help-0

[ocean seal]

Fish

[jolly plover]

Generally have no idea

[upbeat gorge]

Try getting the first few terms and see if you can identify a pattern

[jolly plover]

$1,5,\frac{37}{3},\frac{209}{9}$

[ocean seal]

Fish

[jolly plover]

I can't actually see it

[upbeat gorge]

Try getting more i guess

The more stuff you have, the more evidence you have to guess the common pattern

[jolly plover]

[lone heart]

@jolly plover Has your question been resolved?

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[vapid shuttle]

Am I going about this in the wrong way? Because I don’t really see what to do next

Looks to me like a system of 4 eqns and 9 variables? I don’t really know how to proceed

[jagged raptor]

where do you have 9 variables

[vapid shuttle]

Sorry I meant 6 variables

[jagged raptor]

also you can do this without writing out and equating a bunch of component vectors

[vapid shuttle]

a1-3 and b1-3

[jagged raptor]

ig as a hint i'll say if axb = 0 then ||axb|| = 0

[vapid shuttle]

Well

Tell me if this would be a fair proof or not

If we are in R3

And a*b is 0

Then a and b are perpendicular

And since axb is perpendicular to both a and b

It should be able to exist in R3

But if it is 0

Hmmm

I thought I had something going there

Guess not

[jagged raptor]

here starting with this

how can you rewrite ||axb||

[vapid shuttle]

Well I could compute it

With arbitrary a1-3 b1-3

[jagged raptor]

without that

[vapid shuttle]

I’m not sure

[jagged raptor]

||a x b|| in terms of ||a|| and ||b|| and the sine of the angle between them

was that covered

[vapid shuttle]

I have that

Yes

[jagged raptor]

okay so ||axb|| = ||a||*||b||*sin(theta)=0

[vapid shuttle]

Yes

[jagged raptor]

now we use this

[vapid shuttle]

So the angle theta in between them

Must be 90° (or 270?)

So sine just goes to 1 or -1

[jagged raptor]

yes

meaning for ||a||*||b||*sin(theta) to be 0 one of the vectors lengths has to be 0 (since sin(theta) isnt zero)

[vapid shuttle]

Yes

Okay I see

Thank you

Should’ve just looked for that formula to begin with

[jagged raptor]

yeah its kinda hard if you dont have it in mind to begin with

[vapid shuttle]

.close

[lone heart]

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[mental flame]

pemdas

use it

[shy dove]

it's an image that changes depending on who sees it. in other words, a troll

[alpine sable]

$$ \frac{35}{5} + 10 = 17$$

[ocean seal]

Brandon H

[mental flame]

ye i was giving the benefit of doubt but im rethinking it now lol

[alpine sable]

it doesn't change depending on who sees it

[shy dove]

Brandon ↑

[shy dove]

it does

for me it's 42/6 + 4

[tawny condor]

It only changes if you click on it

Oh

[alpine sable]

Theres a difference between 35/5 + 10 and 35/(5+10)

[shy dove]

I know that image

[tawny condor]

Damn ok maybe it does

[shy dove]

trust

xD

[royal socket]

How does that work what

[tardy stag]

it's a server that gives each person a different math problem image

[royal socket]

I'm just impressed with the troll at this point

[alpine sable]

what?

[royal socket]

A server??

[tardy stag]

anyway 22 / 2 + 23 = <@&268886789983436800>

[tawny condor]

💀

[alpine sable]

[royal socket]

I see that one toi

too*

[shy dove]

twins :D

[tardy stag]

yeah the url itself is https://txnor.com/mathchallenge/?width=809&height=242

[royal socket]

Interesting

wtf I got redirected to a yt video titled discord sex hack

[tardy stag]

yeah

about 3 minutes in i think? it goes over this one

the server can do a lot

[real gazelle]

I don't think that arithmetic checks out

[lone heart]

Channel closed

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[tardy stag]

modular arithmetic

moderatular

[real gazelle]

LOL

[lone heart]

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[hazy rune]

When i was doing rational root theorm method to solve for x, I got this: $x(x+3)(x-4)(x-1)=0$

[ocean seal]

GamingGods

[hazy rune]

So does that mean the possible awnsers are 0, -3, 4, and 1?

[mental flame]

yeah

[hazy rune]

cant I just multiply the x on the left with something like x+3 which would be $x^2 + 3$ ???

[ocean seal]

GamingGods

[mental flame]

i wouldnt phrase it as "possible" answers though, 0, -3 4 and 1 are the answers to x(x+3)(x-4)(x-1)=0

no, you would distribute the x to all terms in the parenthesis

[hazy rune]

x = square root of -3

[hazy rune]

oh right i forgot

Let me check the answer now

BRuH

the awnser was 1,4,-3

there was no "0"

What did I do wrong

[mental flame]

could i see the original problem?

[hazy rune]

I got it right up to $(x-1)(x^3 - x^2 -12x)$

[ocean seal]

GamingGods

[mental flame]

.

[hazy rune]

Original was $x^3 - 13x -12$

[ocean seal]

GamingGods

[mental flame]

ohh you might have just done polynomial division wrong

you shouldnt end up with 4 answers

[hazy rune]

both of us got this, but when I factored it I got x(x+3)(x-4)(x-1) but they got (x+3)(x-4)(x-1)

[mental flame]

send your work

[ruby current]

you can immediately tell that you’re wrong because expanding will give you a degree 4 polynomial

[mental flame]

yeah 3rd degree polynomials can only have 3 solutions

[hazy rune]

ok hold on

[mental flame]

due to the fundamental theory of algebra

[hazy rune]

Ok so

Below the red line, I did it 2 times

One time, I got 4 awnsers

The second time, I got 3 awnsers

[mental flame]

,rccw

[hazy rune]

What expanding bruv

First I used rational root theorm to get plus or minus 1,2,3,4,6,12 divided by plus or minus 1

Then synthetic division

The first synthetic division didn’t work

[ruby current]

Expand this

[hazy rune]

And the second synthetic division uses -1 and it worked.

[hazy rune]

?? Wdym?

[mental flame]

could you explain where you got x^3-x^2-12x?

[wary stream]

You realize when you do this, you start one power less

You don't start with x^3

It starts with x^2

[ruby current]

[wary stream]

The last value is the remainder

[hazy rune]

Damn I forgot

[hazy rune]

Thanks for coming in for the save

[ruby current]

and also, your factor should be x+1, not x-1

[hazy rune]

Right

Thanks

Thank you @mental flame @wary stream and @ruby current

.close

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[lime scaffold]

how would u solve

[lime scaffold]

6+sqrt2/sqrt2

[young finch]

wdym solve

simplify?

[lime scaffold]

yeah

[young finch]

you could split the fraction and rationalize

[lime scaffold]

can u like show me

[young finch]

or just rationalize

[lime scaffold]

like type it or wtv

i have idk its wrong tho

[young finch]

no

[lime scaffold]

i just need to see it be done

[young finch]

do you know what rationalizing is?

[lime scaffold]

u multiply the demonimator

[young finch]

by what

[lime scaffold]

its oppsite

[wary stream]

You're thinking of conjugates

[young finch]

$\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{b}$

[ocean seal]

Arctic

[young finch]

this is rationalizing

[lime scaffold]

holy moly

[young finch]

essentially multiplying by 1

[young finch]

yeah

[lime scaffold]

how should i apply it to this case tho since its addition

[young finch]

same way

numerator can be anything

just multiply the entire numerator by it

or split the fraction first

[lime scaffold]

is the answer 3sqrt2 +2

i give up atp

.close

[lone heart]

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[young finch]

plus 2?

you got 3sqrt2 right

[lime scaffold]

uea

yea

3sqrt2 +2

[young finch]

howd you get +2

[lime scaffold]

sqrt2 *sqrt 2=2

when u multiply the sqrt 2 by each term in the numerator

[young finch]

divided by 2

or else it woul dhave been 6sqrt2

[lime scaffold]

but where does the +2 go if u already used it for the 6

i phrased that so badly

like i already used the 2 in the denominator so how does that cancel the +2

[lone heart]

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[normal sky]

anyone have an idea of what I'm doing wrong:
A street light is at the top of a 16 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 feet from the base of the pole? I keep getting 24/5

[finite flax]

line 3.

you got an "S" variable, not a 5.

[royal socket]

"dd" is so cursed

[normal sky]

i think i did it right tho?
I had s/6 = (s+d)/16, cross multiply
16s = 6s + 6d, minus 6s from both sides
10s = 6d, divide both sides by 6
d = 10s/6 = 5s/3

[royal socket]

Ah

right I see the problem

notice how s is the length of the shadow

But how fast the tip of the shadow moves is given by d(s+d)/dt

not ds/dt

ds/dt is the rate at which the shadow is lengthening

[normal sky]

but why are those not equal?

[royal socket]

Because the woman is moving

The woman's movement also contributes to the shadows movement

because the shadows tip movement = shadow's lengthening + woman's movement

[normal sky]

ah ok i think i get it?

[royal socket]

so the value you should be calculating is ds/dt + dd/dt

[normal sky]

that was correct, thank you!

.answered

1 second

.close

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[lucid phoenix]

what is the min of this?

[lucid phoenix]

nvm i figured it out

.close

[lone heart]

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[lucid phoenix]

.reopen

[lone heart]

✅

[remote heron]

why did you reopen?

[lucid phoenix]

given trapezoid abcd (ab//cd), the area of trapezoid abcd is 180 cm2, ac=24, cd=26, ad=10. find ab

[lucid phoenix]

cause of this

[lucid phoenix]

so can ya help me with this?

[remote heron]

no

[lucid phoenix]

oof

imma wait for some1 else then

[pliant ingot]

how is ac parallel to cd?

can you post a picture of the question?

[young finch]

start by drawing a picture if one isnt provided

[little hazel]

it means ab || cd probably

to start u have all the side lengths of ACD, can u figure out some properties of it

?

[lucid phoenix]

yeah oops

[lucid phoenix]

u draw altitude AH?

[little hazel]

ya

[lucid phoenix]

okie

now whats next?

[little hazel]

so now find AH basically

[lucid phoenix]

how?

[young finch]

trig im assuming

[pliant ingot]

If you have ac and ad, finding ab means that you don't have all side lengths, you have a diagonal

[little hazel]

no u don't need trig

u just use the fact that ACD is a right triangle

to find AH

[pliant ingot]

huh? Why would acd be a right triangle?

[little hazel]

i think im trippin

but ACD is a right triangle by its side lengths, no?

CD is a base of trapezoid?

then u find AH by finding area two ways

then u do area of trapezoid formula = 180 then done?

[pliant ingot]

obviously not to scale

[little hazel]

u r labelling the sides wrong

[pliant ingot]

but adc isn't a right triangle

[little hazel]

ac=24, cd=26, ad=10

[pliant ingot]

it didn't make sense anyway

my point is that there's no right triangle between 3 corners

[little hazel]

there is once u label the sides correctly

[pliant ingot]

show me

[little hazel]

it's not ab = 24 it's ac = 24

[pliant ingot]

,rotate 90

[pliant ingot]

like that?

[little hazel]

close

c and d should be swapped

i'll send diagram

[pliant ingot]

got it

lol thanks

[little hazel]

ya np

[pliant ingot]

so you get the height using similar triangles

and use the area formula to solve for AB

[little hazel]

ya

or just do

area of ACD = 1/2(10)(24) = 1/2(h)(26)

and then area formula for AB : )

[pliant ingot]

x/10=24/26 is basically the same once you cancel stuff

[little hazel]

ya it is just a little faster than doing the similar triangles and more intuitive

[pliant ingot]

I think the OP left though lol

just us

[little hazel]

maybe

with this info OP should be able to solve problem no

now

[lone heart]

@lucid phoenix Has your question been resolved?

[lone heart]

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[lucid phoenix]

like this

[lucid phoenix]

nope

i just put in an equation into my calculator

but i cant solve that equation tho

[little hazel]

what's the equation

[lucid phoenix]

lemme remember

damn i forgot

180=1/2√(56700/676)(26+x)

ah wait here it is

[little hazel]

how did u get the sqrt(56700/676)?

[young finch]

,w compute sqrt(56700)

[ocean seal]

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[young finch]

irrational :(

[little hazel]

bc idt it's correct

[lucid phoenix]

some formulas

[little hazel]

is that for the height?

[lucid phoenix]

dont question it

its indian maths

[little hazel]

ok

[lucid phoenix]

so just help me solve the equation

[little hazel]

well idt the equation is correct

we have to find the height first

[lucid phoenix]

its part of the equation alr

[little hazel]

notice that the height of the trapezoid is the same as the height of ACD

ya but is it corerct

let me check

ya it's not right

[lucid phoenix]

wait

[little hazel]

wait

u meant 57600 right

[lucid phoenix]

lemme check my eq again

[little hazel]

if it's 57600 instead then it's right

[lucid phoenix]

i need to check if its act 57600

yep its 57600

ok continue

the answer is 13 i think

[little hazel]

ok so sqrt(57600/676) can be simplified

[lone heart]

@lucid phoenix Has your question been resolved?

[lone heart]

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[alpine sable]

is there a reason we dont do anything with the 7x

swear its one of the exponent rules

idk

[mental flame]

they factored out sqrt(7x)

thats why it looks like they did nothing with the sqrt(7x)

[atomic stirrup]

Well try and find the sqrt(7x)

[alpine sable]

well yeah you obvioulsy cant simplify it anymore

i was just wondering what happened to the other sqrt 7x

[mental flame]

.

[alpine sable]

how so?

like lets say i run into a problem like that

how do i know i can just factor it out like that

[mental flame]

8sqrt(7x)-17sqrt(7x) = sqrt(7x)(8-17)

or actually you can think of sqrt(7x) as a varible

let u=sqrt(7x)

we have 8u-17u

[alpine sable]

oh

-9u

[mental flame]

combine like terms we get -9u

yep

[alpine sable]

that makes sense now lol

alright ty!

[mental flame]

np :)

[alpine sable]

.close

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[plucky cipher]

hlp pls

<@&286206848099549185>

[lone heart]

@plucky cipher Has your question been resolved?

[wild umbra]

i think

no one helping you

because its of physics

[gleaming ridge]

Allright, I got this

Let's ofc start with (a), the easier one

it's very simple, what did you try?

[lone heart]

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[bronze creek]

Hi, how can the middle term of 2n² + 3n -629 be splitted
Thank you.

[zealous lichen]

???

what do you mean

[wild umbra]

he mean

splitting the middle terms

[buoyant storm]

hint: 629=17*37

[bronze creek]

But it doesn't add up to 3

17 and 37 that is

[subtle birch]

find all factors of 2*629

[bronze creek]

Ok

[subtle birch]

or you can complete the square if you find that dfficuilt

$2n^{2}+3n-629=0⇒\left(n\sqrt{2}\right)^{2}+2\cdot\left(n\sqrt{2}\right)\left(\frac{3}{2\sqrt{2}}\right)+\left(\frac{3}{2\sqrt{2}}\right)^{2}=629+\frac{9}{8}$

[ocean seal]

B-eard

[subtle birch]

imo factoring would be better tho

[bronze creek]

Yea that seems easier

Or the discriminant formula could be used?

[subtle birch]

you could use the quadratic formula but that would lead to long calculations

[bronze creek]

Yea and id only get like 2 minutes for the question which involves statistics also

[subtle birch]

sqrt(9+4*2*629) is not an easy task to find unless you have a calculator

[buoyant storm]

you want to factor $2n^3+3n-629=2n^2+3n-17\cdot37=(2n+a)(n+b)$ where $a,b$ are not necessarily positive constants

[ocean seal]

orthogonal

[bronze creek]

Yes

[buoyant storm]

expanding, $(2n+a)(n+b)=2n^2+(a+2b)n+ab=2n^2+3n-629$

[ocean seal]

orthogonal

[buoyant storm]

continue

[buoyant storm]

also my bad, should be 2n^2 blah blah not 2n^3 blah blah

[lone heart]

@bronze creek Has your question been resolved?

[lone heart]

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[ebon hornet]

ill be honest i have no idea where to start

[lone heart]

@ebon hornet Has your question been resolved?

[subtle birch]

Let the side length of larger pentagon be x

@ebon hornet

and consider this triangle

you can find all the angles

and then proceed to use sine rule to find side length of smaller pentagon in terms of x

[ebon hornet]

ah ok

let me try it

thanks

[subtle birch]

wait you don't even need sine rule

hint: ||isosceles triangle||

[ebon hornet]

wait

so that means that those sides there are equal

but whats the area of a pentagon

actually it doesnt matter

bc if you are halving something

the only answer that makes sense is 4

[subtle birch]

Yes

the side length of larger pentagon is double the side length of smaller one

[ebon hornet]

ok thanks

.close

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[stable grotto]

we have trapezoid with bases 22cm and 12cm and legs 17cm and 21cm. find height

[tardy stag]

make a sketch first

[stable grotto]

is not this true ?

[subtle birch]

That is true

[stable grotto]

so i get answer

but why my book says that the correct answer is 8 ?

[subtle birch]

,w sqrt(17^2-(13/5)^2)

[ocean seal]

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[subtle birch]

smth must be wrong

[tardy stag]

,w 17^2 - x^2 - (21^2 - (10+x)^2) if x = 13/5

[ocean seal]

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[stable grotto]

So my book is wrong yes ?

[tardy stag]

bit confused by what's going on on the bottom

[subtle birch]

,w 17^2-(10-x)^2=21^2-x^2

[ocean seal]

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[subtle birch]

,w sqrt(21^2-(63/5)^2)

[ocean seal]

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[subtle birch]

same answer

[stable grotto]

Ok then book is wrong

[tardy stag]

the main thing i can think of is that they expected you to draw an acute trapezoid

but idk if that's even true

[subtle birch]

84/5 is the answer with acute trapezoid

[stable grotto]

Ok tnx I will be back with another geomerty question 🙂

.close

[lone heart]

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[rough salmon]

x = 4 (mod 11)
x = 6 (mod 13)

[rough salmon]

How can i find x

[mortal trellis]

brute force. or chinese remainder theorem

[outer lark]

if you have a=b(mod m) then that implies m | (a-b) you can use this to setup and equation and solve for x I believe

[rough salmon]

I dont quite understand

If i rmb correctly, i think theres a way to make the mod same, but im not sure

[rough salmon]

Nvm its the Chinese remainder theorem

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[lone heart]

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[hollow brook]

Hi guys

[outer lark]

(x-4)/11 = (x-6)/13 gives x=(-7) which satisfies

[lone heart]

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[indigo anchor]

hey where did I mess up i’m trying to find the x ints

[gray isle]

$(a-b)^2\redneq a^2 - b^2$\
you didn't square properly ,
nor can you erase that $\frac{1}{500}$

[ocean seal]

ℝam()n()v

[indigo anchor]

AHHHH

you’re right

i need to stop doing that

thank you

[young finch]

what

[indigo anchor]

yep

freshman fail again

aight

.close

[lone heart]

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[gray isle]

\.freshman

[keen plinth]

frash

[young finch]

idc anymore my nights ruined

[gray isle]

you saw nothing

[young finch]

ramonov can you put periods in between those paranthesis

[alpine sable]

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

Why's not working

[indigo anchor]

wait

(a - b)^2

is (a - b)(a - b) right

[gray isle]

yes

[indigo anchor]

or is it (a - b)(a + b)

ok

cool

[young finch]

do you want to reopen

[indigo anchor]

nah that’s all

[gray isle]

recall definition of squaring

[young finch]

thats (a-b)^2

[indigo anchor]

.reopen

[lone heart]

✅

[young finch]

wait

[indigo anchor]

nvm we reopening

so which one is it

[young finch]

none

I forgor

[indigo anchor]

@gray isle ples

[winter light]

What do you need?

[young finch]

oh

[indigo anchor]

we need to know if

[young finch]

x^2-4

is x+2)(x-2

same for every other constant subtracted

[indigo anchor]

(a - b)^2 is (a - b)(a - b)

or (a - b)(a + b)

[young finch]

x^2-c = (x+sqrtc)(x-sqrtc)

I blanked so hard just now

[winter light]

What does squaring mean? Multiply something by itself, right?

[indigo anchor]

ye

[winter light]

Then, which do you think is correct?

[indigo anchor]

• and - ?

is it

yes or no is ok

[winter light]

$\left(a ± b\right)^2 = a^2 ± 2ab + b^2$

$\left(a + b\right)•\left(a - b\right) = a^2 - b^2$

This is what you should remember

[young finch]

first line needs to have a ± before 2ab

[ocean seal]

Alberto Z.

[young finch]

yeag

[indigo anchor]

😭😭😭😭😭😭😭

yes or no

please

[young finch]

what do you think it is

[acoustic olive]

What yes or no?

[young finch]

give an answer

[winter light]

But what? I don't understand your question

[indigo anchor]

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

[acoustic olive]

Yes

[indigo anchor]

thank you

[winter light]

Yes, of course

[indigo anchor]

.close

[lone heart]

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[lone heart]

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[young patio]

I kinda just cut the big hexagon into tiny triangles. Is that the way to approach this question?

[young patio]

All good, I'm sure someone else may know,

[median oar]

That works

[young patio]

So there's no evidently better way of doing it? Cutting it up will work?
I get the answer (D) 9,

[median oar]

I would personally try to find the area of the triangle on the bottom right

The one slightly bigger than the green triangle

[young patio]

That would just be whatever the height might be, times base and halved. What would I do afterwards?

[median oar]

Oh dear

Nah I’d first set the green triangle to area of 1

Then I’d fold the corners into the green hexagon to realise the triangle is 1.5x the hexagon

Then I’d realise the big hexagon is 6 of the small triangle

So that’ll be 6*1.5 = 9

Which is what you got

[young patio]

Ah, that feels more efficient,

Thanks for your help!

.close

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[warm tartan]

O is the incentre of triangle ABC. Prove that angle BOC=90+angle BAO

[upbeat gorge]

Focus on the angles with vertex O

[lone heart]

@warm tartan Has your question been resolved?

[lone heart]

@warm tartan Has your question been resolved?

[lone heart]

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[winter seal]

Can anyone explain sequences and series??

[winter seal]

I'm like completely lost from the beginning

like wtf is that

[upbeat gorge]

Sequences in general are a list of numbers following certain patterns or rules

[wind cloak]

A sequence is an ordered set of numbers that are related to one another put simply, while a series is a sum of a sequence

[upbeat gorge]

Series refers to the sum of terms of sequences

Yeah

[wild umbra]

well

ig shes not asking

for the definition

its a whole chapter here

like which part are you lost in?

[wind cloak]

It's a whole chapter everywhere

We only assumed and answered what OP asked for

[wild umbra]

yeah

but i dont think so

shes meant that

now shes gone

[lone heart]

@winter seal Has your question been resolved?

[gray isle]

rules / patterns aren't necessary

[winter seal]

Im lost on how to like

approach the whole idea

I know what a "sequence" is and I know what a "series" is

but idk how they can be related to math

I'm looking at calc.2 for dummies rn

to see infinite series

and I'm kind of getting it but it looks really overwhelming

[lone heart]

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[noble halo]

hi can someone help me with this question

i have no idea on how to do this by showing working

[proper vine]

Ok

I can help you for a little since I’m doing something but basically

11-h=1

h is 10

Nvm hats not a 10 scratch that

[noble halo]

11?

[winter light]

Why H is 11?

[white crater]

wait whered u get H 11

[proper vine]

I seen the H as an 11

Mb

[white crater]

💀 haha all good

so for this you'd say

[proper vine]

Didn’t see the line in the middle

[winter light]

Try to express the areas as base×height

[winter light]

And then use the fact that H - h = 1

[noble halo]

how is that?

[proper vine]

Ya and u can then solve using method of sub or elimination

[white crater]

given

[noble halo]

is it that we substitute H to h+1

[proper vine]

(H)(y-2)=15
(h)(y)=20

[noble halo]

ok got that

[proper vine]

Ya then you plug in instead of H

[noble halo]

but what does the question mean by 'forming a quadratic formula?

[rancid trail]

You need to frame an quadratic equation

[proper vine]

(h+1)(y-2)=15
(h)(y)=20

[noble halo]

yes i got that

i expanded top and use substitution and ended up getting 2h-y=3

can u tell me if im correct or wrong

[somber eagle]

you can just get y in terms of h or the other way around, eg y = 20/h

then sub that into the first equation to get the equation in terms of just y

[proper vine]

Ya you’ll have to sub the anwser with the value you got in one of the original formulas and you’ll get the other value

So you can find the values of x and y

[somber eagle]

bruh

where is x

[proper vine]

h should be 4 and y should be 5

[noble halo]

ok thanks

[proper vine]

I didn’t mean x I meant h

[noble halo]

ill try do the working steps

[proper vine]

Ya

Just make sure those are your anwsers for each

[noble halo]

sure

oh i got it

[noble halo]

This shopuld be fine right?

[proper vine]

It’s wants you to form a quadratic equation

[noble halo]

i formed one there

[proper vine]

Oh ya I see

[noble halo]

i got 2h^2-3h-20=0

correct right

[proper vine]

We’ll that should work

[noble halo]

ok then

[proper vine]

Ya

[noble halo]

thanks for the help

[proper vine]

Np

[lone heart]

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[untold zealot]

I have a recursive math question. I dont know how to solve this question. I started with writing it as r^(n+2)+r^(n+1)-2r^n=-3^n to be able to simplify it to r^2+ r - 2 = ((-3^n)/(r^n)) I then turned ((-3^n)/(r^n)) to a constant k. So that we have r^2+ r - 2 = k but im lost what to do next

[lone heart]

@untold zealot Has your question been resolved?

[naive crystal]

Idk if it works but, -3 = -1-2 = -a0-a1 since you're replacing a_n with r^n then you might also want to replace -a0-a1 with -r^0-r^1

[floral needle]

do you need help

[naive crystal]

Yk, only 1st degree recurrence are easy to solve...

[floral needle]

ok

[zealous lichen]

I think there's formulas for linear recurrence

[untold zealot]

I only know how to solve these

[zealous lichen]

oh wait it's not linear

[naive crystal]

Yes replace a_i with x^i

[untold zealot]

Im not sure

[naive crystal]

Yes cuz it's linear

[untold zealot]

Can I use pq formula to solve it ?

I cant see why it would make a difference by changing -a0-a1 with -r^0-r^1

[naive crystal]

Ye that might not work...

[untold zealot]

hmm I used chat gpt trying to find the answer and they used PQ but I dont trust chat gpt with math so yeah. But do you know a way to solve it ?

[naive crystal]

A_2 = -3

$r^{n+2}+r^{n+1}-2r=(-r^2)^n$

Img ain't loading..

[ocean seal]

_basudev

[naive crystal]

This helps perhaps

[ocean seal]

_basudev

[untold zealot]

I found out a different way to do it much less complicated as what I tried before now I just need to solve c1 and C2 and then it's done 👍🏻 but thanks for the help

.close

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[subtle gate]

how is the answer this

and not just -3072

[gray isle]

why do you think its just -3072

[subtle gate]

oh

my bad 💀

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[alpine sable]

are these two triangles equal

because when I try to solve for F in both of them I get different answers

for instance when I try to solve for F in the left I get -> mg/f = (sine theta * Fn)/(cos theta) = mg/tan theta = F

however when I try to do the same on the right I get-> f/mg = (sine theta * fn)/(cos theta * fn) = tan theta * mg = F

mg/tan theta != tan theta * mg

even though they both are equal to F

where is my mistake

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

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[strong schooner]

Every point with integer coordinates in the plane is colored with one of n colors. Prove that there exists a rectangle in the plane whose vertices are all colored with the same color.

[tacit arch]

is this question from a discrete math class or a probability theory class

[lone heart]

@strong schooner Has your question been resolved?

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[potent flame]

Is this correct?

[keen pasture]

Yes

[potent flame]

thank you

.close

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[weak ridge]

.close

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[obtuse solstice]

can someone explain this? I don't understand it at all

[tacit arch]

Have you taken differential equations yet

And or linear algebra

[obtuse solstice]

today was lecture 1, this is part of reference material

[obtuse solstice]

no not yet

[pale kestrel]

its a little absurd to do this stuff before linear alg

[obtuse solstice]

the course structure is differential equations -> matrix algebra -> linear algebra -> numerical methods for my fall semester

[pale kestrel]

raised_eyebrows at that...

this will really make sense once uve done linear alg

[zealous lichen]

so if two solutions are off by a constant, i.e. y1=cy2 then they are linearly dependent

[obtuse solstice]

so it's just different particular solns from the same general soln or?

[zealous lichen]

basically yes

it only works for two solutions though

you might need to consider n solutions later

so you should probably learn some linear algebra to fully understand it

[obtuse solstice]

hmm if I have time I'll read about it, or just wait till we get to the chapter

thanks

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[vapid steppe]

for part c,

cant you substitute f''(c)=-2

[zealous lichen]

umm just do that

I don't see why not

[vapid steppe]

,w lim as x->c of (-2ln(x^2-c^2+1))/(x^2+5x-cx-5c)

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[zealous lichen]

don't cheat lol

[vapid steppe]

.close

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[chilly idol]

i dont know what the ext means?

[alpine sable]

Typo

[chilly idol]

ok

well i know that -3 is not equal to x

0 cannot be equal to x

-2 can be equal to x

[alpine sable]

So you see no options match

[chilly idol]

yeah

[alpine sable]

So either it is a typo error , or the answer is Not listed

[chilly idol]

ya

im gonna choose not listed

[alpine sable]

You hadn't yet?

[chilly idol]

nope

[slender gull]

What's those points for?

[chilly idol]

its a warm up

ok thanks for helping

.close

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[paper sage]

A homogeneous wooden door with a width of 1.2 m and a mass of 16 kg hangs on frictionless
hinge on one side. We shoot a bullet with mass 15 g and speed 600 m/s in the middle
the door, where it gets stuck. What is the angular velocity of the door after impact?

[paper sage]

I don't understand how to begin

[tardy stag]

sketch or diagram is always a good start

[alpine sable]

A diagram would help

[paper sage]

okay I will try

[alpine sable]

Okay so I gtg but what you have to do is
Find Velocity of Door by conservation of momentum
Use that velocity of door with the distance between where the bullet hit and hinge of door = 0.6m (given)
Use formula for angular velocity

[paper sage]

L=mvr?

[alpine sable]

That's angular momentum?

You have to find angular velocity

[paper sage]

yes that is by w=L/I

I already finished this question thanks for helping me

.close

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[polar zephyr]

If the length of one side of rhombus is 6 cm. and one angle is 60°, then the area of field in shape of rhombus is

[west girder]

2 equilateral triangles

[polar zephyr]

Yeah but can u explain in brief

[west girder]

The shape is 2 equilateral triangles glued together with side length 6

[polar zephyr]

Ahh

Yea

[polar zephyr]

What's the answer then

[west girder]

Find it

[polar zephyr]

36√3?

[west girder]

Area of one equilateral triangle is s^2sqrt3/2

[west girder]

Yea

[polar zephyr]

aight thanks

[west girder]

[polar zephyr]

No wait

@west girder 18√3

[west girder]

It’s two of them together

[polar zephyr]

/4?

/2 is for height

[west girder]

Oops yeah mb

[polar zephyr]

Yea

Thanks i get it

.close

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[queen jay]

is this right?

cause 3^5=243

[royal socket]

I think so yes

[queen jay]

yay

thanks

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[paper sage]

One way to control the altitude of aircraft when they land is by means of interference between
a direct beam and a ground reflection from a radio transmitter emitting the wavelength 90 cm.
At what height h should you place the transmitter antenna to create a signal maximum in a
direction that has a slope of 3.0° above the horizontal? Keep in mind that the radio signal is phase-shifted
180° when reflected to the earth's surface.

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[paper sage]

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[vagrant falcon]

How do I find specific/validate solutions (unit circle) within the intevral of 0,2pi?

[tacit arch]

do you know what $\pi n$ means?

[ocean seal]

riemann

[vagrant falcon]

yea, basically around the circle once

or like 6.28

oh wait

you said pi

[median oar]

Look at the 2nd picture it gives you an idea of how you use the unit circle to find all solutions

[vagrant falcon]

thats half circle

[median oar]

Then pictures 3 4 5 are the algebraic manipulation to get to solutions in the 0 to 2pi interval

[vagrant falcon]

huh

[versed totem]

ah sorry

my bad

[vagrant falcon]

can you guide me step by step so find all solutions for my problem?

[median oar]

Do you understand the steps I’m taking in my problem?

[vagrant falcon]

kind of

[median oar]

What’s the first step

There’s 2 parts to picture 1, but part 1 is already done for you (rearranging the equation)

Part 2 is to draw a triangle

[vagrant falcon]

bascially

7pi/6 is the result of arcsin(-1/2)

[median oar]

That is not what we are doing

Each step is important, don’t go “basically”

[vagrant falcon]

yea i know

im just clarifying something

[median oar]

So first, draw your triangle

[median oar]

Well it’s wrong

[median oar]

That’s not what we are doing

[vagrant falcon]

i did my best to draw

but

yea

[median oar]

Hmm

The triangle needs to always have a base

The base needs to be on the x axis

Something like these ones

[median oar]

No the angle needs to be from the origin

Just draw the triangle without the axis first

[vagrant falcon]

that it?

[median oar]

Man

Your triangle has a side of sqrt2

And a hypotenuse of 2

That’s half a square

At least draw like it

[median oar]

Ok

But we want to label the sides we got

Namely the length of the hypotenuse

This was all I wanted you to draw

[median oar]

A triangle with the given sidelengths

[vagrant falcon]

ok

[median oar]

Have you got this down?

[median oar]

No it should say sqrt2 for 1 side

And 2 on the hypotenuse

We don’t actually care about the last side

[median oar]

What we really need to do is this

We need to recognise the kind of triangle this is

Because that gives us the angle

[vagrant falcon]

its a right triangle

[median oar]

And yeah it should be pi/4 not pi/2

[median oar]

The angle here was wrong

Ok step one complete

Now look at step 2

https://cdn.discordapp.com/attachments/490557019623915520/1136053675320684544/IMG_3087.jpg

This is all the ways we can draw a triangle

There should only be half as many for you since it’s half a square

A bunch of these for you look the same

Keep in mind I am only labeling the known sides

[median oar]

For this I’m labeling only sqrt3 and 3

[median oar]

Don’t worry about the angles

Ok so

There are 2 more on the bottom

Also don’t label the x axis

We should pretend we don’t know what that length is because it is clutter on the drawing

[median oar]

Label the important sides

The x axis is not important

[median oar]

Ok

You’ve correctly realised the hypotenuse is never negative

Which is quite important here

[vagrant falcon]

yea

[median oar]

We should now realise the bottom 2 are impossible!

Because they represent sin(θ)= **-**sqrt2/2

[median oar]

If we were looking to solve this question then we’d use the bottom 2 triangles

[vagrant falcon]

yes

[median oar]

But we are solving for the positive one so just the top 2

[vagrant falcon]

yea

[median oar]

You did also weight the correct angle here, but it’s not labeled right

The 3pi/4 is from the positive x axis to the hypotenuse

Not from the negative x axis to the hypotenuse

[vagrant falcon]

ok

[vagrant falcon]

gotcha

[median oar]

Ok that’s step 2 done

Now we just have the algebra left

Notice our solutions are either pi/4 + 2kpi, k is an integer

[vagrant falcon]

yep

[median oar]

Or 3pi/4 + 2kpi, k is an integer

[vagrant falcon]

ye

[median oar]

https://cdn.discordapp.com/attachments/490557019623915520/1136050605719298078/IMG_3215.jpg

So I write this down

The left side is just whatever is inside the trig function

So for you it would say 2x + pi/4

[vagrant falcon]

i did it like this

[median oar]

Nah

You need the 2

[median oar]

This means the negative ones as well

The triangles on the bottom

We don’t want those

[vagrant falcon]

ok

[median oar]

either you say 2n, n is an integer, or n is an even number

They mean the same thing but we need to add 2pi’s every time

Oh wait

[vagrant falcon]

ok

[median oar]

You’ve already done that

You’ve solved for the x

For each case

[vagrant falcon]

yes

[median oar]

Now it’s the easy part

[vagrant falcon]

yay

[median oar]

Just plug a bunch of n’s in

So go n=0, write your 2 x values

Go n=1, write your 2 x values