help-41

[tough mica]

(2x3 + 2x4,1 - x3, x3, x4) = x3(2,-1,1,0) + x4(2,0,0,1) + (0,1,0,0)

S1 = <(2,-1,1,0),(2,0,0,1),(0,1,0,0)>

WTf is this?

[winged panther]

???

Given that (2, 1, 0, 1) is a common solution to both linear systems, it means that (2, 1, 0, 1) satisfies the two equations:

ax_1 + x_2 + 3x_3 + 2x_4 = 1
x_2 + bx_3 - 3x_4 = -2

Plug (2, 1, 0, 1) to get a = -1, b = 0

Now all that's left is to check that the two linear systems indeed have infinitely many solutions

[amber wasp]

@tough mica Has your question been resolved?

[tough mica]

b is not zero

[winged panther]

... Why

Does plugging (2,1,0,1) in x_2 + bx_3 - 3x_4 = -2 not give b = 0

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[tough mica]

.reopen

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[tough mica]

b can be 2 3 4 5

all of them satisfies (2,1,0,1)

[tough mica]

because b is free I think

[amber wasp]

@tough mica Has your question been resolved?

[winged panther]

??????? WHAAAAT

[tough mica]

idk, this exercise is so nasty

[winged panther]

Do you see the 4 given linear equations? One of them is $x_2 + bx_3 - 3x_4 = -2$

[grizzled pagoda]

Arya

[winged panther]

Can you plug in (2,1,0,1) in this equation?

[tough mica]

1 + b*0-3*1 = -2

1 + b(0) -3 = -2

[winged panther]

Yeah so b is free mb

Therefore you have (a, b) = (-1, k) where k is any real number

But you have to check to make sure S_1 and S_2 together have infinitely many solutions

[tough mica]

what does it mean that S1 and S2 have infinitely many solutions in common

[winged panther]

Btw just a quick question, what topic are you doing currently?

Is that matrices?

[tough mica]

linear algebra

it sucks so much

the exercises are nasty

[winged panther]

Bruh

Use determinants

The equation can be rewritten as:

[tough mica]

is non square matrix

determinants is for square matrices tho

[winged panther]

The matrix is 4x4 with coefficients of x_1, x_2, x_3, x_4 in each equation

[tough mica]

ok

,w det {{1,0,-2,-2},{a,1,3,2},{1,-b,-3,b},{0,1,b,-3}} = 0

[tough mica]

this is so nastyyyyy

[winged panther]

,,\begin{bmatrix}1&0&-2&-2\a&1&3&2\1&-b&-3&b\0&1&b&-3\end{bmatrix}\begin{bmatrix}x_1\x_2\x_3\x_4\end{bmatrix}=\begin{bmatrix}0\1\2\-2\end{bmatrix}

[grizzled pagoda]

Arya

[tough mica]

yeah

[winged panther]

Now for infinitely many solutions, determinant of the 4x4 matrix need ta be 0

And you've arrived at a = -1 so you should just put that in before solving

[tough mica]

,w det {{1,0,-2,-2},{-1,1,3,2},{1,-b,-3,b},{0,1,b,-3}} = 0

[grizzled pagoda]

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[winged panther]

There ya go

[tough mica]

,w det {{1,0,-2,-2},{a,1,3,2},{1,-b,-3,b},{0,1,b,-3}} = 0

[grizzled pagoda]

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[tough mica]

anyways I appreciate the help

.close

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[frosty cliff]

Basically the set of solutions are more than zero dimensional

Sounds stupid but if u think about it that’s literally it

like if all solutions lie on a line then there’s still infinitely many solutions

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[split sail]

Is regression coefficient affected by linear coding

[split sail]

<@&286206848099549185>

[gusty rampart]

Use this and don't instantly ping helpers @split sail

[noble niche]

can you close this, he made another one

[split sail]

I've been waiting for 1o minutes?

10*

Close the other one leave this

[noble niche]

do .close

[gusty rampart]

I already closed the other one

[split sail]

Alright

[gusty rampart]

I thought you posted that at 50 minutes past my mistake

[noble niche]

oh ok

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@split sail Has your question been resolved?

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@split sail Has your question been resolved?

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[river spade]

I honestly have zero clue how to begin this

[grim forge]

What do you understand by an inflection point

[river spade]

when it changes in concavity

[grim forge]

Yep

Another way to phrase it is when the second derivative changes sign

[river spade]

would i need a calculator at all for that?

[grim forge]

No

So firstly let's look between x=-5 and x=-4

From -5 to -4 does f'(x) increase or decrease

[river spade]

it decreases

[grim forge]

Good

[river spade]

ohhh

i see

[grim forge]

Yep

[river spade]

decrease, increase, increase, decrease

ty for the help

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[ruby dome]

If a differential form admits primitive, then is it exact?

[ruby dome]

that is, if I can find a primitive one can I say that it is exact?

Without even seeing the domain

<@&286206848099549185>

[sinful tusk]

yes im there

speak

child

[ruby dome]

There are the questions above

[sinful tusk]

wdym

[ruby dome]

if a differential form is exact I just need to find the differentiable function f such that the differential of the function coincides with the differential form of df=w?

[brisk nexus]

why this ?

[ruby dome]

@sinful tusk

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@ruby dome Has your question been resolved?

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@ruby dome Has your question been resolved?

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@ruby dome Has your question been resolved?

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[viscid mesa]

hello having a tricky time understanding the logic of this question

[viscid mesa]

I feel like this has to be true

[sterile nymph]

imagine that A_1 is the natural numbers

A_2 is the natural numbers without 1,

A_3 is the natural numbers without {1, 2}

and so on

what is the intersection of these sets?

[viscid mesa]

infinite?

[sterile nymph]

Which specific natural number is still in the intersection?

[viscid mesa]

Won't A_1, A_2,... eventually catch up

[sterile nymph]

well, if the set is infinite, then there must be at least one natural number in the set right?

but any natural number you claim is in this set is eventually removed from the set by the intersection

so actually, this intersection is not just finite, but empty

yet each of the sets has infinitely many elements

Sorry, this is probably a lot

[viscid mesa]

Digesting

I am thinking now that if A_n is starting infinitely to the right, then I guess there can't be an intersection

[sterile nymph]

I mean, there is an intersection, it's just empty

[viscid mesa]

right

Okay, thank you I will think about this for a bit and may come back

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[viscid mesa]

.reopen

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✅

[viscid mesa]

sorry, okay isn't the intersection just A_n?

Like just working with smaller sets A_1, A_2, A_3, the intersection of these sets is A_3 which is infinite

[sterile nymph]

yes.

but what is A_inf?

for each finite n, you get A_n which is infinite

[viscid mesa]

right

[sterile nymph]

We can think of it like this, all finite n, we have the test "if m >= n then m is in A_n"

what if we apply this logic to our hypothetical A_inf?

"if m >= infinity then m is in A_inf"

Obviously this is an abuse of notation somewhat

but there is no such m in the naturals for which that holds.

[viscid mesa]

Hmmmm okay, I think it clicked a little. I think my biggest confusion is how a pattern that can happen infinitely can be broken as soon as infinite is introduced if that makes sense

Should be good now though, thank you very much

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[sterile nymph]

To be fair, it's quite confusing

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[obsidian shoal]

$x^4 + 81$

can someone help me factoring this? i tried to do something but it failed

[grizzled pagoda]

Simon James B

[sterile nymph]

x^4 + 81 = x^4 + 3^4 = (x^2 + 3^2 i) (x^2 - 3^2 i) = (x + 3 i^2)(x - 3 i^2)(x + 3i)(x - 3i) = (x + 3)(x - 3)(x + 3i)(x - 3i)

[obsidian shoal]

$x^4 + 81 = x^4 + 3^4 = (x^2 + 3^2 i) (x^2 - 3^2 i) = (x + 3 i^2)(x - 3 i^2)(x + 3i)(x - 3i) = (x + 3)(x - 3)(x + 3i)(x - 3i)$

[grizzled pagoda]

Simon James B

[obsidian shoal]

i am so very sorry but i do not understand

[sterile nymph]

which step?

[obsidian shoal]

all😭

what i?

[sterile nymph]

$i = \sqrt{-1}$

[grizzled pagoda]

OmnipotentEntity

[obsidian shoal]

we never learned this

[solar gust]

ah

[obsidian shoal]

srry

[solar gust]

its not your fault xd

don't be sorry

[sterile nymph]

wait, did I make a mistake?

[obsidian shoal]

no idea

[sterile nymph]

I think I also just simply made a mistake

yeah that should have no solutions in the reals.

[solar gust]

as omni said you can't factor it in the real

[sterile nymph]

yeah my bad, let me fix my factorization

[obsidian shoal]

they never said in what to factor

wait so if not in the reals.

means in none below it

[solar gust]

yeah

[sterile nymph]

\begin{align*}
x^4 + 81 &= 0 \
x^4 + 3^4 &= 0 \
x^4 - (-1) 3^4 &= 0 \
x^4 - 3^4 i^2 &= 0 \
(x^2 + 3^2 i) (x^2 - 3^2 i) &= 0 \
(x^2 + 3^2 \sqrt{i}^2) (x^2 - 3^2 \sqrt{i}^2) &= 0 \
(x^2 - (-1) 3^2 \sqrt{i}^2) (x^2 - 3^2 \sqrt{i}^2) &= 0 \
(x^2 - 3^2 i^{5/2}) (x^2 - 3^2 \sqrt{i}^2) &= 0 \
(x + 3 i^{5/4}) (x - 3 i^{5/4}) (x + 3 i^{1/4}) (x - 3 i^{1/4}) &= 0
\end{align*}

[woeful sable]

I'm surprised they are having people factor these without knowing the concept of imaginary numbers first.

[obsidian shoal]

it should be (x^2 +3 * sqrt2x +9)(x^2 - 3 * sqrt2x + 9)

[grizzled pagoda]

OmnipotentEntity

[sterile nymph]

oh, ah, that's what they want.

[obsidian shoal]

they gave this as homework

[sterile nymph]

So this is completing the square.

haha, yeah I guess that makes more sense.

[woeful sable]

I mean, that's what I'd give to students after they learned the concept of imaginary numbers

[sterile nymph]

ok ignore what I just wrote.

[woeful sable]

What's the point of giving homework if the answer is gonna be having no real solutions

😂

Atleast have some solutions

Imaginary or not

[solar gust]

the real point is to "how say" that there is no real solution

[woeful sable]

well I guess

But that's kinda boring

[obsidian shoal]

it should be (x^2 +3 * sqrt2x +9)(x^2 - 3 * sqrt2x + 9) it is what they say

[solar gust]

which is completely useless

[solar gust]

do you have the whole thing or just the answer ?

[obsidian shoal]

this is the whole answer that they marked as correct for the given expression x^4 + 81

i do not have how they solved

[sterile nymph]

well, we can assume that there is a solution, and then figure out what it must be, I suppose.

[obsidian shoal]

there was a bot to check if (x^2 +3 * sqrt2x +9)(x^2 - 3 * sqrt2x + 9) = to the initial expression

i do not remember how to do it tho

[sterile nymph]

like, you assume there is a solution of the form (x^2 + ax + 9)(x^2 + bx + 9) then work out what a and b must be.

[solar gust]

,w is (x^2 +3 * sqrt2x +9)(x^2 - 3 * sqrt2x + 9) = x^4 + 81 ?

[grizzled pagoda]

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[obsidian shoal]

bruhh

[solar gust]

you meant sqrt(2x+9) ?

[obsidian shoal]

no no i said that because the result said it is not always equal to

[sterile nymph]

3 sqrt(2) x

not 3 sqrt(2x)

[obsidian shoal]

what if

x^4 = (x^2)^2 and 91 = 9^2 and then use a^2 + b^2 ?

[sterile nymph]

If you expand (x^2 + ax + 9)(x^2 + bx + 9) you get x^4 + (a+b)x^3 + (18 + ab)x^2 + 9(a+b)x + 81

then you have a+b = 0 and 18 + ab = 0

[obsidian shoal]

nav i give up

[sterile nymph]

this gives you a = -b, and 18 = a^2

so a = sqrt(18) = 3 sqrt(2)

Therefore (x^2 + ax + 9)(x^2 + bx + 9) = (x^2 + 3 sqrt(2) x + 9)(x^2 - 3 sqrt(2) x + 9) = x^4 + 81

[obsidian shoal]

i give up i am moving to functions. Thanks for trying tho but this problem assigned to us is way more advnaced then we are so i have no idea why they gave it

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[split sail]

Hello

[keen pawn]

👋

[split sail]

for the following venn diagram

the part shaded in orange

[gentle mason]

A-AnB?

[split sail]

i understand we can write it as P(A)- P(A intersection B)

would P(A-B) also yield the same result ?

i was trying to prove P(A union B) = P(A) + P(B) - P( A intersection B)

[split sail]

i used this result and it worked

not sure if it correct though

[proven vapor]

yes, A - B and A - (A intersection B) are the same sets

[split sail]

thanks

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[split sail]

Q=0.18 (L√(SA))/(³√D)=25

[split sail]

This was on a Mazda CX=90

We have 3 mechanical engineers, and one electrical trying to figure it out

[amber wasp]

@split sail Has your question been resolved?

[winged quest]

Trying to figure out what? There's no question here.

[split sail]

It's a bumper sticker there has to be some sort of "witty" resolution

.close

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[river brook]

is this sufficient to prove the incenter exceter lemma?

[river brook]

( kinda sloppy if anythings unclear i can explain )

also i took the wrong order in quadrilateral IBIaC

but other than that

<@&286206848099549185>

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[split sail]

Hey, how do i simplify this to make it easier to solve

[split sail]

Apparently i need to do long division on it but i am kinds confused on that

[quick ridge]

so you don’t know long division?

or synthetic division

if you don’t then you could always add 0 to the numerator and split the fraction

-1+1

$\int \frac{x^2-1+1}{x+1} , dx$

[grizzled pagoda]

knief

[quick ridge]

see where that gets you

[split sail]

Ahaaa

I think i get it now

i can expand (x^2-1) and that would have x+1 so i can just take those 2 out right?

[quick ridge]

then what’s left

[split sail]

so (x-1) + 1

[quick ridge]

nope

[split sail]

Wait

[quick ridge]

that would be x

[split sail]

$\int \frac{(x-1)(x+1)+1}{x+1} , dx$

[grizzled pagoda]

Mystic

[quick ridge]

yes

[split sail]

would it not be like that and i cancel the top and bottom out?

[quick ridge]

write out what you mean

[split sail]

$\int (x-1)+1 , dx$

oops

[grizzled pagoda]

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

[split sail]

Like that you get the idea

[quick ridge]

what

but you can’t do that

again that’s just x

[split sail]

Ahhh wait

wait

$\int \frac{x^2 - 1}{x-1} + \frac{1}{x-1} , dx$

[grizzled pagoda]

Mystic

[split sail]

like that?

[quick ridge]

there ya go

but it was x+1

not x-1

[split sail]

Oh yeah

[quick ridge]

also do backslash comma

not just comma

to add a little space

[split sail]

Aha

Ok

$\int \frac{x^2 - 1}{x+1} + \frac{1}{x+1} , dx$

[grizzled pagoda]

Mystic

[split sail]

So like that?

[quick ridge]

yes

now simplify

and you’ll have two easy things to integrate

[split sail]

i can alreadyintegrate 1/X+1 as ln|x+1| just gotta simplify the other

Alright so I integrated x-1 into 1/2x^2-x and the other into ln|x+1| so the final answer would be 1/2x^2-x+ln|x+1|

Let's go

[quick ridge]

nice

[split sail]

Perfect tyvm

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[cursive wasp]

I am wondering what the best way to take notes is for a class like real analysis or modern algebra? 1 book divided into sections (definitions, proofs, ...) or like perhaps using a plastic file divided into sections idk

[honest dagger]

"best" falls on you. Imo, studying and knowing the why of things really helps. In math, the task is to apply your knowledge. So rote facts on paper will do you no good. Take note of its utility and try to form your own thoughts on what you are learning

[cursive wasp]

I 100% agree with knowing the why, but having a neatly compiled list of definitions and theorems comes in handy when you stuck on a proof and you wondering what else you could try

[honest dagger]

correction: sometimes raw facts can be helpful, if you need to refer to them constantly, say the name or details of a theorem

[honest dagger]

Then have two sets of notes: comprehensive and organized, and then a "fast-facts" sheet

[cursive wasp]

👍 Fair enough thanks

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[solar gust]

Derivative and where it is 0

[mossy eagle]

I'm doing algebra actually so I don't think I need to find the derivative

[plush fjord]

try cross multiplying

you will get a quadratic and then you can determine the range for y from the discriminant

[amber wasp]

@mossy eagle Has your question been resolved?

[mossy eagle]

I did that but there's an issue

@plush fjord

[plush fjord]

what did you get

[amber wasp]

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[mossy eagle]

.reopen

[amber wasp]

✅

[mossy eagle]

That's the method I was following

That step ehich I circled I reduced the equation by 4 rather than -4

So i got -4y²+3y+1>=0
(-4y-1)(y-1)>=0

So the -4y-1 changed the sign in my table which changed my solution set

@plush fjord

[plush fjord]

yeah this is mine as well

it looks like you simplified it? I used the quadratic formula from the unsimplified version

which was

$16y^2 - 12y - 4 <=0$

[grizzled pagoda]

clair

[mossy eagle]

It is -16y²

My quadratic was -16y²+12y+4>=0

[plush fjord]

it is the same thing

just divide by -1

[mossy eagle]

Solving it like that gives me a wrong range of values why

[mossy eagle]

I thought its the same thing

[plush fjord]

did you change the inequality?

after you get the inequality with y you just simply solve

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[light ore]

rq can we cancel terms if the denominator continus with a plus/minus sign?

[light ore]

ex 3/3(2)+2

[split sail]

you can cancel the top 3 with the bottom 3 leaving you with 1/(2)+2

[light ore]

Oh ty

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[frank sage]

not sure how to set this up
book says it’s newtons law of cooling but it didn’t help out when i read through it

[sterile swan]

Whats the question??

[frank sage]

it’s i think setting up the equation

A = (something)e^kt kinda deal

[celest cove]

gotta do separation of variables

[frank sage]

yeah so far i’m at ln(100-A) = t

would it just be 100-a = e^t

[celest cove]

thats a lil off

on both sides

[frank sage]

would it not be dA/(100-A) = kdt?

[celest cove]

the integral of 1/(100-A) isnt ln(100-A)

but also the integral of k wouldnt be t

and youre missing the integration constant also

sorry my internet is being annoying

[frank sage]

it’s saying that k is a constant

and the integral of a constant in terms of anything would be that term

[celest cove]

if i differentiate ln(100-A) i get -1/(100-A)

just a lil different

[frank sage]

or am i missing something

[celest cove]

the integral of a constant k wrt t would be kt+c

think to how if i differentiate 5x im left with just 5

then go in reverse

[frank sage]

if the k was a 1 and you integrate 1dt it would be t right because the derivative of a constant with a variable would be a constant, and it doesn’t specify we have more than one k so i’m just confused why it wouldn’t be just t

but it also kinda makes sense now that i read it

[celest cove]

its true if k=1 but we dont know that

k is just k

[frank sage]

ok i have 100 - A = C + e^kt now

[celest cove]

not quite, the left also had an issue

[celest cove]

and the removal of the ln here is a tad off but we'll get there

[frank sage]

alr

[celest cove]

how do we integrate 1/(100-A)

[frank sage]

by parts

wait

no

u sub my bad

[celest cove]

you indeed can do that

do you recall that the derivative of ln[f(x)] is f'(x)/f(x)

[frank sage]

not really

[celest cove]

ah, alrighty then we can go ahead with the u-sub

[frank sage]

unless it was an inverse trig function i just assume 1/anything is ln

[celest cove]

often we can manipulate the expression into this form, then reverse it to the ln, but the u-sub does that in a more worked way

eg i know the derivative of 100-A is -1

so i may rewrite 1/(100-A) as -[ -1/(100-A)]

then jump to -ln(100-A)

[frank sage]

why is it a double negative

like

-[-

[celest cove]

because i need the numerator to be the derivative of 100-A in order to reverse to the ln without doing a u-sub

which happens to be -1

1=-(-1)

[frank sage]

oh it’s just chain rule

[celest cove]

yeah, u sub is essentially the reverse of the chain rule

in essense anyway

[frank sage]

it would be - integral 1/u

so -lnu

[celest cove]

yup

[frank sage]

my bad my bad

[celest cove]

so -ln(100-A)=kt+C

[frank sage]

so then would it be -100 + A if i distribute the negative or should i not bother

after i get e^kt + C

[celest cove]

you cant distribute into the ln im afraid

[frank sage]

i meant after i put an e

[celest cove]

youll also never have this

[frank sage]

isn’t e^c still a constant

or is it multiplication

[celest cove]

yeah but it would be multiplied not added

[frank sage]

yeah

i just remembered

[celest cove]

ideally you should multiply the -1 across first or youre going to just get 1/(100-A)=Ce^(kt) which is fine but just an extra step

[frank sage]

bc then it splits into e^kt * e^c

okay so i have -(100-A) = C*e^kt

[celest cove]

not quite

[frank sage]

what did i miss

[celest cove]

youve pretty much said that if -1=x then -e^1=e^x

but it would be e^(-1)=e^x

you left the -1 out when you raised both sides

[frank sage]

the negative one from the -ln?

[celest cove]

yeah

[frank sage]

oh i forgot to do that

[celest cove]

you did -e^ln rather than e^(-ln)

you can just multiply by -1 before you do it

then you just have ln(100-A)

[frank sage]

and then it would be dividing by C in the end? bc it would be -kt - C or am i getting ahead of myself

[celest cove]

a bit ahead of yourself

e^(-C) and e^C are both constants

can replace either with just C

[frank sage]

so it would be C*e^-kt?

[celest cove]

yup

=100-A

[frank sage]

okay that makes so much more sense

then i have my initial t = 0 from the problem

if A = A_not

[frank sage]

i’m getting A = -(100-A_0)e^-kt

i think i’m close but i might’ve missed a step

[celest cove]

hm so C=100-A0

youre missing the other 100

A-100=-(100-A0)e^(-kt)

just need to add it over

[frank sage]

oh

wait

that 100 seems like it just spawned in from how i did it

i did 100-A = Ce^-kt
A = -Ce^-kt + 100
set t = 0 and A = to A_0
A_0 = -C + 100
C= 100 - A_0
A = -(100-A_0)e^-kt

[celest cove]

you just forgot the 100

you replaced C but made the 100 go poof after doing so

[celest cove]

compare the second line to the last

[frank sage]

oh

the + 100 disappeared

[celest cove]

yup

[frank sage]

silly old me

[frank sage]

thank you so much i might’ve ripped hair out if i had to look at the problem longer

.close

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[little frigate]

!help

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[little frigate]

So umm

[grim forge]

So the area the inequality covers should be all the points which have y greater than 2

Firstly let’s draw the line y=2

You can just sketch this on a notepad

[little frigate]

Ok

Ready

[grim forge]

Draw the line y=2

What can you say about all the points above y=2

[little frigate]

Not sure

[grim forge]

All the points above y=2 have a y coordinate greater than 2 yes?

[little frigate]

Ye

[grim forge]

As such to graph the inequality y>2 what should be the area shaded?

[grim forge]

Welp ig we’ll let otori cook

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[glad pulsar]

do i consider r1=r2 here

[glad pulsar]

<@&286206848099549185>

[wintry slate]

r1 and r2 are PK and QK?

[glad pulsar]

yes

im stuck here

[wintry slate]

Then no, r1=r2 only if the chord is perpendicular to the axis

[glad pulsar]

ok then how do i solve this

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[glad pulsar]

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[solemn escarp]

Let ABCD be a trapezoid as in the image. If P,Q,R bisect AD,BC,BC respectively, S,T split CD into 3 equal side lengths, AB = 4, BC = 6, AD = 3
a) Find the area of QST
b) If the star shaped polygon PIQERFSGTH has an area of a, and pentagon EFGHI has an area of b,
Find a + b

[solemn escarp]

ive got the answer to a) as 3

but i have no idea how to start b)

[native glade]

How you got side DC

Wait nvm got it

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[split sail]

How did they get the f? (Histogram - Statistics)

[split sail]

<@&286206848099549185>

[vocal isle]

?

Ban incoming.

[prisma ember]

f stands for frequency

[vocal isle]

!noping

[amber wasp]

Please do not ping individual helpers unprompted.

[split sail]

How did they get that?

[prisma ember]

look at the histogram

do you have a picture of the histogram?

[split sail]

i didn't draw it during the discussion sorry

[prisma ember]

then youre out of luck

[split sail]

but do you know how did they get f?

[prisma ember]

f is the height of the bar

[vocal isle]

!15min

[amber wasp]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[split sail]

like the numbers 2, 3, 7, 3, 2?

[prisma ember]

yes

[split sail]

What do you mean height of the bar?

[vocal isle]

Height of 'a bar' represents its frequency.

[prisma ember]

heres an example i pulled from google

age 20-30 has frwquency 2

[split sail]

How did they even get that

Do they just count 1 2 3??

[prisma ember]

thats based off the data

[split sail]

Okay here the question that was only given is 5, 15, 12, 16, 8, 10, 22, 16, 13, 15, 12, 12, 14, 19, 17, 16, 18, 22

My step 1 is 22 - 5 = 17

my step 2 is S^k >_ n

[prisma ember]

that part is a coincidence

[split sail]

my step 3 is i = 17/5

[split sail]

finally my table is that

for step 4

I don't know how did they get the frequency

with just those 3 steps

[prisma ember]

you count the number of data points you have in those ranges

for example in the range 5-8, you have one 5 and one 8

so there are 2 points in the range 5-8

hence the frequency f of 2

it would be easier if you sorted the list

[split sail]

5 - 8 is 2

9 - 12 is 3

13 and 16 is 4

HA THEY ARE WRONG

Its not 7 it's 4

wait hen 17 - 20 is also 4

along with 21 - 24 is also 4

dang

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[vocal isle]

Decide whether the following is negative or positive:
tan(9^9^9^9^9^9)

[solar gust]

,calc tan(9^9^9^9^9^9)

[grizzled pagoda]

Result:

NaN

[solar gust]

damn

[amber wasp]

@vocal isle Has your question been resolved?

[vocal isle]

,w tan(9^9^9^9^9^9)

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by anshul6368

[sterile swan]

,w tan(a^b)

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by .bonk__

[amber wasp]

@vocal isle Has your question been resolved?

[chilly jackal]

where does this question come from..? I doubt any question bank would ask you to evaluate at such a large number

also

I believe that its -ve

because your huge number (if I havent messed up) is congruent to 3 mod 6, so it can be definitely written as 2kpi + 2.something, meaning that you get tan(2.something) which is definitely less than 0

assuming ofc that ur number is inputtted as radians

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[vocal isle]

.close

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[vocal isle]

thanks

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[civic spindle]

back

[civic spindle]

F

alr

can someone explain why Xdiv2pow2=perfect square root of the original number?

so if you did it with 9 or 7 youd get anumber with factors that perfectly square?

[vocal isle]

Hi

?

Show your work and the question correctly.

[civic spindle]

???

that is

tho

like

if you had 7 and you wanted to get a number which has two factors that multiply by each others to make 7

youd devide by 2

so 3.5

^2

[vocal isle]

I know you are a troll.

[civic spindle]

proof?

[civic spindle]

i will give source

https://www.youtube.com/watch?v=P8W2M0jq2Qs&list=PLybg94GvOJ9FoGQeUMFZ4SWZsr30jlUYK&index=40

also over 5 ppl answered prior to you

but the answers were extremely technical and nobody attempted to simplify or explain it directly without long arduous proofs

except for 1 or 2 people but idk where there answers were bc they got lost in the crowed

[cunning birch]

is this about (x+a/2)^2 formula

[civic spindle]

i think so

thank you someone who isnt an idiot sob:

[civic spindle]

alrr can you explain how and why it works?

https://media.discordapp.net/attachments/1015759484708208792/1045835008012714017/a_3a6633f55a6525e045c618e50b97782a12.gif

[cunning birch]

well

it's about quadratic identity

(x+y)^2 = x^2 + 2xy + y^2

you just expand the left hand side

[vocal isle]

You are also A(aperture)

[cunning birch]

to see that's what you get

so

taking y = a/2

(x+ a/2)^2 = x^2 + ax + a^2/4

this way

if you have to factor something like

x^2 + 6x + 5

notice that taking 6 = a

we get

(x+3)^2 = x^2 + 6x + 9

[civic spindle]

i cant understand that

because its spoken in math language but i cant process that mentally because even when I did know what you were talking about it was still really hard to match that to my understanding

tbh

this is one of the only times

where ive felt like understanding a problem is actually nearly impossible

because

1. Understanding it looks like itd take a rediculous ammount of permutations and calculations to check whether or not it were true and in what way

2. the context is extremely vague and not mentioned anywhere so its very hard to gain info on it.

3. it looks very kinetic like i cant solve it because it doesnt make sense outside of something that is true just because it is

to be fair

if i can identify it as a self-evident truth rather than an abstract contextual concept

it technically may negate as thered be no point in learning it

as it'd be something only true because it is true rather than because it has an idea behind it

hmm, either way, its very difficult to find help here, and that guy did specify the name of it

too many hooligans in this world smh

half the time im too pained to even talk properly tbf but i dont worship my ego like most

erm

thinking:

may as well just continue i guess

unless someone comes here

SHRUG

painful!!!!

this is the worst type of problem

there is nothing enjoyable about this

well anyways i may as well start

https://tenor.com/view/divider-globe-world-space-sci-fi-gif-2881399313018008423

[pseudo crescent]

https://www.youtube.com/watch?v=T0HyWIFbsHQ

[civic spindle]

1. In context of quadratics where the goal is to find roots (values that make the equation/function/set of operations equal to zero) they factor to find the value of X by itself that'd make

no

no im serious

that

that thing

is like

hell

that explanation was so so overcomplicated i had no idea what was going on

[pseudo crescent]

the video?

[civic spindle]

and ik it is cuz i alr understand quadratcs its literally just the way they are wording and symbolically representing it that causes me the problem

thats why im always so angry at ppl who try to help me

its not that i agree with my pain its just that it hurts me when i try to understand and someone comes to me with symbols and representations that ruin understanding then leave without ever trying to adjust

shrug

ima just continue with what i was doing

1. In context of quadratics where the goal is to find roots (values that make the equation/function/set of operations equal to zero) they factor to find the value of X by itself thatd make the equation zero (root)

btw i have a much more verbal/chain of concepts-based understanding which is why when I see numbers alone it basically makes it extreme diff to read

2. During factoring, they may do things such as intentionally adding things to the equation to forge a factorable equation if it isnt possible to factor.

processing limits smh

to even think about this problem takes everything

whatever

3. In quadratics the factored multiplication looks like (x, 4) (x, 9) where 4*9 and X*X both recreate the whole numbers prior to factorization and 4*x and x*9 attempt to recreate the middle number.

to be honest people who do math here probably are aliens in the sense they have completely different life experiences the way im explaining this here is wholly alien to the way theyve explained it tbh

i wonder how they think tbh, thinking in chains of numbers and expressions seems impossible to me but I guess they have far more complex models of everything so nothing appears difficult to them leading to extremely difficult explanations becuase people probably forget why and how things were hard or why a person of lesser intelligence would be stuck on something they could generalize through via principles

thats probably it

they have truly clear and deep models of things meaning they dont struggle because they can generalize and think in clarity. The jump in processing hypothetically could be over 5 times higher because getting stuck can lead to days of time loss vs someone who can clearly see through any problems steps immediately and generalize on top of that.

shrug

either way i should continue, philosophical ponderings are for later but I do need to find an answer for my anger otherwise it might eternally stress me out which is why its hard for me not to mention this since nobody would take it seriously. The answer might come from within tbh as again one can might only be capable of relying on the infinite maelstrom of aspiration within ones soul to see through such things.

moving on...

4. But, sometimes you cant do that, you have to remove or add things to the equation which works because the goal is to solve for X when you add five thats five on top of the original answer of zero its from zero.

a

[civic spindle]

5. ```A perfect square is when the factors are the same for a problem so (x, 4) (x, 4) or (x,4)^2 meaning you can sqrt that to get (x,4) which also sqrts the answer which lets say was 5 now 5sqrt = 2.23

```This is useful because x +4 = sqrt5 to x=sqrt5-4 which turns the problem into a linearly solvable equation rather than guessing answers to the equation.```

[amber wasp]

@civic spindle Has your question been resolved?

[civic spindle]

6. Finding a number that contains factors that may generate a perfect square may be extremely difficult; however, there was a method proposed titled the "(x+a/2)^2 formula"

indevelopment

ight

[civic spindle]

https://tenor.com/view/tv-static-screens-multiple-pink-gif-19540733

[civic spindle]

finishing:

indevelopment

indevelopment

i would have to gtg right as i got to the good part smh

saved

https://th.bing.com/th/id/R.10d46a51c85f94f3debfa869ff6b107d?rik=FxsDckiS3PErUw&riu=http%3A%2F%2Fbgfons.com%2Fuploads%2Findustrial%2Findustrial_texture689.jpg&ehk=ixfChzS3bYyKBQudrrft3A2XuHNY9Ps7blONywbNnwA%3D&risl=&pid=ImgRaw&r=0

7 The question of why exactly dividing a number by two and powering it generates a number with two factors

(FOR EDITING)

.close

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[barren swan]

Hey guys I have a question if I’m finding the percent change of 370 to 444 would the awnser be .2?

[celest cove]

how did you get that?

ah change

[barren swan]

?

Percent change

I subtracted 444 by 370

And got 74

I divided it

And I got .2

[celest cove]

you have the decimal right

but thats not the percentage

[celest cove]

yup

[barren swan]

So it 20%

[celest cove]

thats not a percentage though

[barren swan]

Or 2%?

[celest cove]

perfect

20% increase

[barren swan]

Oh ok

Thank you

Hey

I’m getting confused about this problem

Can I get some help?

I have to find the percent change of 275 to 418

I subtracted 418 to 275

And got 143

But when I am dividing it

The number just never stop?

Hey can I get some help with this question I’m confused

[naive pivot]

should be 55 not 65

[barren swan]

Where

[naive pivot]

after first subtraction

[barren swan]

55?

Wait ima restart real quick

[naive pivot]

the 13 should turn to 12

[barren swan]

The “1375” ?

Oh

That 13

I think I did my subtraction wrong

[naive pivot]

ye

[barren swan]

Wait but

Don’t when you subtract

And u borrow another number

Don’t u just add that ont?

Ontop*

[barren swan]

Ohh

But can’t we just keep the 5

On the 0 and 5 beneath?

[naive pivot]

itll be 0-5 which won't give a positive number

[barren swan]

Oh

Ah it ended now

I got .52

Then x by 100

Gives me 52 percent

Ty

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[fresh hatch]

what are you stuck about here, and what is the translation?

[grizzled pagoda]

938c2cc0dcc05f2b68c4287040cfcf71

[tough mica]

how to find the lines

any ideas

Look. here is the trick

since every P in L' is at a distance 2 to the plane Pi

and particularly L' intersects L

then there are some points in L which are at a distance 2

to plane Pi

@fresh hatch

anyways, lets find a lambda and then the possible points which are at a distance 2 to Pi

also, notice that L is orthogonal to L'

in particular the direction of L is orthogonal to L'

and notice that if every P in L' is at a constant distance 2 to Pi then what does that mean? that every L' is parallel to the plane Pi

in particular it means that there are two possible L'

one from below one from above

anyways lets find the direction of the lines L'

so, because the lines L' are orthogonal to the line L. in particular their direction vectors must be orthogonal

the direction vector of L is (2,1,-1)

and the direction vector of the two lines L' is (a,b,c)

so (a,b,c) is orthogonal to (2,1,-1)

and because the lines L' are parallel to Pi they must be perpendicular to the normal of Pi :)

think about it this way @fresh hatch a two lines floating within your ceiling, being parallel to your ceiling and the plane being the floor of your house

the normal of the plane is orthogonal to the lines floating in your ceiling because

the floor is parallel to the ceiling

think about the normal of the plane

it has to be orthogonal to the direction of L'

anyways, (a,b,c) = (1,-2,2)×(2,1,-1)

,w (1,-2,2)x(2,1,-1)

[grizzled pagoda]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tough mica]

the direction of lines L' is (0,5,5)

now to find both points we need to find the points in L that are at a distance 2 to Pi

particularly because we know every P in L' is at a constant distance of 2

and we know L n L' have a nonempty intersection

there must be one or two points in L which are at a constant distance of 2 to the plane Pi

lets find them using the point to Plane distance formula

[winged heart]

Could u help me when I make a form?

[tough mica]

open your own channel and people will help u I swear

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[shy sage]

N is natural number set, P is prime number set, and R is real number set. If f : P -> N, g : N -> R and h : P -> R with g(x) = x + 1, f(x) = (3x² - 6x - 27)/(x-1) and h = ( f • g), then the value of x so that h(x) = 9 is....

[shy sage]

How to solve?

[silver maple]

so you're basically solving f(g(x))=9

g(x)=f^(-1)(9)

so 9=(3x^2-6x-27)/(x-1) then solve the quad equation and do shennanigans?

[shy sage]

Where does this equation comes from?

[silver maple]

i took the inverse on both sides

not 100% sure if its legal but thats how i naturally solve these types of equations

[shy sage]

I don't really understand...

[silver maple]

uhhhhhh

do you know how inverses work

[shy sage]

How to find the inverse of that f(h(x)) = 9

[shy sage]

Yeah i only know the inverse like f(x) = x + 1 then f^-1(x) = x - 1

[silver maple]

that works too (thats g(x) tho)

[shy sage]

So I only need to solve f(g(x)) = 9?

[silver maple]

yeap

[shy sage]

So that's so simple

Thanks anyway

.close

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[green inlet]

instead of giving a counterexample for AB

i took the transpose of AB which is just BtAt, and since Bt = B and At = A, so traponse of AB = BA

and then I said BA may or may not equal AB

[fluid wren]

So, what's your question?

[green inlet]

my proof is also correct ye?

[fluid wren]

What you are saying is correct, but if you want to be rigorous, you still need to provide a counterexample

You can't conclude "AB may not equal BA given A, B are symmetric" without a counterexample

[green inlet]

ye i get what you mean. thank you

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[patent dock]

idk how to solve 32

[patent dock]

i simplified to -(sinx+cosx)=0

[silver maple]

so sinx+cosx=0

now we want to do this

sinx=-cosx

tanx=-1

but wait! we must note when cosx=0 since that is where we would be dividing by zero (and thus we must extract those solutions)

[patent dock]

so is the answer any x except 0

[silver maple]

no

theres two parts to this: 1. not including when cosx=0, 2. when tanx=-1

[winged panther]

Why would tan x = -1 include cos x = 0 solutions??

[patent dock]

so when tanx=-1 it works?

[silver maple]

its a solution, yes

[gloomy tide]

it doesn't, but it's good to check

[silver maple]

im saying to not include them because then the equality would fail as we would have division by 0

[winged panther]

No its not.. and it's not necessary as cos x = 0 => sin x = 0 here which cannot be

So just checking for tan x = -1 is sufficient

[patent dock]

so the answers are 3pi/4 and 7pi/4

[winged panther]

Nvm

[silver maple]

yes

[gloomy tide]

yes 7π/4

[silver maple]

??

7pi/4 is in the 4th quadrant...

[silver maple]

but next time

please note the extraneous solutions

you may end up getting something similar, and dividing by cosx is risky since you could potentially be dividing by 0

[silver maple]

thats why i noted this earlier

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[silver mantle]

Can someone help me with probability

Shit someone helped me with a difficult question yesterday and I forgot it

[winged quest]

Search the server for your own comments to find it again

[silver mantle]

🥲 that's gonna take so long lmao

I forgot what time it was also

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[split sail]

i didnt even understand the question

could osmeone explain it pleas

e

[shadow stump]

i would expect that you should calculate the two probabilities

then presumably bet on the one which has favorable odds (better than 50%?)

[split sail]

okay i will try

for the second

would the probability be

$\frac{1}{36} + \frac{1}{36^2} +... + \frac{1}{36^24}$

[split sail]

?

@shadow stump

[shadow stump]

how do you get that expression?

[split sail]

they said double 6 atleast once

so

sorry bad internet

for the first case

💀

ah yes

so for the first case

getting double six just for once of the 24 throws

so it would be 1/6 * 1/6

oh

[zenith heron]

When you're dealing with the probability of, for this case getting atleast 6, you should do P(atleast one 6) = 1 - P(no 6)

[split sail]

i thought that would make the expression quite complex so went on with this

which is wrong

[zenith heron]

This is a very low number

[split sail]

i forgot to multiply it by 5/ 6

[zenith heron]

Just do this, the probability of getting two 6 in two dice is 1/36

So the probability of getting not a double 6 is 35/36

And you have 24 trials

[split sail]

would be expression be

[zenith heron]

So the probability of not getting a double six is (35/36)^24

[zenith heron]

And you want to get atleast one double 6 so just subtract it by 1

So the equation would be 1-(35/36)^24

[split sail]

oh

i get it

thanks

.ckose

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[obtuse bolt]

How should I try to relate the length of de to the length of ab

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@obtuse bolt Has your question been resolved?

[frozen valve]

I think you could connect ADE together and find the lengths of the triangle using sin cos tan, then connect with the length in the middle AD

[obtuse bolt]

<@&286206848099549185>

[frozen valve]

what I told you would work

the big triangle angle is 180

where a is 90

c = 30

b=60

bc would be 60 because 2acb would be 60

e to d would 3meters

connect AED into triangle

find length of hypotenuse with sin cos tan rule

and then find the opposite length

with same rule

b angle is already 60

which the other half of triangle would be 30

find the hypotenuse with the b angle and length AD

[obtuse bolt]

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[winged panther]

.reopen

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✅

[winged panther]

Just a quick second

@obtuse bolt

You can extend BA to O such that ACO = ACB.
Now, note that CO/CB = OA/AB . Also, OA/DE = OB/EB

=> AB = (OA•BC)/OB = 2(OA•BE)/OB = 2•DE = 6

You should probably check this ^^"

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[high stag]

Hey, I'm confused about the middle section when they went between these two lines. I haven't seen d used in that way before, and what happened to d(theta)? Where did the 1/2 come from?

I'm assuming the cos(theta) has been replaced by d(sin(theta)) but I'm still confused

[quick ridge]

cos theta dtheta was replaced with sin d(sin theta)

[high stag]

Well. I guess I'm actually confused about how they went from the second line here to the numerical answer

[quick ridge]

it’s like letting u = sin theta

then du = cos theta dtheta

du = d(sin theta)

[quick ridge]

what 1/2 are you referring to?