]

Alright

Thankss

[topaz sinew]

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[topaz sinew]

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[tight sage]

how do i solve 2| x+3 | -1 = 0.25| 4x-4 | +6

[topaz sinew]

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

[tight sage]

1

[mint crescent]

Are you familiar with how to "break down" the absolute value as a piecewise function?

[tight sage]

not sure how

[mint crescent]

there's quite a few ways

what's your preferred way of thinking of absolute value

[tight sage]

thats the first time im doing 2 absolute values that are = to eachother

so i dont even know where to begin with that

[mint crescent]

ok

[mint crescent]

once we do this, it will be a lot more simple

[mint crescent]

but to do that, I need you to answer this

[tight sage]

wait sorry im confused by the question

😭

what did u mean

[mint crescent]

what does absolute value make you think of

[tight sage]

like inequations and the =< > signs

[mint crescent]

I meant the absolute value itself, not so much the exercises you've done involving it

are you find with thinking of absolute value as a distance

[tight sage]

yeah sure

[mint crescent]

specifically |x| being the distance from x to 0 (such as on a number line)

[tight sage]

yeah

[mint crescent]

alright

Without absolute value, how would you express the distance between x and 0 if:
i. x is positive
ii. x is 0
iii. x is negative

[tight sage]

x > 0

x >= 0 x < 0

[mint crescent]

uh

do you know what I'm asking

[tight sage]

sorry its a bit difficult for me cause all the stuff i see is in french

but here its english

i talk english but the notions i learn are taught in french

[mint crescent]

Sans valeur absolue, comment exprimeriez-vous la distance entre $x$ et $0$ si : \ \
i. $x$ est positif \
ii. $x$ est $0$ \
iii. $x$ est négatif \

[thorny flame]

Civil Service Pigeon

[tight sage]

if x is positif

distance = x

if x was positif then whatever number on the right side of 0

[mint crescent]

ouais

[tight sage]

would be the distance of x

[mint crescent]

Et pour le ii et le iii

[tight sage]

if x was 0 then the distance is 0

[mint crescent]

ouais

[tight sage]

if x was negative then the distance from 0 to x would be negative

[mint crescent]

non - la distance est toujours positive ou nulle.

[mint crescent]

Ici, est-ce que tu as utilisé la logique de soustraire le nombre de gauche sur la droite numérique de celui de droite?

[tight sage]

no i just thought that

[mint crescent]

oh

[tight sage]

yeah

[mint crescent]

tu peux alors utiliser cette idée pour répondre au point iii

[tight sage]

well for iii if we want to find the distance if x is negative then we have to do the -x + the x of the right side

[mint crescent]

"the x of the right side"

qu'est-ce que cela veut dire?

[tight sage]

the positive

[mint crescent]

si x est négatif, quel nombre est le plus petit: 0 ou x?

[tight sage]

0

[mint crescent]

quel nombre est le plus petit: $0$ ou $-5$?

[thorny flame]

Civil Service Pigeon

[tight sage]

-5

[mint crescent]

ouais

[mint crescent]

tu veux réessayer?

Allô?

[topaz sinew]

@tight sage Has your question been resolved?

[tight sage]

@mint crescent ah sorry

it got a bit late for me and i been doing maths for like 2 hours straight sorry for not replying but yeah

[topaz sinew]

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[crude quail]

if f prime is decreasing, the slope is getting more negative?

[maiden plank]

yes

[crude quail]

oh, so then i guess gh doesnt work

wait so how can f prime be getting mroe negative for both concave up and conace down?

[maiden plank]

(curves shown are for f, not f')

[crude quail]

ohhh i see

so in both instances the slope of the tangent line is getting more negative

[maiden plank]

yep

[crude quail]

at first i was thinking g to h makes sense, but must the answer be concave down?

c to d would work then, no?

[maiden plank]

CD is most likely the region being asked for, yes

[crude quail]

what does it mean if f prime is negative ?

[crude quail]

since this is what it means if it is decreasing

[maiden plank]

what is f'?

what is its definition?

[crude quail]

the rate of change of f at an instant

[maiden plank]

and how does that translate onto the graph?

[crude quail]

ohh so slope of tangent is negative

thank you!

[maiden plank]

nps

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[rigid cloak]

I am making a mandelbrot plotter need some help with the math

[rigid cloak]

f(z)=z^2+c

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[chilly quiver]

why is this mgsintheta

we're trying to find the weight along the hypotenuse so shouldnt it be mg cosec theta?

mg is the length of the perpendicular

[maiden plank]

weight always acts directly downwards, but the spring balance only measures the weight component parallel to the inclined plane

[chilly quiver]

so essentially the weight component along the hypotenuse

which would be mg x hyp/opp

which is mg x csc theta

[placid swift]

No what

[maiden plank]

doctorstrange could check me on this, but I'm p sure this is the correct force triangle

[placid swift]

Yup that looks correct

Now you could send that force vector to the object and it'll still be mg sin(theta)

[chilly quiver]

ohh

.close

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[full dove]

can someone please explain how are we getting the solution (second picture)? the third picture is my work but i do not know how to proceed further

[full dove]

wait, do we just do c' for the remaining inputs?

but their provided solution still does not make sense

[coral sonnet]

is that a Kmap youve drawn in the 3rd pic

[full dove]

yeah

to minimize the expression

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@full dove Has your question been resolved?

[full dove]

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[spare monolith]

Using Kmap to find an expression, is to circle all the 1’s in the table.

You see three circles (rectangles) that covers all 1’s, that is the Boolean equation of the component.

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[full dove]

yes, that is what i did?

i do not get their solution

the second picture

[spare monolith]

I’m not familiar with MUX, so I aren’t of help here.

[topaz sinew]

@full dove Has your question been resolved?

[full dove]

is okay, thank you still

[sterile finch]

I think there is an error in their solution.

I_0 represents m0 and m1 so it should definitely be 1.

[full dove]

what about I1? i doubt he would make two mistakes in a single slide

[sterile finch]

I1 is for m2 and m3, neither of which is selected.

So I1 should be 0.

[full dove]

but we have B'C', does that not mean it should be A'B'C' + AB'C'?

from the minimized expression (picture 3)

[sterile finch]

One moment.

Let me format this better.

[full dove]

okay

i will be afk for a few minutes

[full dove]

okay this makes sense but

[sterile finch]

I think your error was trying to group it this way.

[full dove]

what if we first minimize the expression and then try to work with it? like, for instance, we are getting A'B' + B'C' + AB

[full dove]

why? we have to group each 1, right?

[sterile finch]

You group pairs but each pair needs to be unique.

[full dove]

no?

over-lapping is allowed

this is from the same presentation

[sterile finch]

So you made the group AB', right?

[full dove]

yeah

[sterile finch]

Well, that group consists of m0, m1, m4, and m5.

[full dove]

i am sorry, i do not follow you

[sterile finch]

Whoops, I did that wrong.

[full dove]

wait

[sterile finch]

The grouping you did was for AC', which is not a valid selector.

[full dove]

it is A'B'

not AB'

[sterile finch]

(A + A')·B'C'

It's too early in the mornig for me.

That's that red group.

[full dove]

yes, this makes sense; and that is what i am talking about

yeah

so then, why did we only do AB'C' and not A'B'C'?

i am not sure if i am making a lot of sense right now

is just a little confusing

[sterile finch]

Let's see. We can disregard A so you just have B'C'.

[full dove]

but then how would we represent that using a multiplexer? because here, our 'select lines' are A and B

[sterile finch]

And you can disregard B' as well because it's a selector.

[full dove]

eh, is that allowed?

[sterile finch]

If A and B are the selector, you only concern yourself with the remaining C.

Since it's just f(A,B,C).

AB determines the pairs and C decides who stays and who goes.

[full dove]

i did not know you were talking in that sense, but yes, i get it

(please ping me if you are responding)

[topaz sinew]

@full dove Has your question been resolved?

[full dove]

@sterile finch i am sorry for pinging, but are you still here?

[sterile finch]

Yes.

I wrote this up for you.

Using different selectors, it shows the different I_n.

[full dove]

in the first case, what if we take A and B as select lines?

ah wait, that is the fourth one

[sterile finch]

Third, there are only three cases.

[full dove]

yes, sorry

okay so

this makes sense but

can i tell you what i do not understand?

[sterile finch]

Sure.

[full dove]

okay so, i get this and the very first diagram you sent. what i do not get is what is wrong with my approach; like, why does it not make sense? if we have B'C', we can write it as AB'C' and A'B'C', right? i want to understand this because we are required to simplify the expression(s) using a k-map, and if i were given that question in the exam i would 100% do A'B'C as well.

does that make sense?

because what if we were working with more than three variables, say, five? it would be very tedious to write each term out

[sterile finch]

Yes, it would be tedious.

[full dove]

also, my k-map is correct, right?

so it really does not make sense

[sterile finch]

Your k-map is setup for BC-selectors, not the AB that was used in the solution, using the format I showed in the image above.

[full dove]

oh

wait i am stupid lmao

right

well, thank you. sorry i over-looked that part

[sterile finch]

I hope this clarified things for you.

[rancid jasper]

Fudge I’m curious

Is this ECE related

[full dove]

yeah

[rancid jasper]

R u a first year or naw

[full dove]

it did, thank you. do you mind helping me with one more doubt i have?

[rancid jasper]

I’m tryna learn ECE skills as a first year that’s why

[full dove]

i am pursuing master's

[sterile finch]

Just ask.

[full dove]

what exactly do they mean by this?

(the red part)

@rancid jasper sorry, i am not currently accepting friend requests.

[full dove]

(PA stands for parallel adder)

i do not understand how they derived that

[sterile finch]

N-bits creates a 2*N logic gate delay as stated by the second line. t_pd is the propogration delay from the first line. Multiply them and you get the total delay, T_delay.

[full dove]

ah

okay, thank you!

[sterile finch]

yw

[full dove]

.close

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[topaz sinew]

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[bitter sparrow]

Hello, I am struggling with the concept of image in relation to functions and composite functions.

[bitter sparrow]

This is the question in particular. I was going to ask in class today but my professor is out of town and we are not having class today. I do not want to wait till Monday to get the answer.

[ruby tree]

You don't need to find the expression of h(x) for that

Well, let's do it in two ways

Now that you have h(x), you can see it's a quadratic, can you find its minimum?

[bitter sparrow]

Yeah, I can in theory, let me try do that. Been a hot minute since I did that to a quadratic (about 11 years)

Okay:
x = -b/(2a)
x = -4/2
x = -2

y = h(-2) = ((-2)^2)+4(-2)+5
y = 1

Minimum is at (-2,1)

I think that is correct.

[ruby tree]

Yes it is

Now you know that h(x) can never be less than 1

Can it be any value greater than 1?

[bitter sparrow]

Yes, infinity many real numbers.

[ruby tree]

Right

"infinity many real numbers" doesn't really matter (it's not sufficient), but that's a detail

The point is that for any value y >= 1, you can find an x such that h(x) = y

Do you agree with that

[bitter sparrow]

Yes, for any y E R, where y >= 1, there exists a x E R such that h(x) = y.

I need to download a Greek keyboard on mobile. New phone

[ruby tree]

(you can just say "in")

Ok so that's the image of h

All real numbers greater than or equal to 1

In interval notation: [1, +inf)

[bitter sparrow]

So that is what an image is? A sort of statement describing the values of y for which you can find x values to plug into the function to get those y values as a product?

Or is it just (1, +inf)

[ruby tree]

Yes, it's the set of possible outputs of the function

[bitter sparrow]

Ahhhh okay.

[ruby tree]

[1, +inf), closed at 1

[bitter sparrow]

Thank you very much.
I am satisfied with this answer.

[ruby tree]

Now, instead of computing h(x) and finding the minimum of that quadratic, you can just use f and g

What is the image of g?

[bitter sparrow]

2?

Wait

[2,+inf)

[ruby tree]

Are you sure?

g(x) = x + 2, right?

[bitter sparrow]

Ahhh wait hold on no

I misread my own wack equation. I need to learn latek.

It is for all real numbers no?
(-inf, +inf) in R

[ruby tree]

Yes, the image of g is simply R

What's the domain of f?

[bitter sparrow]

R

[ruby tree]

Right, and what's the image of f?

[bitter sparrow]

It is a quadratic, so minimum of f is 1,

[1, +inf) in R

[ruby tree]

Right because x^2 is always non-negative, so add 1 and you get all values >= 1

So we have g's output, all of R, going into f, which also accepts all of R, and outputs [1, +inf)

That's all you need to know: h's image is [1, +inf)

[bitter sparrow]

Thaaat is cool.

Okay, I think I understand image completely now. It is the set of all outputs in the codomain of a function for which we can find inputs in the domain

[ruby tree]

Yes

[bitter sparrow]

Bless this server's existance.

[ruby tree]

If you had for example f(f(x)), you could say that f's image is [1, +inf), f(1) = 2, and f is increasing on [2, +inf), so 2 is the minimum of f(f(x))

That makes the image of (f o f), the composition of f with f, [2, +inf)

[bitter sparrow]

I understand. Thank you for your help!

[ruby tree]

You're welcome

.close if you don't have any more questions

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[fierce jay]

hii can someone help me w this question? i got log10(1/2) but idk if its correct cuz 1. it isnt rational and 2. its negative so i doubt it ;(

[quasi depot]

why should it be rational

[maiden plank]

you're adding irrationals, 99 times out of 100 you're getting an irrational

[quasi depot]

and yeah negative is problematic as all of the terms are positive

[fierce jay]

ohh fairs

[quasi depot]

do you mind showing your work?

[fierce jay]

can you guys still check my answer tho

[maiden plank]

sure. how did you get your answer?

[fierce jay]

uhh its kinda silly but ill show u 😭

[shut obsidian]

Use log rules

[stoic halo]

How do you use integration in anything in motion

[shut obsidian]

!occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[shut obsidian]

Open a new one, for example #help-4

[fierce jay]

here's what I did, it all js seemed to cancel out BUT again not sure how correct it is

my working isn't proper cuz i had alr half solved it in my head when I first did this so 😔

[quasi depot]

how did you cancel things to 1/2

[maiden plank]

so

the top started at n and bottom started at n-1

presumably the series has an equal number of terms for both top and bottom

[quasi depot]

bottom should end with n+196 also

[maiden plank]

but they both somehow end at n+197

[shut obsidian]

It might be easier to do it discretly first

[fierce jay]

OHH YOU ARE RIGHT

[shut obsidian]

Instead of the other way around

[fierce jay]

yeahh fairs

[thin girder]

Also do you really need to use n=3 instead of just writing the numbers?

[fierce jay]

also, another question except idek how to start w this one can anyone gimme a hint

[quasi depot]

wait are we done with the other question?

[fierce jay]

not really i js followed what everyone does 💔

[fierce jay]

yes we are

WE ARE RIGHT

i dont have an answer key

i cant check

;(

[thin girder]

No the answer should be lg(100)=2

[fierce jay]

howw

[shut obsidian]

Can you do this

[thin girder]

Idk how you got 1/2 tho, explain that

[shut obsidian]

$\log \l(\frac 32 \cdot \frac 43 \cdot \frac 54 \cdots \frac{200}{199}\r)$

[fierce jay]

ITS WRONGG i js did it in my head thats why

[thorny flame]

Kepe

[thin girder]

The fraction simplifies to 200/2

[quasi depot]

both denominator and numerator have n, n+1, ...., n+196 so these cancel out leaving (n + 197) / (n-1) = 100

[fierce jay]

yeah ill try again

[shut obsidian]

Now draw in diagonals that indicate the cancelling

[fierce jay]

okk i see it

[fierce jay]

tyy

omg im lowk speeding thru questions i should give it time 😭

I HAVE 2 days to decide if i wanna sign up for this math competition or nah so im tryna do as many problem sets as i can to see if i uh

should do it or nah

its not looking good

[clear python]

Doesn't hurt to participate

[thin girder]

Try spliting the exponents in terms of 3 and 5

[thin girder]

I mean you dont lose anything so why not

[fierce jay]

ok so I got this far and im lost again what 💔

[fierce jay]

ur righttt imma do it

[thin girder]

Yeah that works

Group the 2¹¹ and 3^11^2 so that they have exponent 11

[fierce jay]

oh i js realized the notation is incorrect

3^11^2 and (3^11)^2 arent the same thing

lmao

[thin girder]

Yeah kinda needed brackets

[fierce jay]

i got lazy and i js put it in my calculator

found the 11th root

I GOT y=18

YAY

that seems like a very

correct

answer to me

[fierce jay]

but how do u do that?

they have different bases

[thin girder]

(2×3²)^11

This is what i mean

[fierce jay]

ohh thats what u meant

alright

ok my faith in my mathematical ability is halfway restored

ill sign up for it

tyy <3

[thin girder]

👍

[fierce jay]

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[full dove]

i need help simplifying this

[full dove]

this is probably wrong but here's my work

it is not completed yet

I feel this is not how you're supposed to do it

please ping me if you're responding

[topaz sinew]

@full dove Has your question been resolved?

[ruby tree]

I'm going to use ~ for not, ^ for xor, | for or, and nothing for and
Let u = xy
(uz)^u = uz~u | (~u | ~z)u = ~uu | ~zu = ~zu = xy~z

Then you're left with (xy~z)^(yz)

@full dove

[full dove]

let me write that down

oh, right

so all of that was not necessary

[ruby tree]

It's just a shortcut

[full dove]

can we take "common" (i do not know if that is the correct term in english) from two terms that are being xor-ed? like, is ab ^ bc = b(a ^ c)?

[full dove]

why did you think of taking xy as u and not something else?

[full dove]

also, i know it is redundant but is my work correct so far?

[ruby tree]

I believe that's true

[ruby tree]

Because xy was in both terms

I could have taken yz and worked with the other pair

[full dove]

i see

[ruby tree]

I think you got the very last term wrong

[lost bough]

what do we have here

differentiation?

[ruby tree]

Yeah it's true, but you need to be careful to remember the correct distributive properties

[lost bough]

oof nope

[full dove]

what do you mean?

[ruby tree]

https://lukas-prokop.at/articles/2021-08-24-on-distributivity-of-arithmetic-and-boolean-operations
AND is distributive over both OR and XOR, but XOR is not distributive over anything

[full dove]

eh

i see

[ruby tree]

(so (a^b)(a^c) is not a^(bc))

[full dove]

makes sense

[ruby tree]

Also OR is distributive over AND but not XOR...

I wouldn't bother remembering this; just that OR and AND distribute over each other

[full dove]

is there a way to solve such kind of questions without boolean algebra? there's k-maps, sure, but here we have terms both in SOP and POS forms

(also, i am trying to solve the whole thing but i do not think my answer is coming out to be correct)

[ruby tree]

Karnaugh maps should be fine

There's also https://en.wikipedia.org/wiki/Quine–McCluskey_algorithm but I haven't learned it

[full dove]

we have that in our syllabus but it has not been taught yet, hence the professor will not accept it

[full dove]

so we first convert POS terms into SOP (or vice-versa) and then solve the whole thing?

[full dove]

also, what answer are you getting after solving this?

[ruby tree]

Well, you could make a big truth table

[full dove]

is confusing to work with

if you are talking about making one with both POS and SOP terms

[ruby tree]

(xy~z)^(yz) = (xy~z)(~y|~z) | (~x|~y|z)(yz) = (xy~z) | (~xyz | yz) = xy~z | yz = xy | yz

[ruby tree]

8 lines for every combination of x,y,z, then (xyz), (xy), and (yz), then the xor of one pair and then the final result

[full dove]

xy~z | yz = xy | yz
how so?

[full dove]

are we talking about truth tables?

[ruby tree]

Well either z is false in which case it's xy, or z is true in which case it's yz

[full dove]

which law is this?

[ruby tree]

None in particular I'm just taking a little shortcut

xy~z | yz = y(x~z | z) = y(x | z) = xy | yz, right?

[full dove]

yeah, makes sense; sorry

[full dove]

this please?

[ruby tree]

X Y Z | XYZ | ... | XYZ^XY | ...

0 0 0    0            0
0 0 1    0            0
0 1 0    0            0
...
1 1 0    0            1
1 1 1    1            0

[full dove]

oh

(this did not load before?)

but yes, wait, that makes it so much easier

so we would get (xyz') + (xyz) + (x'yz) + (xyz) from this, right?

we simplify that further

or no?

[ruby tree]

Idk I haven't calculated

[full dove]

xy | yz
from this

anyway, i get your point

i find this approach a lot easier

[ruby tree]

I mean sure, but xy + yz is much simpler

[full dove]

i was talking about min-terms

like if we only constructed a truth table

[ruby tree]

Ok sure you'd get (xyz') + (xyz) + (x'yz)

[full dove]

yeah

all right, thank you so much!

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[honest forge]

we have a triangle ABC, a circle (O) with diameter BC intersecting AB, AC at D, E. Call the point that which BE and CD intersect H and the point where AH and BC intersect F. Through D draw a line that is perpendicular with BC that intersects (O) at D1. Prove that E, F, D1 are collinear.

[clear python]

Do you have a diagram

Even a rough one

[honest forge]

Excuse the sideways img

[smoky sparrow]

,rccw

[thin girder]

Ok so there are a few ways which we can use to prove colinearity
Which do you think we should choose

[honest forge]

Maybe angles since there are many equal ones?

[thin girder]

By angles you mean the sum =180 right?

[honest forge]

yeah

[thin girder]

Ok so we need to prove BFD +BFD¹+DFA +AFE =180

2 infos that can help us is that BFD =BFD¹ and that BFD +DFA =90 degrees

[honest forge]

and F1 = F2 too i think

[thin girder]

Yes using the assumption and these infos, THAT is what we need to prove

Now can you see any angle thats also =F2 (hint is look at FHEC)

[honest forge]

i think it's HEF but i can't prove it just yet

[topaz sinew]

@honest forge Has your question been resolved?

[thin girder]

Nope its not
Can you prove that HEFC is a cyclic quadrilateral, meaning its 4 points are points on a circle?

[topaz sinew]

@honest forge Has your question been resolved?

[topaz sinew]

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[merry helm]

for b

[merry helm]

why 9x area of ABC

and not 3x

[thin girder]

Ratio of 1/3
The sides grow by 3
The height of the triangle also grow by 3

=> grows by 9

[outer laurel]

if you have a square with side length 1 and double the side length to 2, what happened to the area?

[merry helm]

4x

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[bleak cosmos]

help me solve this my friend challenged me to solve this

[fallow heart]

Is this a txt? If so, write it as a normal message, not as an attachment

[thin girder]

Thats actually a virus, trust

[fallow heart]

Yeah I'm really suspicious too

[bleak cosmos]

wait i'm sending image

[thin girder]

Now why can i open this with hecking Google wallet

[bleak cosmos]

that's also a image i renamed it

[thin girder]

Since when was google wallet a thing even

[mighty geyser]

,r

[bleak cosmos]

my friend gave me this question on whatsapp and challenged me to solve this

he is actually topper of my class

he aces every exam

[acoustic pecan]

a start would be finding p+q as a sum

[smoky sparrow]

yes, you can do (1st term of p + 1st term of q) + (2nd term of p + 2nd term of q) + ...

[bleak cosmos]

actually i'm in 10th grade and i don't know that much

[bleak cosmos]

but thankyou for you idea

[blazing karma]

First year of highschool?

[bleak cosmos]

my math is weak but geo is good

[molten crown]

do you know log properties

[bleak cosmos]

a little

[molten crown]

if you do then this problem is simple

[blazing karma]

Alr whats the capital of Australia

[bleak cosmos]

like 10 is the base in question

[bleak cosmos]

geo means geometry

[blazing karma]

Oh

[molten crown]

Do you know that log base 10 of 2^j can be written as j * log base 10 of 2

[bleak cosmos]

yes

[molten crown]

then simplify what p and q are

[fallow heart]

The key knowledge required for this exercise are:

• log properties
• Gauss formula

@god.chess

[molten crown]

whats gauss formula

[fallow heart]

1 + 2 + ... + n

[molten crown]

oh wait thats its name?

[fallow heart]

Yup

Even though it's just a particular case of sum of arithmetic sequence

[bleak cosmos]

(n+1)n/2

[molten crown]

yes

[fallow heart]

Awesome, you'll need it towards the end

[molten crown]

find p and q in terms of n and add them

remember that log a + log b = log ab

[bleak cosmos]

okay

(p=n\cdot \log _{10}2)

[thorny flame]

GODSWORD

[bleak cosmos]

(q=n\cdot \log _{10}5)

[thorny flame]

GODSWORD

[fallow heart]

Not quite

[blazing karma]

Here the p and q notations are wrong it should be p_n and q_n

[fallow heart]

Show the steps you did

[bleak cosmos]

i'm dum b

i got p + q = (n\cdot \log _{10}2+n\cdot \log _{10}5=66)

[thorny flame]

GODSWORD

[bleak cosmos]

(n(\log _{10}2+\log _{10}5)=66)

[thorny flame]

GODSWORD

[bleak cosmos]

oh wait

[bleak cosmos]

ye

that means i get n(log10(2 * 5)) = 66

[molten crown]

nah

[fallow heart]

Are you ignoring the summation Σ ? 😬

[bleak cosmos]

n(log10(10))=66

[fallow heart]

No no stop @bleak cosmos

[bleak cosmos]

i did it like this

[fallow heart]

And that's wrong 😅

[molten crown]

How did you get this

[fallow heart]

You forgot the summation

[fallow heart]

@bleak cosmos

[bleak cosmos]

oh log also have power as j

[bleak cosmos]

yes

(p=\sum _{j=1}^{n}\log _{10}(2^{j}))

[thorny flame]

GODSWORD

[bleak cosmos]

(p=\log _{10}2\cdot \frac{n(n+1)}{2})

[thorny flame]

GODSWORD

[fallow heart]

There you go, awesome 👍

[blazing karma]

It is better to do the adittion p+q first

[bleak cosmos]

you mean this?

[fallow heart]

Huh?

[bleak cosmos]

(q=\sum _{j=1}^{n}\log _{10}(5^{j}))

[thorny flame]

GODSWORD

[bleak cosmos]

same for q

[fallow heart]

Good

[bleak cosmos]

(q=\log _{10}5\cdot \frac{n(n+1)}{2})

[thorny flame]

GODSWORD

[fallow heart]

So now you can solve p + q = 66

[bleak cosmos]

(log10 5 * n(n + 1)/2) + (log10 2 * n(n+1)/2) = 66

n(n+1)/2(log10 5 + log 10 2) = 66

n(n+1)/2 (log10(2*5))=66

n(n+1)/2 (log10(10))=66

n(n+1)/2 = 66

n^2 + n = 132

n^2 + n - 132 = 0

n^2 - 11n + 12n - 132 = 0
n(n-11) + 12(n-11) = 0
(n+12)(n-11) =0

so n either equal to -12 or 11

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[bleak cosmos]

thank you guys for you help

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[sharp wyvern]

what have you tried

[faint pollen]

Is this correct?

[prisma mesa]

Nope

[sharp wyvern]

no

this break is wrong

[prisma mesa]

4x^2 - 1 is (2x+1)(2x-1)
not 4x^2 + 1

there is no formula for sum of squares

[sharp wyvern]

technically you can use i

a^2 - b^2 = (a+b)(a-b)

[prisma mesa]

no useful formula

for this purpose

[sharp wyvern]

technically you can take the real part of the answer , but yeah not practical

[sharp wyvern]

any other thought?

[prisma mesa]

does the 1/(4x^2 + 1) at least look remotely similar to some well-known integral?

[broken saddle]

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

[thorny flame]

ImOakley

[prisma mesa]

this doesnt really matter as that part is already wrong

[prisma mesa]

Yeah, nice, perhaps try some substitution to bring it closer to this

[faint pollen]

What substitution

[topaz ledge]

We change variables

[prisma mesa]

what do you think?

How can you change that 4x^2 to x^2 (or u^2)?

[faint pollen]

Like i say y^2=4•x^2

[prisma mesa]

sure, u could start with that. Can you simplify it tho?

[faint pollen]

y=2x

[prisma mesa]

yep, thats it

do the sub and youll be pretty much done

[faint pollen]

Like this?

[prisma mesa]

you should be dividing / multiplying by something when changing dx to dy

[faint pollen]

To what?

[prisma mesa]

when you do substitution and you want to change dx to dy, you gotta divide by dy/dx

[sharp wyvern]

you considered
y = 2x

but then you changed dx into dy
implying-
dx = dy

isnt that concerning

[prisma mesa]

sorry, i meant divide here

[faint pollen]

Like this?

[sharp wyvern]

uhh..

[faint pollen]

Or like this?

[sharp wyvern]

( both are wrong , im not sure whats the difference excpe the colon in one) ,
lets try this
y = 2x
can you differentiate both sides

[faint pollen]

(y)'=1

[sharp wyvern]

thats incorrect

f(x) = kx
f'(x) = ?

[faint pollen]

k

[sharp wyvern]

so if
y = 2x
y' =?

[faint pollen]

2

[sharp wyvern]

can i say
(because y' = dy/dx)
dy/dx = 2
=> dy = 2 dx

or
=> dx = dy/2

[sharp wyvern]

bingpot

just get rid of this imposter

[faint pollen]

Ok🤣

[sharp wyvern]

always remember this while substituting
you can call this a variable change too

correct this too and also conclude this in x as y wasnt given in the question as your final answer

[faint pollen]

Is this correct?

[sharp wyvern]

where did the /2 go 💔

[faint pollen]

Where should it be /2

😭

[sharp wyvern]

when you did
dy/2

why didnt you cancel 8 into a 4

[faint pollen]

I didnt write 4

[sharp wyvern]

didnt*

sorry

[faint pollen]

Okok

I didnt know i could do that

[sharp wyvern]

look at these two lines,

[sharp wyvern]

i mean yeah, dissapearing the 2 is kinda considered rude
infact just killing off numbers is very disrespectfull to math 🥀
yes you got it now , do you understand why

[faint pollen]

Yes

🤣

Thank you

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[sharp meteor]

from whats there

I assume the set is A1

but because it says all sets are infinite

makes me assume that the statement in true

[strange whale]

"the set" ?

[sharp meteor]

sorry not set

like

what i mean is

if i was to look at A1

i would have everything because everythiong else has a part of A1

idk how to define that properly sorry

[strange whale]

ok but if A2 is strictly inside than A1

that intersection won't be A1

[soft sorrel]

What’s

[sharp meteor]

would it be union?

[strange whale]

the union of all of these would be A1 yes

[sharp meteor]

hm i sort of get it

[strange whale]

but here the sets are getting smaller and smaller

[sharp meteor]

but its not clicking

[sharp meteor]

yep this makes sense

[strange whale]

so the intersection gets smaller and smaller in a sense

the more sets you intersect

[sharp meteor]

so there woudl be an intersection between A1 and A2, and A1 and A3 and so on right?

[strange whale]

the things that are in all of the A1, A2, A3 ...., that's the intersection

[sharp meteor]

ahhh

yh it gets smaller

[soft sorrel]

Where are you from

[sharp meteor]

so this would not be infinite

but tend towards 0

?

[sharp meteor]

why is this relevant

I am from spain

[strange whale]

well it depends really

for both statements

[soft sorrel]

I from Poland 🇵🇱

Ok

[sharp meteor]

how so

[strange whale]

the shrinking could plateau out in a sense

like take An = [0, 1+1/n] (the interval)

[sharp meteor]

ah i see

alright ty

how can we prove these statements?

is it just case of logic?

[strange whale]

we ain't even finished with the previous one

[sharp meteor]

😭 oh bruh

[strange whale]

I did say could

[sharp meteor]

it wont increrase tho

[strange whale]

you don't prove stuff with just one example

I gave you a nice case there

[sharp meteor]

this is what the exercise wants

[strange whale]

what if there are un-nice cases ?

[sharp meteor]

ive just started analsyis 😭 why is this so tough

[sharp meteor]

wdym?

[strange whale]

if you change this just a little bit you can get a counterexample to your statement

just saying

[sharp meteor]

i was thinking

hm wait

if we leave it at n

[0,n]

then An would be an empty set?

since its infinite?

[strange whale]

no they all contain [0,1]

so the intersection is at least [0,1]

[sharp meteor]

no as in

A1 = {1,2,3...} A2 = {2,3,4...}

then An would be empty

[strange whale]

well An would be {n, ...}

but the whole intersection would be empty yes

[sharp meteor]

An is infinite

[sharp meteor]

yea this si what i mean

so intersectionw wuld be empty

which is a counter

[strange whale]

ok but that doesn't fit the condition in the question tho

A1 >= A2 >= A3 ....

[sharp meteor]

how does it not?

A2 is a subset of A1

[strange whale]

wait crap

yeah

ok good

[sharp meteor]

what was ur idea on changing this?

to reach a contradiction

[strange whale]

[0, 1/n]

[sharp meteor]

wait that would mean it would tend to inf

[strange whale]

no

[sharp meteor]

wait what

[strange whale]

what's "it" then ?

you ain't being clear

[sharp meteor]

the intersection

[sharp meteor]

mb

[strange whale]

well what do you think the intersection is ?

[sharp meteor]

its whats in all A1 to An

oh wait

nah im not sure wym

[strange whale]

what's the intersection of all the An = [0, 1/n] ?

is my question

[sharp meteor]

0?

[strange whale]

wdym by 0 ?

empty set or the set with 0 in it ?

[sharp meteor]

latter

[strange whale]

alright why is it so ?

[sharp meteor]

waitt

it will be 0 and 1

cause A1 will have 0, 1 A2 will have 0, 1, 1/2 ?

right?

[strange whale]

no

[sharp meteor]

[strange whale]

A1 has all real numbers between 0 and 1

[sharp meteor]

ah yea sorry

[strange whale]

A2 has all real numbers between 0 and 1/2

1 ain't in A2 already

[sharp meteor]

ohh

[strange whale]

gbye

(to 1)

[sharp meteor]

ahh

i see

so it would be 0

from that case

my understanding was wrong

[strange whale]

if you find only one set among the A_n which doesn't have x, then boom x is not in the intersection

[strange whale]

it's the other side of this coin

[sharp meteor]

yep this makes sense

so the case we are talking abt

0 would be included in all sets right?

[strange whale]

right

the harder thing is showing everything else ain't in the intersection

[sharp meteor]

hm

what kind of proof is this?

cause i dont havce any clue

[sharp meteor]

on how to approach this

[strange whale]

something a bit like this

but more general

like take an x in [0,1] (that's the only things we have to look at really)

and for that x find an n such that x ain't in An

n will depend on x most certainly

like the smaller x is, you expect to take bigger and bigger n's to exclude x

[sharp meteor]

what abt just x =1

waut not

no no

cause A1

[strange whale]

when I say take an x, I mean it's arbitrary

[sharp meteor]

ohh

[strange whale]

math speak sometimes...

[strange whale]

well in (0, 1] actually

we know 0 is already in the intersection

[sharp meteor]

yea

no other xs will be in the intersection

but to prove that uh

can u say that as x gets smaller, like u said, the greater the value of n, hence the "smaller" the chance of an x that intersects?

idk how to phrase it

[strange whale]

that's an intuition for the proof

not an actual proof

[sharp meteor]

hm

[strange whale]

like a proof would be actually giving an n as a function of x as I said

it's very eps-delta like really

[sharp meteor]

what is that 😭

[strange whale]

you said you were doing analysis right

[sharp meteor]

yes

started it yesterday

[strange whale]

you ain't seen limits yet

ah ok

[sharp meteor]

basically learning pre req rn

[strange whale]

yeah I mean the counterexample you gave w/ integers is pretty f-ing good

[strange whale]

well when you'll see eps-delta you'll remember our convo lol

[sharp meteor]

yh it was somehat similar to the example the book gave so i tried to use that asw

[sharp meteor]

oh lol

ty tho

[strange whale]

but you can try to struggle w/ it a bit to spoil yourself

[sharp meteor]

with the eps-delta?

[strange whale]

no as in just bash your head against that problem

by spoiling I mean essentially discovering the spirit of eps-delta proofs all by yourself

not reading ahead

[sharp meteor]

hm, im assuming eps-delta is smth being infitesimly small

mb cant spell

very small rate of change

[strange whale]

well eps-delta is how limits are defined

so yeah looking at v small stuff

[sharp meteor]

so we say as x becomes very small

yh icl idk what to say

instead of using colloquial language lol

[strange whale]

for all x (in (0, 1] here), there exists an n such that x is not in An

[sharp meteor]

that would be the beginning

[strange whale]

that's just the objective

[sharp meteor]

yep

yea, im going to sleep on that idea

[strange whale]

but stated a bit more formally

[sharp meteor]

ill try tomorrow morning and come back here

lemme just write the problem again and ss

set An = [0,1/n] show that the intersection of A1,A2,A3,A4...An is just the set contaning 0:
for all x (in (0, 1] here), there exists an n such that x is not in An

does trhat suffice?

[strange whale]

yea

[sharp meteor]

alr tysm

.close

[topaz sinew]

Channel closed

Closed by @sharp meteor

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wise ivy]

im confused about the red box

how come we're just assuming P(k) is true?

because isnt that what we want to prove

since k is arbitrary

[acoustic pecan]

we're proving that its true for all natural numbers

[wise ivy]

right

[acoustic pecan]

the inductive step is assuming its true for some k

then using that to prove its true for k+1 on that basis

[wise ivy]

but that "some k" can be arbitrary, right? so i dont understand how its not proving it for "all k" since k can be anything

[acoustic pecan]

k is any some natural number

we have the base case here to be 1

the induction shows 'if true for k, then its true for k+1'

so by 1 being true, 2 is true, so 3 is true, so 4 is true,... so all natural numbers have it be true

the k being arbitrary is the point, its a general rule

the base case is what makes it 'real'

[wise ivy]

I'm still a little confused, sorry

in the hypothesis, our assumption is that P(k) is true for any natural k, right?

so doesn't that literally already prove what we want, because k is allowed to be anything, which is what we want.
so, from my understanding, aren't we just assuming what we want to prove?

[acoustic pecan]

the k in P(k) isnt any natural number, but some natural number

its not a rule, but a case

[wise ivy]

oh shoot

i see

ah that makes sense

ookok tysm

.close

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Channel closed

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[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pure magnet]

Help

[pure magnet]

piecewise functions

[celest swallow]

What’s the question

[pure magnet]

I needa lock in bc ive been dling my PIQ’s

[worthy storm]

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[pure magnet]

HELP me lock in

[celest swallow]

Is that the question?

[ocean pewter]

The graph has two components, correct?

[pure magnet]

Yes

[celest swallow]

Answer if you want:
||f(x) = {1/3x + 3 for x <=1
{6 For x > 1||

I think at least

Have you tried solving it?

What’sthe hard part?

[pure magnet]

i dont get it bc i havent been in that class due to applying for colleges

Im at a B right now

[celest swallow]

Want explanation?

[pure magnet]

It was wrong😅

I gotta do 5 of these

[celest swallow]

My answer?

[pure magnet]

Yes

The “for” part was x<-3 and x>1

[celest swallow]

Oh

It’s been like 20 minutes so you can ping a helper

<@&286206848099549185>

[dry path]

help Sora code

[pure magnet]

are you a helper

whoever is a helper can you dm me

[ocean pewter]

Mb, was receiving help myself

What does the left half of the piecewise look like?

[pure magnet]

want a picture of a different question

[ocean pewter]

Sure

[celest swallow]

Bro got assigned to Help 42 😂

[ocean pewter]

Is there a stigma about help 42?

[celest swallow]

“The Meaning of Life, The Universe, and Everything”

[ocean pewter]

What does the graph look like to the left of x = -3?

[pure magnet]

uhh im not sure

[ocean pewter]

It's a line, right?

[pure magnet]

like i said i barely know any of these due to doing my PIQ’s

[pure magnet]

yea

an arrow that goes to infinity

[ocean pewter]

Yeah

Every line has a point on it

And every line has a slope

[pure magnet]

yes

[ocean pewter]

Would you be able to calculate the slope of the line?

[pure magnet]

no i dont know how to really do that

[ocean pewter]

Given two points, the slope if the difference in y divided by the difference in x

[pure magnet]

I promise im not dumb i was so locked in in alg 2 and geometry i got A’s ive just been focusing on college

[ocean pewter]

I'm not saying you are

[pure magnet]

I know im just saying

do you know the answer

[ocean pewter]

Honestly not off the top of my head, but I'd rather work with you to find an answer than spoonfeed it

[pure magnet]

Okay sweet

[ocean pewter]

Name any 2 points on the leftmost line

[pure magnet]

-6,1

And -9,3

[ocean pewter]

Good

What is the difference between the y coordinates?

[pure magnet]

ones up by 2

[ocean pewter]

Good

And the difference between the x coordinates?

[pure magnet]

By -3

[ocean pewter]

Yes

The slope is then just 2/(-3)

[pure magnet]

Yes

[ocean pewter]

Since we found a point while finding the slope, we can find an equation of the line using point-slope form:

$y - y_0 = m(x-x_0), $ where m is the slope, $(x_0,y_0)$ is a point on our line

[thorny flame]

CalculusDude

[pure magnet]

okay, and all of this will give me a function in the end right

[ocean pewter]

This will give an equation for the left line

[pure magnet]

Like this right

[ocean pewter]

When you solve for y (I'm assuming the question wants you to solve for y)

Yeah