help-19

[dawn roost]

Tkssss

Thanks a lot

Close.

[amber schooner]

other way

[dawn roost]

.close

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[keen solar]

Can anyone tell how I can write the range for number 3😭😭😞😞

[keen solar]

f(x) = -(x-4)² + 2

[unkempt lichen]

do you have the maximum value?

[keen solar]

What's that

[unkempt lichen]

of this function

since this function opens downward, it must have a maximum value

hint: look at your vertex

[keen solar]

Well I mean I know its 2 but

How can I write it

[unkempt lichen]

so the max is 2, and it can extend infinitely down there, right?

[keen solar]

{y | y > 2} like this?😭😭😞

[unkempt lichen]

flip that inequality sign

if you write it like this, your graph opens upwards

you're saying that's the minimum of y

[keen solar]

Ohh

[unkempt lichen]

also, don't forget <=

not just <

[keen solar]

{y | y < = 2}

Like that?

[unkempt lichen]

correct, but normally i would not write a range in set notation like this i suppose
but since you wrote the other two like that, yeah

[amber schooner]

to develop the habit of using inequalities, in the future you could note that -(x - 4)^2 <= 0 with equality when x = 4 so adding 2 to both sides gives -(x - 4)^2 + 2 <= 2 with equality when x = 4

[keen solar]

Oh alright

Thank you all for your help

.close

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[lyric hedge]

hi!! i’m in highschool IM2 and i was wondering if someone could help me with quadratic sequences. i get the concept but i dont really know how to solve for the equation if that makes sense 😭 i have a quiz tmr so if anyone’s free pls help!

[crude remnant]

can you post ur q

[lyric hedge]

yes!! give me a sec

[crude remnant]

so i dont know the terms you guys use in class but i would recommend taking a look at the jumps between each pair of y's

[lyric hedge]

so the common differences?

[crude remnant]

whats the sequence of differences

[crude remnant]

ok from this its pretty easy to get future terms and previous terms right?

[white brook]

/transler

[amber schooner]

what

[lyric hedge]

ya i know how to do that i just dont know how to put it into the equation

[mental lotus]

do you know the general form of how a quadratic equation looks like?

[lyric hedge]

my teacher said we can use the standard one which is an^2+bx+c

or the vertex one

[mental lotus]

good

[mental lotus]

so you know that, there is gonna be some specific one value for each of a,b,c such that you get the equation that fits these numbers

that means, if you put the values of (x1, y1) into the equation y = ax^2+bx+c, you will satisfy the equality if the a,b,c are correct

[lyric hedge]

uhh 😭

[mental lotus]

do you get what I am saying here?

[lyric hedge]

kinda

[mental lotus]

feel free to ask if you got stuck somewhere

[lyric hedge]

when my teacher explained he said to solve for b it’s the 2nd common difference - 1st common difference - 3

and for c it’s the whenever x = 0

[mental lotus]

yea, thats the similar idea to what I am saying

I was just planning to explain the idea of what that means

but, sure I can directly show you the equation

[lyric hedge]

hold on ima try to solve it rq

ok i got an= 1x^2 - 4x + 2

does that seem right?

[mental lotus]

yep

you can check for yourself by putting in the numbers back into the equation

you know for x1 = 1, you have y1 = -1

[lyric hedge]

yeah

[mental lotus]

and indeed, if you put it in the equation, you get it back

and same for the rest too

[lyric hedge]

ok thank you!

i know how to do the work but i js wanted to double check cuz my teacher sucks at teaching 💔

[mental lotus]

np

[mental lotus]

you might wanna watch this, if you still feel unsure https://www.youtube.com/watch?v=7Isa4P8lISc

[lyric hedge]

omg thank you ur genuinely a lifesaver

[odd edge]

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[fleet oracle]

Photo loading

[fleet oracle]

Pre-Calculus, algebra 2 review

[mental lotus]

you cant divide the factors like that

[unkempt lichen]

for all of them, common factors can include binomials too!

[fleet oracle]

16-21

[unkempt lichen]

in case you didn't know that

[mental lotus]

in 17, you just made the common factors disappear

[fleet oracle]

I am absolute stuck on them ☹️

[mental lotus]

but you did it correctly in 16

[unkempt lichen]

factor the common factors out first

[fleet oracle]

Let’s start with 17, what did I do wrong?

[mental lotus]

you see, in 16, you preserved the common factors, by taking them outside the equation, and combining the rest of the terms together

however, in 17, you made those factors disappear

you need to take them as common factor, and combine the leftovers into a single term

all this needs to be multiplied by (x+1)(x-3) still

[fleet oracle]

( 2-3 )? <- wrong

[mental lotus]

2-3 is not wrong

[fleet oracle]

So (x+1)(x-3)(2(x-3)-3(x+1)?

[mental lotus]

the algebra in this snap is correct, but you needed the extra two factors too thats all

[fleet oracle]

How do I multiply that?

[mental lotus]

this is correct

[fleet oracle]

I know, do I just leave it as it is?

[mental lotus]

you can simplify the last part as you did before

make it -1*(x+9)

so you get -(x+1)(x-3)(x+9)

if you do that, then 17 becomes complete

16 was already good, so no problems there

but you need to redo 18

sadly, that one step you did is also wrong

[mental lotus]

,rccw

[mental lotus]

?

[fleet oracle]

I’m stuck…

[mental lotus]

why is your 2x-3x=5x?

you were right the first time around

[fleet oracle]

Oh I messed up basic prealgebra

I’ll fix it

[mental lotus]

your algebra was all good you just didnt write the two factors thats alll

you didnt have to do this

[wooden python]

what's the original problem & the goal here

[odd viper]

18. Take common (x+1)^3/2

[fleet oracle]

17. Solved

[fleet oracle]

I don’t understand…

What about the fractions?

[odd viper]

Take them common

In 7/2
3/2 also came so take it outside eq will be like
(x^2+1)^3/2 (1+(x^2+1)^2)

[fleet oracle]

Where does the 7/2 go and what’s the +1?

[wooden python]

$(x^2+1)^{7/2} = (x^2+1)^{3/2} \cdot (x^2+1)^2$

[clever fjord]

Ann

[wooden python]

maybe it's easier if we temporarily replace x^2+1 with a single letter like z

z^(7/2) = z^(3/2) * z^2

z^(3/2) + z^(7/2)
= z^(3/2) * 1 + z^(3/2) * z^2
= z^(3/2) (1 + z^2)

@fleet oracle does that make it clearer

[fleet oracle]

It doesn’t…

[wooden python]

ok let's see

let's maybe also remove the fractional exponents and go for a different example

z^4 + z^9 = z^4(1+z^5)

does this make sense to you

(entirely different expression with no direct link to your hw problem but im trying to illustrate a point with it)

[fleet oracle]

Yes it does.

[wooden python]

ok right

this is because z^9 can be written as z^4*z^5 so we can factor a z^4 out of both terms

[fleet oracle]

Can I factor 3/2 out of both terms from question 18?

[wooden python]

and since the first term was just z^4 on its own, after factoring z^4 out it becomes just 1

[wooden python]

not 3/2 itself.

you can (and should) factor out (x^2+1)^(3/2).

[fleet oracle]

I got to go so I unfortunately have to leave the last 4 questions unanswered to get points 😓

I’m just gonna have to submit…

[wooden python]

welp, that's your decision to make

[odd edge]

@fleet oracle Has your question been resolved?

[rough birch]

do you have other questions king?

[fleet oracle]

Too late…already submitted. 😪

[rough birch]

okk, just .close then to close the help channel

[unkempt lichen]

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[unkempt lichen]

and all the best with your marks

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[vale knoll]

Hey

[vale knoll]

Should I be given the last mark ?

Cuz Edexcel cut me off the last mark

[odd edge]

@vale knoll Has your question been resolved?

[vale knoll]

No hasn’t

@rough birch

Anyone

<@&286206848099549185> I

Why am i being ditcheddddd

Guys

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[woeful briar]

@vale knoll hi

[woeful briar]

.reopen

Nvm idk if I can help

.close

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[marble laurel]

What

[odd edge]

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[fickle bluff]

Im not sure what to do on this question. Cofactorization? Can I use the properties of eigenvectors on this?

[fickle bluff]

Am I allowed to do row elimination so my matrix is triangular?

[low locust]

no

row elimination completely fucks up eigenvalues

[fickle bluff]

[low locust]

wait

do you mean on A or A-lambda I

[fickle bluff]

on A so the matrix is triangular then the diagonal is just the eigenvalues right?

[low locust]

you cant do that

[fickle bluff]

.close

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[clear cobalt]

I need help with question C ii

[wooden python]

here is an important detail

[clear cobalt]

Oh ok

Well ik it’s 90

[wanton sierra]

well if that angle is 90 degrees, what are the other angles of that bottom triangle

?

[clear cobalt]

45

[wooden python]

you also know x = h cot(10°) by the same logic as for y

[clear cobalt]

What

[wooden python]

<@&268886789983436800> scam

[clear cobalt]

Bruh what is this

[wooden python]

you also know x = h cot(10°) by the same logic as for y
so you can apply pythagoras

<@&268886789983436800> second time

[clear cobalt]

Please stop that

[clear cobalt]

Thanks

Makes sense now

.close

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[teal jungle]

Can somebody give me a hint on where to go on from here?

Also for the step where I square root both sides to remove the square for cos, how do I know where the answer is positive or negative?

[crystal charm]

it's both

you get cosx = 3/4 and cosx = -3/4

[teal jungle]

Should I write that out with the plus minus symbol

[vernal yacht]

You’re pretty close from here

[warped glacier]

which quadrant does pi/2 < x < pi represent?

[teal jungle]

Second quadrant

[warped glacier]

yeah, and so will cos be positive or negative in the 2nd quadrant?

[teal jungle]

I’m just not sure where to go from the triangle I drew 🥲

Negative

So it should be cos x= -3/4 right?

[warped glacier]

yeah and then sin will be positive, so that means sin/cos will be negative

[warped glacier]

right, so what's opposite/adjacent in your triangle

[teal jungle]

Root 7 over 3

[warped glacier]

yep, so you just need the negative of that!

[teal jungle]

i dont really understand how tan will give x

[warped glacier]

because in actual fact you have $\frac{\sqrt{7} / 4}{-3 /4} = - \frac{\sqrt 7}{3}$

[clever fjord]

south

[warped glacier]

it's literally SOHCAHTOA

but instead of 3 you have to write -3

[teal jungle]

ill get tan x = -sqrt 7 / 3, but how do i find x? i cant get an exact value from this

😭

[warped glacier]

read the question again.....

[teal jungle]

omfg

im so sorry

[warped glacier]

it happens

no worries!

[teal jungle]

thank you so much

[warped glacier]

.close

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[waxen pivot]

Hi, so I have the following as answer for an exercise I'm making. How can you say that the two angles are the same?

[waxen pivot]

The alpha in red and the alpha in black

[zealous reef]

cause they have the same sine

[waxen pivot]

Ok so how can you say that

[zealous reef]

sine a = opposite/hypotenuse

whcih has been defined to be 1/5 for this angle

and the red angle also has a sine of 1/5 so they're the same

can you send the whole problem though

it's hard to say without knowing what these mean

[waxen pivot]

Yes, give me a minute, I will translate

A bundle of protons is brought to high speed and curved in a magnetic field like shown in the picture. In the zone where there is no magnetic field, the bundle goes straight. Then the bundle comes in a zone of 5 cm width where there is a magnetic field. It follow the circlular path with r=25cm. Then it leaves the magnetic field and goes straight again. The angle between the path before and after the magnetic field is called a. Which of the following is valid.

[waxen pivot]

And this is the solution, so I'm wondering how you can say that a is the same as that angle in red

[odd edge]

@waxen pivot Has your question been resolved?

[waxen pivot]

Could someone help me out please?

[crystal charm]

you can ping helpers

[waxen pivot]

Ok

<@&286206848099549185>

[odd edge]

@waxen pivot Has your question been resolved?

[waxen pivot]

Damn this sucks, I guess not then

.close

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[glossy basin]

@waxen pivot I am not sure if this is correct this is what I think.

The green is the beam of protons.

The beam left the rectangle box tangent to the arc, so angle between green beam and red radius = 90°.

Let the above black line outside the box denote the old path of beam and the below black line be parallel to the above black line and be drawn at where proton left the box.

We will later prove how our given alpha (angle between proton beam after leaving field and the above black line is equal to the angle between proton beam leaving field and the below black line, so for now just accept the angle alpha is same as shown in figure.

If we join the green beam which left the magnetic field to the centre of the arc, such that the yellow line is parallel to the above 2 black lines, we get this triangle in the figure.

If we let theta be angle of arc, then we get by triangle theta is the angle on right between green and yellow lines.

Now, between two parallel lines (below black line and yellow line), green is transversal and so angle theta = angle alpha.

[digital shore]

nvm

[glossy basin]

no, this will prolly close automatically after some time

[waxen pivot]

Thank you!

[waxen pivot]

Yeah this makes sense, appreciate it. Worth the wait

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[proud vine]

guys so i completed high school 1 and a half years back, and now im in uni, doing engineering. the sem 1 maths module is what i did in 11th and 12th grade. and well honestly rn im kinda cooked cause i have a class test in week 8, and rn we're in week 4. any tips on how i can improve?? i have limits and derivatives, continuity, and integration

[crystal charm]

you practice them

[unkempt lichen]

do you have particular questions you are stuck on?

[proud vine]

alrighty

[crystal charm]

also maybe it's best you go ask this in #study-discussion

[crystal charm]

I mean it really does come down to practice a lot

[proud vine]

pretty much i get stuck when it comes to limits

[unkempt lichen]

if you are stuck on some questions due to not knowing techniques, you can post them here as examples to practice some of the common techniques

[crystal charm]

and also watching videos to understand concepts

[proud vine]

i'll do that then

[unkempt lichen]

but for open-ended questions like yours, it might be better to ask in discussion channels

[proud vine]

sure, i did but i didnt get replies yet, but thank you guys

[unkempt lichen]

alright, good luck!

[odd edge]

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[trim sentinel]

For a positive integer n, define n! = 1 × 2 × · · · × n. Call a quadruple (a, b, c, d) of positive integers red if 1 < a < b < c < d and a!b!c! = d!. We say
(ai, bi, ci, di) < (aj , bj , cj , dj ) if di < dj . Arrange all red quadruples in ascending order. If
the fourth quadruple in this list is (a4, b4, c4, d4), compute c4 + d4

[sand oak]

do you have an idea how to do the brute force approach

as in enumerate each red quadruple in order

[odd edge]

@trim sentinel Has your question been resolved?

[trim sentinel]

brute force is not tidy

[sand oak]

then do you know any instance of a red quadruple. doesnt even have to be the first one

[ashen yew]

1!3!5!=6!

oh mb a>1

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[marsh bridge]

Does anyone know how I can figure out these values easily?

[amber veldt]

there are 2 fundamental approaches to these types of problems

one is, to add one equation to another. the other, to subitute one equation into another

for the former example, what happens if you add the equations together?

for the latter example, what happels if you write the first equation as A = -B ?

if you need an explanation of WHY you're allowed to do things like add equations to one another LMK

[sinful grove]

When you have something of the form 1/(x-a)(x-b) with a ≠ b and want to write it as A/(x-a) + B/(x-b) then there’s a very easy and direct approach which avoids solving a system

Notice that if 1/(x-a)(x-b) = A/(x-a) + B/(x-b) then we can recover A by multiplying both sides by x-a and plugging in x=a

And vice versa for B

So ur rule for getting A is 1/(a-b) while for B it’s -1/(a-b)

[marsh bridge]

Ah I forgot about solving systems, went back to review that and now I’ve remembered how to solve it

.close

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[wanton bison]

no translation no trabajo

[pastel orbit]

ah? didn’t you get help with this last night?

[lone elbow]

symmetric: xRx ∀x∈A

antisymmetric: xRy , yRx => x = y

symmetric: xRy => yRx

[wanton bison]

last night was so hard, the hungover made him forget

[lone elbow]

transitive: xRy ,yRz => xRz

[unkempt lichen]

i mean...

[lone elbow]

equiv rela = reflexive + symmetric + transitive

[wanton bison]

Actually, it's good for you

[lone elbow]

poset = reflexive + antisymmetric + transitive

[wanton bison]

you get to farm helpful XP again

[unkempt lichen]

no thx, it's midnight here

[lone elbow]

pre order = reflexive + transitive

[mental lotus]

no way you are letting that stop you

[lone elbow]

all the relations on A that are symmetric and antisymmetric are such that

xRy => yRx
xRy, yRx => x=y

[unkempt lichen]

i need to touch grass tmr or i will eat grass for the resto of the month LOL

plus i already answered part a more or less to completion

[lone elbow]

so basically the relations which have only reflexive pairs

[wanton bison]

real just refer to last help channel

[lone elbow]

basically, all relations in A that have pairs that are solely reflexive are symmetric and antisymmetric by nature

for a relation to be of partial order and an equivalence relation we need

partial order = reflexive + antisymmetric + transitive

equiv rela = reflexive + symmetric + transitive

both = reflexive + symmetric + antisymmetric + transitive

[wanton bison]

do you have a question btw or are you just active recalling

[lone elbow]

1. xRx ∀x ∈ A
2. xRy => yRx
3. xRy, yRx => x=y
4. xRy, yRz => xRz

[lone elbow]

lets say I have a relation R that has only reflexive pairs, it is antisymmetric and symmetric by nature, but is it transitive?

what I mean is , R = {(1,1),(2,2)} for example

and A = {1,2}

2R2, 2R2 => 2R2

1R1, 1R1 => 1R1

[unkempt lichen]

so... have you answered your own question?

[lone elbow]

is hard to put it in a proof

untitively both hold for relations which have all self loops

i think

like My answer will be

,, \mathcal{R} \subseteq { (a,a) \mid a \in A}

[clever fjord]

Renato

[lone elbow]

thats the answer for i)

if you notice @wanton bison

i used subseteq and not equality

because the relation is not necessarily reflexive

in the case of ii) it needs to be reflexive

recall reflexive = xRx ∀x ∈ A

wirh emphasis in the FOR ALL

so in the case of ii) the answer is
R = {(a,a) : a ∈ A}

not subseteq

[wanton bison]

I think you can prove it by writing down the definitions of symmetric and antisymmetric

[unkempt lichen]

(what i said earlier)

[wanton bison]

Try to apply them on the definition of transititvy

[wanton bison]

Oh, didn't read

[unkempt lichen]

nono it's fine

[wanton bison]

I was trying to figure it on my own too

[unkempt lichen]

but just emphasizing

[lone elbow]

where?

[sturdy cape]

So, this relation is literally the = relation

[wanton bison]

last help channel

[unkempt lichen]

your prev help channel

[lone elbow]

oh you mean in another day

[sturdy cape]

Transitivity would mean that x = y and y = z imply that x = z; is this true @lone elbow?

[lone elbow]

yes, the equality relation is transitive

so?

[sturdy cape]

Also, this is a subset of this?

[lone elbow]

whats your point

[lone elbow]

what u mean by this the first time you said this

[sturdy cape]

R is a subset of the equality relation?

[lone elbow]

i guess so, I just didnt thought of it that way

but is good that you guys have creative ideas

[unkempt lichen]

a) only asks for sym and antisym

[lone elbow]

yeah, we can find a lower bound

like, the subseteq set builder I showed is smaller of a set than the set of equality relation

[lone elbow]

ahh you mean for ii)?

[sturdy cape]

I only meant to ask for a clarification of this

[lone elbow]

we would need to know if the equality relation is exactly an equivalence relation and a poset

[wanton bison]

set

[lone elbow]

its subseteq because i) is symmetric and antisymmetric and not necessarily reflexive

what I mean is

let A = {1,2,3}

take R = {(1,1),(2,2)}

this relation is sym and antisym

but is not reflexive, becase 3 doeant relate with 3

@sturdy cape sorry if I am not being formal enough, this is my first week of university dude

[sturdy cape]

tbf, you're the one adding a lot of words here

I'm guessing the one remaing is (ii), then - not (i)?

[lone elbow]

ii) is exactly R = {(a,a) : a in A}

[sturdy cape]

yep

[lone elbow]

but I am having trouble with

last question

one relation on A is it possible its not sym and not antisymmetric?

[sturdy cape]

So, first establish what R being not symmetric means

Similarly for R being not antisymmetric

[lone elbow]

xRy but not yRx

xRy, yRx, but not x =y

[sturdy cape]

"R is symmetric in A" means "$\forall x,y \in A, xRy \iff yRx$"

[clever fjord]

Waes (Wires)

[sturdy cape]

What's the logical negation of this

[lone elbow]

idk how to negate <=>

[sturdy cape]

P iff Q - i.e. both P and Q are true, or both P and Q are false

So how would we negate this

meaning-wise? (we'll deal with the symbols later)

[lone elbow]

first we negate P and Q

[wanton bison]

When you are doing counter examples, you try to find one example so that the thing you claim doesn't work, so simply you find some x,y such that xRy iff yRx is not true

[lone elbow]

like

[sturdy cape]

@lone elbow By "negate" - I mean, what would I actually mean if I said that this relation is false?

[wanton bison]

you don't have to focus on iff

[lone elbow]

assume it's not symmetric, then the hypothesis for antisymmetric doesn't hold

[wanton bison]

I would think of it as $\neg (\forall x,y) P \equiv (\exists x,y) \neg P$

[wanton bison]

like my point was you dont have to unravel iff into other logical operations

[lone elbow]

R is not symmetric in A means ∃x,y in A, such that , xRy <=> yRx

this?

[lone elbow]

like P = (xRy iff yRx)

[sturdy cape]

Not quite

[sturdy cape]

@lone elbow

[wanton bison]

yes

[sturdy cape]

"I take an umbrella with me IF AND ONLY IF it is raining" - what would be the opposite of this?

[lone elbow]

R is not symmetric in A means ∃x,y in A, such that , ~(xRy <=> yRx)

[lone elbow]

it's not raining therefore I don't take my umbrella?

[lone elbow]

its raining iff I take an umbrella!

???

[sturdy cape]

What scenarios does the original statement mean?

[lone elbow]

I only take my umbrella if its a rainy day

[lone elbow]

symmetry

xRy <=> yRx

[sturdy cape]

Bruh, I mean which combinations of "it is raining", "it is not raining", "I take an umbrella" and "I don't take an umbrella" are true?

[lone elbow]

either both true or both false

[sturdy cape]

Right

[wanton bison]

when it rains it pours

[lone elbow]

🌧️

[sturdy cape]

So if I want to say the exact opposite of this

then...?

[lone elbow]

I can't dude

I dont understand, I can't dude

I can't dude

[sturdy cape]

"It's raining and I haven't got an umbrella"
"It's not raining and I have an umbrella"

[wanton bison]

ooposite = negation?

[sturdy cape]

Do you see how that's the logical negation of this statement?

[lone elbow]

(one true, one false) and (one false, one true)

[sturdy cape]

There we go

[lone elbow]

sort of, is hard dude

[sturdy cape]

So how would we negate xRy iff yRx?

We've got the "there exists x,y in A" part, that's correct

"such that..."?

[lone elbow]

(xRy, ~(yRx)) and (yRx, ~(xRy))

[wanton bison]

(maybe) words are better than formalism

[lone elbow]

(xRy but not yRx) and (yRx but not xRy)

[wanton bison]

so you actually understand

[sturdy cape]

There we go

(now, in practice, if xRy but not yRx, we can swap x and y around to get "yRx but not xRy")

[wanton bison]

it should be an OR?

[sturdy cape]

So we can simplify this to
"R is not symmetric in A precisely when $\exists x, y \in A$ such that $xRy$ but not $yRx$"

[clever fjord]

Waes (Wires)

[lone elbow]

swap x with y and you get both are true

what I mean is. both say the same thing

I think?

[lone elbow]

we are assuming that x =y, meaning we are assuming reflexivity?

[wanton bison]

like i meant even if (one true, one false) is true, that suffices for the equivalence to fail, regardless if (one false, one true) was true or false

[lone elbow]

I am getting confused dude

[wanton bison]

A iff B is true when both A, B were true or both false.
So A iff B can only be false if A was true and B was false OR A was false and B was true, you cannot have an AND because that's impossible

I actually don't understand why we focused so heavily on the logic

You know how to disprove things or how counterexamples work

[lone elbow]

yes, but we negated the or

first it was a or

then we negated the iff, we got an and

[lone elbow]

is it false or true the proposition? that not antisymmetric and not symmetric is possible

[lone elbow]

can a relation on A be not antisymmetric and not symmetric?

[wanton bison]

It's not a logical or, it's a xor

But anyway let's move on

[lone elbow]

true

[lone elbow]

how?

we got stuck in propositional logic territory

[wanton bison]

You literall say either

Either one is true (other false) or vice versa

[wanton bison]

I think yes?

[lone elbow]

do you have any example?

[wanton bison]

symmetric and antisymmetric implied transitivity irrc so consider the contraposition

[lone elbow]

I am trying to get intuition

[lone elbow]

contrapositive?

[wanton bison]

{(1,2),(2,1),(3,4)}

[wanton bison]

yes

[lone elbow]

you mean p => q consider ~q => p

[wanton bison]

No

not q implies not p

[lone elbow]

this relation is symmetric and not antisymmetric

[wanton bison]

oh bruh i confused it with reflexive

you can still argue with the contrapositive

[lone elbow]

how

oof I am so confused rn

[wanton bison]

symmetric and antisymmetric implied transitivity, right?

[lone elbow]

yes

because then R ⊂ {(a,a) | a in A}

[lone elbow]

x R y => y R x
x R y , y R z => x = y
x R y, y R z => x R z

[wanton bison]

Maybe the | relation

[lone elbow]

what does that symbol mean? divisibility

[sturdy cape]

yes

[wanton bison]

nvm that doesn work either

[lone elbow]

the waes dude reading silently

[sturdy cape]

I went to eat ffs

[wanton bison]

ok the contraposition may imply it's true but that doesnt prove there exists an example necessarily?

[lone elbow]

from my perspective is like this dude

[wanton bison]

Ok then let's add (3,4)

[lone elbow]

r = {(1,2),(2,1),{3,4)}

this?

[wanton bison]

Yes

[lone elbow]

it's not reflexive, not symmetric not antisymmetric and not transitive I think

assuming A = {1,2,3,4}

[wanton bison]

Yea this example is stronger than the contrapositive

[lone elbow]

wdym?

it's still possible we find a counterexample

[lone elbow]

oh I see

[wanton bison]

The contraposition implies it's true but that doesnt show it exists actually one

Like vacuously true

[lone elbow]

the answer is, yes it is possible

[lone elbow]

we found one example that is not symmetric and not antisymmetric

[wanton bison]

because of your example though

yes

[lone elbow]

the question is

is it possible a relation in A not be symmetric and not antisymmetric

the question is yes

[wanton bison]

the question to the question

[lone elbow]

answer*

we found one relation that is not symmetric and not antisymmetric

which shows that it is possible

[wanton bison]

do u have a breakdown

need to recollect your thoughts

[lone elbow]

we finished dude

right?

I think this is good enough dude

[wanton bison]

yes dude

(3,4) just destroyed the symmetry property

[lone elbow]

I appreciate it 😭 😭 😂 😁

.solved

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[edgy frigate]

I need help

[mystic nova]

Make a diagram first

[edgy frigate]

I have no idea

[mystic nova]

Do you know how to draw a prism

Triangular one

[edgy frigate]

Here

(C’MN) is (alpha)

[mystic nova]

Yeah I know

[edgy frigate]

I dont have any ideas ahhhhh

[mystic nova]

Do you expect where MN is so that it divided the prism into 2 equal parts

[edgy frigate]

I cant

[mystic nova]

Oh

[edgy frigate]

Thats is problem

If i denote v1 n v2 as the volume of 2 parts

We got if v1 = v2

=> v1/v = 1/2

v2/v = 1/2

Whats next

What should i do next

[mystic nova]

hold on gimme a min

[edgy frigate]

Sure

[mystic nova]

Could you show me the original question?

Even if it's not english

[edgy frigate]

Sure

Wait a min

Here

Last question

25

U done?

[mystic nova]

Idk, my brain freeze

Ng VN đk

Nhớ là làm bài này 1 lần rồi nhma quên cách làm

2h sáng não đơ rồi

[edgy frigate]

Yehh

Lmao v:)

T cần đc giải cứu

[mystic nova]

Thử tạo 1 hình lăng trụ tứ giác đi, cho mặt DD'C'C song song vs AA'B'B

Vẽ hình rồi gửi sang đây xem thử, h đang k có bút ko có giấy k có gì cả

[edgy frigate]

T nghĩ chỉ cần dựng mặt phẳng alpha r chứng minh là đc mà ta

[mystic nova]

Chịu, gần 3 tháng rồi k làm mấy bài ntn

Dạo toàn làm max min

[edgy frigate]

Đây

[odd edge]

@edgy frigate Has your question been resolved?

[odd edge]

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• Show your work, and if possible, explain where you are stuck.
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[shy robin]

Tried integrating $\int_0^4 x\sqrt{1+2x}, dx$ and got $\frac{68}{3}$, is this true?

[clever fjord]

Bakoles

[marble laurel]

Hmm

Could you show your work?

[shy robin]

i used integration by parts for the first time so i dunno if it's right

[marble laurel]

Hmm

[mystic saffron]

du is not that

[marble laurel]

Indeed

It's half of the actual du

[mystic saffron]

yes

[shy robin]

really?

[mystic saffron]

yes

[shy robin]

i thing i remember a formula for d/dx sqrt(x) is 1\2sqrt(x)

[mystic saffron]

yes but you dont have x

you have an composed function

[shy robin]

so what do i do

[mystic saffron]

chain rule

you do chain rule

[shy robin]

oh

okay

i don't get what is the u and v part of sqrt(1+2x)

[mystic saffron]

what do you mean?

currently your main worry should be differentiating sqrt(1+2x)

[shy robin]

u'v + uv' no?

[mystic saffron]

no

this isnt f*g

you are stating the [f(x)*g(x)]'

[shy robin]

okay

[mystic saffron]

do you know chain rule?

[shy robin]

nope

[shy robin]

mhm

so i did the first part

and got 1\2sqrt(1+2x)

and

g'(x) is 2?

[mystic saffron]

nope

[mystic saffron]

yes

you are going the good road

[shy robin]

alright

f'(g(x)) should be d\dx sqrt(1+2x)

right?

[tall ivy]

Also why not just u = 1 + 2x, a more simple integral to solve

and a simple 1/2 after differentiating

[shy robin]

is this right or wrong

[mystic saffron]

that also works, but I guess he just want to use ibp

[tall ivy]

Yeah

If you're talking about the substitutes differential

[tall ivy]

Ahhhh I see, well integration is diverse, I'll respect his approach too

[shy robin]

mmmm

whats' the difference between d/dx [f(g(x))] and f'(g(x))

looks the same to me

[tall ivy]

yeah ig

It should be the same

[shy robin]

okay screw it, i'll use u = 1+2x

[tall ivy]

Wait I'll solve with your approach

[shy robin]

alright

[tall ivy]

I'll send it, one sec

[shy robin]

du should be $\frac{1}{\sqrt{1+2x}}$?

[clever fjord]

Bakoles

[mystic saffron]

yes

[shy robin]

okay

great

[mystic saffron]

keep going

[shy robin]

so final answer

64/3

cause the 16/12 becomes 16/6

i think so

am i right?

[tranquil willow]

1+2x=u² substitution will work??

[fast basin]

Hi, I’m judy. I just started studying again after 3 year. I’m not smart at all. I’m from Norway and I’m studying economics degree. We have math and I hope I can ask you guys whenever I don’t understand anything. I’m very social person so studying it’s not really easy for me after the 3 years. 😭

But I really need to get this degree

[shy robin]

i didn't think of that

[tranquil willow]

By parts?

[shy robin]

yes

it's better to do u substitution?

i could try that

[tranquil willow]

Think so

[tall ivy]

Are you sure about your answer?

I'm getting 298/15, solved it with your approach @shy robin, differentiation was lengthy but integral was fairly simple

[shy robin]

ugh

[tall ivy]

1 + 2x = t would simplify the differential too

[tranquil willow]

It would come smoothly

[shy robin]

i dunno what i did wrong tho

[tall ivy]

Let me check

[tall ivy]

By parts in definite gets complicated

[shy robin]

ah

i used the indefinite formula ig

[tall ivy]

Also there's no 1/2 after differentiating

It's just 1/sqrt(1 + 2x), you missed chain rule, you differntiate (1 + 2x) too and multiply it to the numerator

[shy robin]

yeah i changed that

[shy robin]

here

[tall ivy]

So that 2 from differentiating 1 + 2x would remove the 1/2 from d/dx (sqrt(f(x))

Ahhhh okay

[shy robin]

then it's the outside part that is wrong?

[tall ivy]

I'll solve using ibp too then, let's see if something works

[shy robin]

so far i did this

[tall ivy]

Here you go @shy robin

[tall ivy]

Didn't show the differentiation because I have done that here, I just took dx = t dt

[tall ivy]

Wait are you still using ibp or are you just doing it normally?

[shy robin]

i use u substitution this time

[shy robin]

yeah i see it now

thanks

[tall ivy]

Anytime mate

[shy robin]

so i got here

[tall ivy]

Try it with IBP too

You'll learn from your mistakes

[shy robin]

yeah i'm kinda stuck

[tall ivy]

Sure mate, whenever you got a question you can ask here 🙂

[tall ivy]

What is it?

[shy robin]

how do i integrate this

[tall ivy]

u⁵/10 - u³/6

[shy robin]

okay thanks

[tall ivy]

And just add the limits, I think you're doing it correctly if my memory serves me well

[shy robin]

i hope so

[tall ivy]

You got it?

[shy robin]

i'm doing the substraction

[tall ivy]

Yeah yeah take your time

[shy robin]

yeah i got it

298/15

[tall ivy]

Ayyyyyy

[shy robin]

thanks a lot

[tall ivy]

Great mate

[tall ivy]

Anytime

[tall ivy]

Now solve it with IBP too, and if you get stuck then counter-check with this

[shy robin]

alright i'll try that

[tall ivy]

Learn from your mistakes

[tall ivy]

Great!

[shy robin]

maybe later cause i'm about to fall asleep

[tall ivy]

Yeah sure, but remember to do this, it adds confidence to this approach

[shy robin]

quick question

[tall ivy]

And it's not a bad approach at all

[tall ivy]

Yeah?

[shy robin]

i'll be year one student at infosec very soon

and i wonder if i need to study discrete mathematics

by myself

[tall ivy]

Ahhhh I'm unaware of that, I'm an undergraduate too

[shy robin]

alright good luck 👍

[tall ivy]

Same to you mate

[shy robin]

.close

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[harsh frigate]

could someone check my work?

[mystic saffron]

you need to use the angle between the arrow and a horizontal line at the start of the arrow

if you draw a flat line above the 60 arrow then another down to the 120 angle you will have drawn a right angle triangle

then you can find the angle between the 60 arrow and the new line at the top. that angle will be 30 degrees

thats the angle you use to seperate the vector into horizontal and verticle components

the 100 line doesnt need to be seperated into directions because its already horizontal

[odd edge]

@harsh frigate Has your question been resolved?

[harsh frigate]

wait so which part would i have to redo, would you be able to draw it out for me?

[odd edge]

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[mystic saffron]

the purpose of splitting a vector into horizontal and verticle ones is that its easy to add vectors which point in the same direction

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[reef sinew]

Where's the mistake?

[reef sinew]

.close

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[graceful parrot]

Find the domain of f(x,y) = log(x^2-y^2)

[graceful parrot]

Would i set the inside function to x^2 - y^2 > 0

i dont think log can be 0

wait can it

,graph logx

,w graph logx

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by dark945850

[graceful viper]

it cannot

[graceful parrot]

ah okay

x^2 - y^2 > 0 so my presumption here would be correct

,calc log(0)

[clever fjord]

Result:

-Infinity

[graceful parrot]

So then I would acknowledge that x^2 > y^2

yea?

i think

idk

[vivid wyvern]

Yea ig

[fair prism]

x^2-y^2>0 should be good enough

[vivid wyvern]

U can get rid of the square as well

[graceful parrot]

okay

[graceful parrot]

im still not sure how to write this in terms of my domain answer

sorry 🙏

[fair prism]

The domain is all x and y such that x^2-y^2 > 0 or in set notation ${(x,y) \mid x^2-y^2>0}

[clever fjord]

Light

[quasi falcon]

you can basically solve the inequation to find the whole domain, you will get a set of values of x in terms of y > or viceversa

[graceful parrot]

ah okay

[quasi falcon]

the logic of x^2 > y^2 is basically correct yeah

[graceful parrot]

it doesnt ask me for it, but how would i find range for this?

unsure of where id start if i wantt o find that

[quasi falcon]

It isnt exactly a "range" in the usual sense

[graceful parrot]

oh okay

[quasi falcon]

the lowerbound* for x will be a function

[graceful parrot]

and thx

[quasi falcon]

btw, you can drop the squares too and use abs.

[graceful parrot]

alright thx

.close

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[odd edge]

@lone elbow Has your question been resolved?

[narrow crypt]

[ W = \vec{F} \cdot \vec{s}]

[clever fjord]

k

[narrow crypt]

@lone elbow "Find the work done by a force F=(8,-6,9) that moves an object from point P=(0,10,8) to point Q=(6,12,20) along a straight line. The distance is measured in meters and the force in newtons."

is this correct

[narrow crypt]

cool

[narrow crypt]

.

[lone elbow]

s?

@narrow crypt

[lone elbow]

,w sqrt(6^2+2^2+12^2) * sqrt(8^2 + 6^2 + 9^2)

[clever fjord]

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[lone elbow]

,w sqrt(6^2+2^2+12^2)

[clever fjord]

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Requested by renatoxbz

[lone elbow]

,calc 46*4

[clever fjord]

Result:

184

[lone elbow]

,w sqrt(8^2 + 6^2 + 9^2)

[clever fjord]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by renatoxbz

[lone elbow]

?

@narrow crypt @narrow crypt @narrow crypt @narrow crypt @narrow crypt

dude what does s vector mean?

,calc cos(0)

[clever fjord]

Result:

1

[lone elbow]

,w ( {6,12,20} - {0,10,8}) . {8,-6,9}

[clever fjord]

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Requested by renatoxbz

[lone elbow]

.solved

[odd edge]

Channel closed

Closed by @lone elbow

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[odd edge]

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[tawdry glacier]

hii can i get help. generally forgot how to caculate area

[smoky nacelle]

what area do you need to calculate?

[tawdry glacier]

the whole thing

[trail reef]

Can we assume the red line is parallel to the 12m line?

[tawdry glacier]

i believe so

[trail reef]

Okay great

[smoky nacelle]

it's divided into 2 shapes

and we'll calculate area of each one

[tawdry glacier]

i understand that

[tawdry glacier]

its just i forgot how to caculate the area of the triangle

[trail reef]

Yeah it's purple plus red

[smoky nacelle]

the height multiply with base then divided by 2

[trail reef]

But the first step is to find the length green and red lines

[smoky nacelle]

as far as i remember

[tawdry glacier]

what does the 10m represent

[smoky nacelle]

the whole side

that whole side is 10m

[trail reef]

Correct, but since this is a right angle you can just multiply the cathete

Aka the red times the green

Do you know how to find that?

[tawdry glacier]

so 5 times 12

[trail reef]

Yeah

Great lol

[tawdry glacier]

thats 60

then u add the other 60 from the red

isnt that 120

[trail reef]

Waaait

Typo

[smoky nacelle]

then just divide it by 2 i think

[trail reef]

My bad

[tawdry glacier]

OOOOOOOOOOOOOOOOOHHHHHHHHH

i got it

[trail reef]

Multiply by the cathete and divide by 2

[smoky nacelle]

since the base x height is to find the area of a rectangle

[tawdry glacier]

ait

wait

[trail reef]

OOPSIE

[tawdry glacier]

so its 30+60

and boom

[smoky nacelle]

yep

[tawdry glacier]

thankss

how do i exit

[smoky nacelle]

type .close i think

[tawdry glacier]

.close

[odd edge]

Channel closed

Closed by @tawdry glacier

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[odd edge]

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[slow cradle]

does anyone know

how to solve this

i saw a way

but its like

completely random and i have no clue where they got all the rules from

i didnt even know there were rules

[wooden python]

do you know the sine rule, a.k.a. sine law, a.k.a. law of sines?

or ig there is a somewhat different way to look at it even if you don't -- but you'll still need to know your trig basics

@slow cradle do you want:
A) an explanation via the law of sines?
B) an explanation via more basic trig?

[slow cradle]

can you try with law of sines

i tried doing it

but i didnt get the answer

[wooden python]

ok gimme a moment to make a diagram so we can talk about stuff more easily with names

@slow cradle sorry i was away

so here i've labelled the pts for convenience

we're gonna be focusing on angle C here

sine law says $\frac{\sin(A)}{a} = \frac{\sin(C)}{c}$. isolating $\sin(C)$ gives $$\sin(C) = \frac{c \sin(A)}{a}$$ and putting our values in ($a = x-1$, $\angle A = 30\dg$, $c = -x^2+6x-5$) we get $$\sin(C) = \frac{\frac12 (-x^2+6x-5)}{x-1}$$

[clever fjord]

Ann

[wooden python]

so far so good? @slow cradle

[odd edge]

@slow cradle Has your question been resolved?

[odd edge]

Channel closed

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[slow cradle]

.reopen

[odd edge]

✅

[slow cradle]

yes

you can factorise the quadratic and ull get (x-1)(5-x)

[wooden python]

ok wonderful so this lets us whittle the thing down a bit

and we get $\sin(C) = \frac12 (5-x)$

[clever fjord]

Ann

[wooden python]

now the key insight will be to note that we must have $0 < \sin(C) < 1$

[clever fjord]

Ann

[slow cradle]

yes

[slow cradle]

thats because you want double solutions right

[wooden python]

indeed

you want two solutions in (0, 180°)

that's why sin(C) has to be positive

and it also can't be exactly 1 cause that would leave you only with C=90°

[slow cradle]

yh

[wooden python]

yeah so then 0 < 1/2 (5-x) < 1 and you solve for x

https://tenor.com/view/shrimp-as-that-clash-royale-hee-hee-hee-haw-gif-25054781

[slow cradle]

i tried that

i got it wrong

the ans is D

[wooden python]

that so?

oh yeah i guess we actually need x-1 and -x^2+6x-5 both > 0 don't we

or... is that the issue?

[slow cradle]

possibly

that might be the issue actually

but why would it though

yh it os

it works now for some reason

[odd edge]

@slow cradle Has your question been resolved?

[odd edge]

@slow cradle Has your question been resolved?

[odd edge]

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[noble solstice]

there should be a mistake here somewhere but i'm just not seeing it the correct answer is p = -17

[wooden python]

mod(x-1)

!xy

[odd edge]

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.

[noble solstice]

its just saying that the remainder is R when divided by x-1 and -R when divided by x-2

did i usse the mod notation incorrectly? i hate modular arithmetic i swear 😭😭

but i also have to use correct notation when setting up the problem otherwise my brain cant continue solving it

oh wait it should be at the end mb

Sorry i was on my phone, here is the exact original statement

[wooden python]

it is a very odd use of mod imo

better to just say "let f(x) = [that], then f(1) = R and f(-2) = -R"

by the looks of it you miswrote x+2 as x-2

thats where your error's from

[noble solstice]

OH

that took me a while

thanks haha

[wooden python]

better yet tbh, write f(1) + f(-2) = 0

[wooden python]

!nogpt

[odd edge]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[wooden python]

also irrelevant to this q anyway

[noble solstice]

Use another channel :)

[stark agate]

alr

[noble solstice]

Yea thats how the book solved it

I just summed the 2 equations

But I wanted to ask about the mod thing, is it weird to write it that way? Cause my notes on this topic use that notation everywhere lol.
e.g.
P(x) equiv P(c) mod (x-c)

[wanton sierra]

this is the silliest google AI thing ive ever seen oh my god did anyone actually read this

[wooden python]

mmm

[wanton sierra]

i thought that was fake

[noble solstice]

or f(x) is a factor of g(x) iff g(x) equiv 0 mod f(x)

[wooden python]

ok like you can write modular equivalence for polynomials in the same way as for integers ig

i have just never really seen a practical use for that sort of notation in problem-solving

as it is much easier to just express remainders modulo a linear via the polynomial's value at that pt directly

[noble solstice]

I've seen some very interesting functions that are defined this way

OEIS A120372

https://oeis.org/A120372

[noble solstice]

btw this function is really interesting as it equals 0 only if it is between two primes, prove that the function has infinite 0's and uve solved the twin prime conjecture

anyways, thanks a lot :)

.close

[odd edge]

Channel closed

Closed by @noble solstice

Use .reopen if this was a mistake.

[noble solstice]

ive seen worse

@wanton sierra

[wanton sierra]

yeah this is just wrong, but i find it hilarious how the other one said yes, then provided ONLY counter-examples, then backpedaled. how useless is that

[noble solstice]

lol ai will take over the world lol

[odd edge]

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