help-10

[keen bobcat]

i can only get that theyre equal but the condition says they cant be

no calculator allowed

[obtuse pebble]

@keen bobcat Has your question been resolved?

[obtuse pebble]

@keen bobcat Has your question been resolved?

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[timid silo]

.reopen

[timid silo]

Help me !!

[robust sleet]

,rotate

[timid silo]

?

[median dome]

pretty sure you can u sub after diving numerator and denominator by sin^2(x)

[timid silo]

Nahh bro not helping

<@&286206848099549185>

Please help

[median dome]

i just solved it using that

[timid silo]

Send pls

[median dome]

$\int\frac{3+2\cos(x)}{(2+3\cos(x))^2} dx = \int\frac{3\csc^2(x)+2\csc(x)\cot(x)}{(2\csc(x)+3\cot(x))^2} dx$

[warm shale]

kheerii

[median dome]

now just substitute $u=2\csc(x)+3\cot(x)$ to get $du=2\csc(x)\cot(x)+3\csc^2(x)$

[warm shale]

kheerii

[timid silo]

Thanks u mate

[obtuse pebble]

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[obtuse pebble]

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[neat bobcat]

How do u diffrentiate H = 2.5x(2-x)

[wooden cipher]

With respect to?

[empty cypress]

x surely

you can just multiply it out then do power rule, if thats teh case

[obtuse pebble]

@neat bobcat Has your question been resolved?

[neat bobcat]

Wdym

[neat bobcat]

Yea i did that but hte answer is 2 apparntly and i got 1

[wooden cipher]

!original

[obtuse pebble]

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.

[neat bobcat]

Ohh ok

[obtuse pebble]

@neat bobcat Has your question been resolved?

[fallen inlet]

I think you should ping

Help

[obtuse pebble]

@neat bobcat Has your question been resolved?

[neat bobcat]

Ping who brah

<@&286206848099549185>

[obtuse pebble]

@neat bobcat Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[full oar]

what does this question mean

do I just find the reciprocals

[quartz girder]

Do you know the definition of being perpendicular?

[full oar]

like for a, m = -2

yea

[quartz girder]

negative reciprocal right?

[full oar]

doesn’t that depend on the number tho?

like it was already -2

I would change it into 2

[full oar]

Like is it just the reciprocal

?

[quartz girder]

If m = x then m perpendicular = -1/x

[full oar]

oh ok

[full oar]

what if it was m = -x tho?

would it be X?

[quartz girder]

if m = -x then it would be -(1/-x)

Two negatives make a positive right>

[full oar]

yea

[quartz girder]

There ya go that should be enough to solve every question

[full oar]

so it’s just -(1/-x)

[quartz girder]

Which would be 1/x

[full oar]

ok

[full oar]

hâta if m = 0?

*what

do i write undefined?

[quartz girder]

Hmm

I'd think infinity but honestly depends on the teacher

Nvm write undefined

[full oar]

btw could u check this for me?

[quartz girder]

but technically the slope of perpendicular to a horizontal line would have infinite slope

[full oar]

can u double check for me pls?

.close

[obtuse pebble]

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[fathom flicker]

I was hoping someone could double check my proof for this

[warm shale]

Austin

[fathom flicker]

So we can write for all x in R, g(x)=C

then applying the MVT instead to f, tells us for all alpha, beta in R, there exists a gamma such that
f'(gamma)=g(gamma)=((f(alpha)-f(beta))/(alpha-beta)
but, g(gamma)+C for all gamma in R, so in other words,
the slope between any two points alpha, beta of our function f, is always C. So f is linear

[frank dome]

yep, this proof seems pretty correct to me

[fathom flicker]

I am mostly wondering about that last line. Is it reason enough to show that the secant line between any two points always having constant slope implies a function is linear? It seems like it should

Maybe that also requires proof

I don't have a definition for a function being linear so that doesn't necessarily help

[frank dome]

it really doesnt, since if slope is constant, let the slope be m, then the function has to be f(x) = mx + b

the degree of the polynomial has to be 1, otherwise you wouldnt have constant slope

[fathom flicker]

Well

who knows we are dealing with polynomials here?

and also that is sort of what I am trying to prove no?

TY Nalin, I've got it sorted out

.close

[obtuse pebble]

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[obtuse pebble]

@grizzled geyser Has your question been resolved?

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[gaunt walrus]

Math concept problem: Can I visualize a quadratic equation with complex roots?

[gaunt walrus]

I know the curve doesn't intersect with x-axis

But what about complex plane?

Can I visualize it in complex plane?

[empty cypress]

not really

https://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1034&context=mathmidexppap

something like this is okay

but its not what you are thinking of

[gaunt walrus]

Flip the quadratic horizontally over the vertex???

lol

[empty cypress]

<@&268886789983436800>

[gaunt walrus]

So it is possible?

for three dimentional (x,y,i) ?

<@&286206848099549185>

<@&286206848099549185>

Am I correct with the concept?

[obtuse pebble]

@gaunt walrus Has your question been resolved?

[tiny seal]

if you mean what i think you do, youd need 4 dimensions

so, you could, it just wouldnt make too much sense

just do (x,y,i), then you are viewing everything as a slice of 4d space

just vary the other value with time

[obtuse pebble]

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[runic void]

$z_n = Arg \left( -1 +\frac{1}{n} \right)$

[warm shale]

.doc

[runic void]

I'm asked to find the limit of this sequence

$\lim_{n \to \infty} zn = Arg \left( \lim_{n \to \infty}-1 +\frac{1}{n} \right)$ Is this notation correct?

[timid silo]

i see the helper needs help

[runic void]

i'm helpful, not a helper

[timid silo]

how did you get the helpful role?

[royal basin]

you've got some missing underscore issues

[royal basin]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[warm shale]

.doc

[royal basin]

the notation's fine

however it looks like $z_n = \pi$

[warm shale]

AnnGhost

[royal basin]

for all n

[runic void]

i can just take the limit inside the argument right?

sort of felt little off to do so

[runic void]

makes sense

thank you.

.close

[obtuse pebble]

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[lime sphinx]

Is there a formula to find this without drawing the graph

[lime sphinx]

The range

[tiny seal]

for 7?

the radicand must be positive, and we cannot divide by 0

so, if you see
$$\sqrt{g(x)}$$
then
$$g(x)\geq 0$$

[warm shale]

Cycadellic

[tiny seal]

likewise,
$$\frac{1}{g(x)}$$
then
$$g(x)\neq 0$$

[warm shale]

Cycadellic

[tiny seal]

oh the range

[lime sphinx]

Ya

Do i have to draw the graph

[tiny seal]

no

[lime sphinx]

Or can i use formula

What is the formula

[tiny seal]

use
$$\sqrt{g(x)}\geq 0$$

[warm shale]

Cycadellic

[tiny seal]

for d, b, and e, use
$$\frac{1}{g(x)}\neq 0$$

[warm shale]

Cycadellic

[tiny seal]

im not sure how to explain d without the limit, but ill try

$\frac{1}{\sqrt{x}}>0$ can be taken as a rule

[warm shale]

Cycadellic

[tiny seal]

intuitively, (this is a big abuse in notation)
$$\sqrt{x} < \infty$$
$$\frac{1}{\sqrt{x}}>\frac{1}{\infty}$$
$$\frac{1}{\sqrt{x}}>0$$

[warm shale]

Cycadellic

[tiny seal]

maybe somebody else can explain it better, but this should work fine

[obtuse pebble]

@lime sphinx Has your question been resolved?

[obtuse pebble]

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[lime sphinx]

Oh okay

[lime sphinx]

I think i kijda get it

Tyanks

Also isnt the domain for this x is real numbers cuz like i said this and someone else said it was wrong

But im pretty sure its correct

[high lily]

how are you getting all real numbers for this

[tiny seal]

its not because, remember how if $\sqrt{g(x)}$
then $g(x)\geq 0$

[warm shale]

Cycadellic

[lime sphinx]

This is what I did but im kinda confused

Is the vertext bit corect?

[obtuse pebble]

@lime sphinx Has your question been resolved?

[tiny seal]

sorry about the late reply

we dont care about the vertex, but you are very close
we are trying to solve the inequality
$(x+4)(x+2)>0$

[warm shale]

Cycadellic

[tiny seal]

its very important to note though that when x<-4, x+2 is negative, so the ineq should flip when solving for this term

[tiny seal]

but youre almost done with this one

[obtuse pebble]

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[lime sphinx]

Its Okay thanks for trying yo help

[lime sphinx]

Eait sorry i dont get this

[tiny seal]

we have $(x+4)(x+2)>0$

[warm shale]

Cycadellic

[tiny seal]

we can divide (x+2) or (x+4) on both sides

treat (x+2) as you would a negative number

following
$$-a>b$$
$$1<\frac{b}{-a}$$

[warm shale]

Cycadellic

[tiny seal]

treat (x+4) as you would a positive number

which just means we can divide both sides by (x+4) and the inequality does not change

[obtuse pebble]

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[toxic elm]

For this question where is the point E?

[toxic elm]

<@&286206848099549185>

[obtuse pebble]

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.

[dark stirrup]

That's what you are tasked with finding out

[toxic elm]

This is my diagram

Where should I put point E, on which side of point C?

[obtuse pebble]

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[crystal shoal]

need help with this

[crystal shoal]

this is what i came up with

[obtuse pebble]

@crystal shoal Has your question been resolved?

[crystal shoal]

<@&286206848099549185> ping

[obtuse pebble]

@crystal shoal Has your question been resolved?

[warm dome]

What’s the task? Feel free to also post in German

[crystal shoal]

ich soll zeigen, obdie rekursive funktion T(n) in einer der 3 ordnungen liegt per vollständige induktion

[warm dome]

Steht dieses O bei der 1 für die Menge aller n^2?

[crystal shoal]

ja theta(n^)

[warm dome]

Sieht für mich so aus als wäre T(2)=10, was zu keinem passt. Oder hab ich da was übersehen?

[crystal shoal]

zu n^4 würds passen

glaub ich

wäre 16

[warm dome]

Jo dann hab ich das theta falsch interpretiert. Ich Google es mal kurz

[crystal shoal]

theta heißt dass t(n) für große n genauso schnell wächst wie die ordnung

[warm dome]

Ahhhh okay

[crystal shoal]

als eingabe sollen halt nur die zweierpotenzen n = 2^k angeschaut werden meint mein prof, das reicht für diese das zu zeigen

[warm dome]

Jap

[crystal shoal]

ich check halt nur gar nciht was ich machen soll

ich kenn halt Induktion nur anhand dieser kack summenformeln

und das man zeigt n -> n+1

jz is aber n = 2^k

[warm dome]

Ganz ehrlich, ich Checks auch nicht 😅 aber ich versuch’s noch zu verstehen

[crystal shoal]

lieb von dir

ich bräuchte halt nur ein beispiel wie man sowas angeht, aber da findet man im netz gar nix und der prof ist so asi und macht auch nix dazu

[warm dome]

Also die Idee ist dass theta(n^) gleich schnell steigen muss wie t(n)?

[crystal shoal]

umgekehrt

die rekursive funktion wächst für genug große n ind er ordnung theta(1. 2. oder 3.)

[warm dome]

Okay Ich glaub ich hab ne Idee @crystal shoal

[crystal shoal]

das wäre coool 😭

[warm dome]

Ich hab mir grad das hier angeschaut

Wie wäre es, wenn wir konstanten c1 und c2 festlegen und schreiben dass

c1*n^2<=T(n)<=c2*n^2

Und dann zeigen dass das gleiche dann auch für T(n+1) gelten muss

Keine Ahnung ob das klappt

Aber wir sollten es mal versuchen

$$c_1n^2\leq T_2(n)\leq c_2n^2\Rightarrow c_14n^2\leq T_2(2n)\leq c_24n^2$$

Das müssen wir zeigen oder zeigen dass es nicht zutrifft

[crystal shoal]

verstehe

aber n muss ja = 2^k sein

[warm dome]

Ja stimmt

[crystal shoal]

das ist so was mich abfuckt

aber

[warm shale]

FirstNameLastName

[crystal shoal]

wenn n = 2^k ist, dann müsste man 2^k+1 zeigen

[warm dome]

Genau

[crystal shoal]

ich versuch mal den ansatz

[warm dome]

Ich glaube durch das +n geht es nicht for n^2

Weil irgendwann wird T(n) zu groß

[crystal shoal]

ich komm nur auf salat

man müsste ja für jede ordnung zeigen, das es für >= und <= gilt

für jeden fall dann eine vollständige induktion

[warm dome]

Ja

[crystal shoal]

ich glaube das war nicht vom prof so vorhergesehen

man müsste sofort die vollstädnige induktion anwenden und käme dann auf eine ordnung

aber verstehen tue ich deinen ansatz

oh mann

[warm dome]

$$T_2(8n)=4T_2(4n)+8n$$
$$T_2(8n)=256T_2(n/2)+120n$$
$$T_2(4n)=4T_2(2n)+4n$$
$$T_2(4n)=64T_2(n/2)+28n$$
$$T_2(2n)=4T_2(n)+2n$$
$$T_2(2n)=16T_2(n/2)+6n$$
$$T_2(n)=4T_2(n/2)+n$$

[crystal shoal]

check ich nicht xd

[warm dome]

😅

Ich konvertiere einfach alles mal

Um zu schauen ob ein Muster auffällt

[warm shale]

FirstNameLastName

[warm dome]

Vielleicht ist es doch n^2

Man hat das alte mal 4

Und dann + das in den Klammern vom neuen

$$T_2(2n)=4T_2(n)+2n$$

[warm shale]

FirstNameLastName

[warm dome]

$$(2n)^2=4n^2$$

[warm shale]

FirstNameLastName

[warm dome]

@crystal shoal wie findet man denn die Ordnung normalerweise

[crystal shoal]

bei ner normalen funktion kann man den limit berechnen

also sagen wir zb 2n^2 + n + 3 element O(n^2), dan bildet man den grenzwert 2n^2 + n + 3 / n^2 und je nachdem wie das ergebnis ist liegt das in der ordnung

oder man zeigt halt dass es ein c gibt sodas 0 < f(x) < c* g(x)

[warm dome]

Okay ich glaube das lässt sich für c=3 machen

Ach was weiß ich

[crystal shoal]

xd

[crystal shoal]

trotzdem danke für deine mühe

[warm dome]

Kein Ding. Bisschen frustrierend dass ich’s nicht gelöst kriege

Poste die Aufgabe vielleicht nochmal auf englisch in nem neuen Channel, ich glaub hier kriegst du keine Hilfe mehr

[crystal shoal]

ist für mich auch frustrerend weil wir einfach keine hilfe vom prof kriegen

.close

[obtuse pebble]

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[split furnace]

These equations are certain to have the solution x = y = 0. For which values of 'a' is there a whole line of solutions?
ax + 2y = 0,
2x + ay = 0.

[split furnace]

The instructor has marked the answer as a = 0.

But I don't quite get it.
If a=0, the equations become,
y = 0,
x = 0

And they only have a unique solution at (x, y) = (0, 0).

How's it the case then, they have a whole line of solutions?

[minor belfry]

hmm

lemme read

does x=y always?

are we assuming that?

if x=y=0, a can be anything

[split furnace]

Nope, we aren't assuming that.

[minor belfry]

okay im not sure exactly what this question is asking

but

if a = 0

then u get the equations 2y = 0

2x = 0

which means both are always true

if u take a = 1

u get x + 2y = 0

2x + y = 0

if u rearrange

u get x = -2y

-4y + y = 0

-3y = 0

y = 0

a = 5

2x + 5y = 0, 5x + 2y = 0

x = -(5/2)y

if u sub in, y = 0 again

[split furnace]

Yeah, these too intersect at (0, 0)

[median dome]

a=0 is incorrect

you want the two lines to be coinciding basically

[minor belfry]

so i guess for all a values, x = y = 0

[median dome]

no

[minor belfry]

when is it not correct?

which a values?

[median dome]

it is correct, but that's not what's being asked

for there to be a whole line of solutions, the two lines must be coinciding

[minor belfry]

ohh thats what it means

[split furnace]

If I may reframe it for you, the question is - find a value of 'a' such that the system,
ax + 2y = 0
2x + ay = 0,
Has infinitely many solutions.

[median dome]

i.e the two equations are equal or proportional to one another

[split furnace]

Okay

[split furnace]

Yes

[minor belfry]

yes this makes more sense now

[split furnace]

Ohh, then a = 2?

[minor belfry]

sorry when u said line, i didnt realise u meant a literal line

[median dome]

yep

[split furnace]

Hehe, I see

[minor belfry]

yea sorry, a=2 is definitely the answer

its a bit of a silly question

[split furnace]

Nice..
I assumed the instructor's answer was true, and was worried how.

[split furnace]

Yeah

[minor belfry]

ur instructor is silly aha

although

there are 2 solutions

[median dome]

how's that?

[split furnace]

Which ones?

[minor belfry]

rearrange the two equations

y = -2x/a

or

y = -ax/2

[median dome]

oh yea

[minor belfry]

a=2 gives: y=-x

but

[median dome]

+-2

[minor belfry]

what else could there be

[minor belfry]

https://www.desmos.com/calculator/dohio2kl8j

[split furnace]

Oh right. a = ± 2, that's what you guys mean..?

[minor belfry]

try moving the slider, it might be interesting for u to see why this question is like it is

[minor belfry]

yep

[split furnace]

Okay!

[minor belfry]

then ull get y = x

[minor belfry]

its not necessary, but i think visually having an understanding is good, and something like -2 is easy to miss

[split furnace]

True.

[minor belfry]

if u look algebraically, -2x + 2y = 0 => -(2x-2y) = 0 => 2x-2y=0 {u can just divide out the -1}

just something to look out for

im assuming u havent done polynomials yet

ull learn eventually that $x^2 = 4 \implies x = \pm 2$

[split furnace]

😭i have

This was a question from Strang's book on Linear Algebra

[warm shale]

shavet

[minor belfry]

oh wow i see

good on u aha

[split furnace]

I mean, I'm familiar with a good amount of math stuff, like group, rings, fields, metric spaces, etc.
But I suck at visualising and solving things.

[minor belfry]

ah yea thats alright

[median dome]

I would recommend not passing these types of comments to OP.

[split furnace]

Yes

[minor belfry]

yea fair enough lmao

[minor belfry]

sorry ahah, i thought u hadnt done polynomials

[split furnace]

No no, it's fine lol.

I'm redoing a lot of forgotten stuff rn

Anyway, thanks @minor belfry and @median dome !

[minor belfry]

all good lol

[split furnace]

.close

[obtuse pebble]

Channel closed

Closed by @split furnace

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[dense tinsel]

How do I do this

[minor belfry]

what are you trying to do

[dense tinsel]

Find x

Similar figures thing

[minor belfry]

okay so you assume they are similar?

[dense tinsel]

I have to do something with k^2

[alpine bison]

what's your elo?

[dense tinsel]

Book says they are

[dense tinsel]

1400 usually why

[minor belfry]

??

[minor belfry]

bruh

[minor belfry]

okay great

[dense tinsel]

Chess stuff

[minor belfry]

if i have a line thats 3cm

[minor belfry]

i know lol

ohhh because u put gm in ur username

[alpine bison]

i think it was higher because there is GM in your nick...nvm just curious

[minor belfry]

thats why he is asking ahahhha

okay now

if u have a line thats 3cm, vs a line thats 2cm

[dense tinsel]

Oh haha i do it cuz it looks nice

[minor belfry]

whats the reduction in size

[dense tinsel]

Yes sir

1.5

K = 1.5

[minor belfry]

2/3 decrease

wrong way

[dense tinsel]

Dude i did 3/2

[minor belfry]

if u go to 3 -> 2

sorry lemme rephrase

[dense tinsel]

holdup

[minor belfry]

if i have a line thats 3cm

and want to make it 2cm

what do i do

[dense tinsel]

Oh you take off a 3rd?

[alpine bison]

wow dennis do you know italian ..me too

[minor belfry]

yea, same as multiplying by 2/3

[dense tinsel]

Ho capito vai via sto studiando

[minor belfry]

subtracting isnt ideal when u are going between stuff like this, its all about proportions

now

if i have a square thats 2x2 whats the area

[dense tinsel]

4

[alpine bison]

sorry

[minor belfry]

if i halve the lengths of each side what do i get

[dense tinsel]

👍💀

[minor belfry]

if i multiply by 1/2

[dense tinsel]

2?

[minor belfry]

if i multiply each side

by 1/2

[dense tinsel]

Oh wait no 1

[minor belfry]

yea

[dense tinsel]

Ok

[minor belfry]

2 * 1/2 = 1

now

whats the area of that square

[dense tinsel]

4?

[minor belfry]

1x1 square

[dense tinsel]

atleast it should be

oh

1

[minor belfry]

okay now

what do u think the correlation is

[dense tinsel]

alright so the area should be 24

but the book says it/s 16

[minor belfry]

nope its not 24

okay lemme do this again

[dense tinsel]

sorry im a poor learner

[minor belfry]

if i have a square thats 1x1

its okay ahha

dont rush

step by step

[dense tinsel]

ok

[minor belfry]

1x1 square area?

[dense tinsel]

= 1

[minor belfry]

if i triple the length of each side, what is the length of each side

[dense tinsel]

3

and area ia 9

[minor belfry]

okay

cool

u 3x the side

but what happens to the area

[dense tinsel]

it becomes 9?

[minor belfry]

yea

whats the correlation

ill give a hint

what unit do u use for a length vs area

[dense tinsel]

meters?

[minor belfry]

length is meters

whats area

[dense tinsel]

^2

[minor belfry]

yep exactly

[dense tinsel]

shit

wait a sec

[minor belfry]

meters ^ 2

[dense tinsel]

im cooking rn gimme a sec

[minor belfry]

bettt

[dense tinsel]

right well i was cooking but i made some inedible steak

i have no idea what to do with the ^2

[minor belfry]

lets look at the square example

1x1 = 1

3x3 = 9

i triple the side length

[dense tinsel]

wait a second

[minor belfry]

and the area is x9

[dense tinsel]

what is the scale factor

it's 2/3 right?

[minor belfry]

yea

so what is the area scale factor

[dense tinsel]

0.4...

wait no

it's 0.666666667

but we gotta square that

right?

[minor belfry]

bingo

keep it as 2/3

u dont need to change into a decimal

(2/3)^2

,w (2/3)^2

[warm shale]

Results provided by WolframAlpha

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

Requested by shavet.

[dense tinsel]

HOLY SHIT

HOLY SHIT

[minor belfry]

thats an awfull decimal

[dense tinsel]

HOLY SHIT

HOLY SHIT

HOLY SHIT

[minor belfry]

but the fraction is way nicer

[dense tinsel]

LADIES AND GENTLEMEN

[minor belfry]

yes lo

[dense tinsel]

WE MCFUCKING GOT EM

stay here bucko i got more

[minor belfry]

alright brother

[alpine bison]

gg dennis

[dense tinsel]

alright @minor belfry

im all eyes

[minor belfry]

okay

[dense tinsel]

so we know that the scale factor is

[minor belfry]

lets first see what we can work out

[dense tinsel]

holdup

[minor belfry]

whats the scale factor?

here

[dense tinsel]

2.777777777778

we gotta square that tho right

[minor belfry]

okay maybe no decimals

fractions

[dense tinsel]

alright

that would be 5/3?

like we root the top and bottom

or we can't do that?

[minor belfry]

okay why are we rooting

okay so

lets start

[dense tinsel]

idk cuz i feel like simplyfying

[minor belfry]

we are comparing the two areas

we are going from big to small

so the big number on top

25/9

[dense tinsel]

right my simplifying got us nowhere forget it

[minor belfry]

good start

but

u cant sqrt to simplify

like

4/2 -> 2/1

[dense tinsel]

25/9 = 8/x

[minor belfry]

but 16/9 -> 16/9

[dense tinsel]

cross multiplication

[minor belfry]

not quite

lets hang off for a sec

[dense tinsel]

alright

[minor belfry]

lets use the square example

1x1 = 1cm^2

3x3 = 9cm^2

yes?

[dense tinsel]

yes

[minor belfry]

now

if im going from 1 -> 9 what do i do

1 area -> 9 area

[dense tinsel]

you make 1 = 3 and square that

[minor belfry]

nono lets not look at lengths

just compare the square area

1 -> 9

what do i do

[dense tinsel]

1 * 9

[minor belfry]

x9

great

now

i know that my small square has side length 1

but my big square, has side length x

and i wanna work out what to do

before, i was using length to work out area right?

[dense tinsel]

yes'

[minor belfry]

in the previous question

so i was going

m -> m^2

when comparing areas

im going m^2 -> m

right?

[dense tinsel]

what if x = 9

wait not it doesn't

[minor belfry]

in ur question?

[dense tinsel]

forget that i spoke

[minor belfry]

yea it doenst aha

its okay

[minor belfry]

is this true?

[dense tinsel]

well yes

[minor belfry]

okay

well

[dense tinsel]

i think so maybe idk

dude im stupid this math unit is a bitch fr ong

[minor belfry]

when i went m -> m^2, what did i do to the scaling factor

[minor belfry]

its alright we get this bread

[dense tinsel]

your smartass squared it

[minor belfry]

okay

so

if im going

m^2 -> m, (to use area to find length), what do u think i do to the scaling factor

[dense tinsel]

u rooted it

[minor belfry]

yea

so

what do u think u do in ur question

u wanna try and cook

ive got cm^2, and im trying to work out x cm

any ideas?

[dense tinsel]

do we square root 8

[minor belfry]

not quite

ill make a quick question

if i have a square thats 16 cm^2 and 4 cm^2

how do i go from 16 -> 4

[dense tinsel]

you square root 16

and it becomes 4

[minor belfry]

uhh

no sqrt

use multiplication

[dense tinsel]

4 * 4 = 16

16/4 = 4

[minor belfry]

yep divide by 4

[dense tinsel]

dude you wanna go in a vc

[minor belfry]

okay now

nah i dont have headphones on me rn

[dense tinsel]

ok

[minor belfry]

divide the AREA by 4

okay what length square gives u 16cm^2

[dense tinsel]

so i divide 9 by 4 to get x?

which area do i divide by

4

[minor belfry]

and what area gives u 4 cm^2

[dense tinsel]

4

[dense tinsel]

2

[minor belfry]

so

16 -> 4, and 4->2

[dense tinsel]

yes

[minor belfry]

one is cm^2

one is cm

when we went from cm -> cm^2, we squared the scaling factor

so if we go from cm^2 -> cm, what do we do to scaling factor

[dense tinsel]

oh wait

we square

25/9

[minor belfry]

not square

the opposite

[dense tinsel]

OH

it's 1.66666667

so 9/1.666667

nope doesn't work

fuck dude

[minor belfry]

okay so

we have 25/9

right?

the scaling factor between areas (cm^2 to cm^2)

[dense tinsel]

i gtg eat breakfast real quick

keep this chat open

sorry bro

[minor belfry]

all good ahhahaha

gimme a ping when ur back

[dense tinsel]

@minor belfry

[minor belfry]

okay

lets go

[dense tinsel]

damn bro was here the whole tike

time

[minor belfry]

okay ill give u the rule

[minor belfry]

nah im just watching utube on my laptop

when u are converting between different types of units

i.e.

length -> area -> volume

u need to square, or cube, or the opposite if ur doing backwards

cube root, or square root

[dense tinsel]

ok

seeing as im doing backwards

please

tell me what i have to root

i do not understand

[minor belfry]

so

when u did the first question

u had lengths, and looking for area

[dense tinsel]

yes

now im look for length

[minor belfry]

so u find a ratio between the lengths, then scale UP (length to area)

so now uve got areas and looking for length

so u find a ratio between areas (25 and 9)

then scale DOWN

(what would u have to do to change the ratio from area to length?)

[dense tinsel]

so i square ROOT 25/9 (2.777777778)?

[minor belfry]

yep

well

lemme show smth

so

[dense tinsel]

OH

OK

SO I GOT

[minor belfry]

$\sqrt{\frac{25}{9}} = \frac{\sqrt{25}}{\sqrt{9}}$

[dense tinsel]

8/25 = X/9

5/3

or can also be written as 1.666666667

[warm shale]

shavet

[minor belfry]

perfect!

okay now

u can multiply the length u have by ur scaling factor

length * scaling factor = x

[dense tinsel]

and 5/3 being the scale factor?

[minor belfry]

yep

[dense tinsel]

8 * 5/3 = x...

there is no way in hell x = 13.33333333

[minor belfry]

1 sec

ahh

so u did ur ratio wrong

so

when ur going from 25 -> 9

what do u multiply by

25 * something = 9

u can rearrange for that something

something = 9/25

[dense tinsel]

3.24

[minor belfry]

that could be right

i think

lemme chec

[dense tinsel]

0.36

is the answer

there is

absolutely no way x = 0.36

bcuz the answer is 4.8, i just dk how to reach it

[minor belfry]

uhh not the answer i got

okay yea

i got 4.8

okay so

[dense tinsel]

hOW

HWO

[minor belfry]

3/5 is ur scaling factor

[minor belfry]

i dont know what this is

[dense tinsel]

i thought it was 25/9

[minor belfry]

no u got it back to front

[dense tinsel]

forget it

[minor belfry]

look here

read those few messages

[dense tinsel]

oh yeah

[dense tinsel]

ok so to get from 25 to 9 we multiply 25 by 0.36

[minor belfry]

yea

exactly

[dense tinsel]

0.36 = 9/25...

[minor belfry]

keep it as 9/25

yep

now

[dense tinsel]

ok

[minor belfry]

then u do the stuff u do for

[dense tinsel]

oh yeah that becomes 3/5 ok

[minor belfry]

cm^2 -> cm

yep

then thats ur scaling factor

so x = length * scaling factor

sqrt(9/25) = 3/5

[dense tinsel]

x = 8 * 3/5?

[minor belfry]

$\sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5}$

[minor belfry]

yep

[dense tinsel]

AYO

AYO

AYO

AYO

[warm shale]

shavet

[dense tinsel]

NAH BRO COOKED A WHOLE BANQUET

[minor belfry]

ez

[dense tinsel]

IGHT

we got three more buckaroo

:)

#4

Rotate it somehow idk

@minor belfry

what im doing first is

[minor belfry]

three more??

[dense tinsel]

im detailing everything

[dense tinsel]

yeah it's gonna get easier dw

i know some of this stuff better

[minor belfry]

okay go ahead

[dense tinsel]

ill send another pic

wait hold it

alright so

letter A for #4 is easy

i just say they share angle A

now nto letter B

scale factor is...

scale factor is 24

deal?

[minor belfry]

uhh

sorry i was checing phone

1 sec

ahhh

yea

oay

do u have a solution?

[dense tinsel]

B. x = 24

yes or no/

[minor belfry]

are you working out CD

[dense tinsel]

no im saying tha

the scale factor

is 24

right?

[minor belfry]

depends what ur working out

[dense tinsel]

i did 16/x = 4/6

cross multiplication

and then division

[minor belfry]

are we working out the area of the big triangle?

[dense tinsel]

now we are

[minor belfry]

okay

[dense tinsel]

wait a second

waiiiit a second

[minor belfry]

do u have the actual solution?

[dense tinsel]

no

but

[minor belfry]

okay

[dense tinsel]

i got 24

and no i dont know

[minor belfry]

not quite

i think its smth else

[dense tinsel]

is 24 the scale factor or the area of the entire triangle

it's gotta be the scale factor right

[minor belfry]

okay so what do we have from both triangles

and what are we missing from one triangle

[dense tinsel]

dude

can you tell me whether 24 is the scale factor

[minor belfry]

no i dont think it is

what do we have that we can compare between triangles

[alpine bison]

gmdennis i have to go, good luck for your chess career! ciao

[dense tinsel]

ciao fra

[dense tinsel]

hmm

they're both similar

well

we know that length AD = 10

[minor belfry]

aka, side length of big triangle

[dense tinsel]

and that CD and BE are parallel

[minor belfry]

what else do we have

[dense tinsel]

?

[minor belfry]

not what im looking for

a number

something to compare

we have big side length = 10

[dense tinsel]

triangle ABE = 16cm^2

[minor belfry]

okay

anything elsle

one more thing

something to compare from small triangle to big triangle

[dense tinsel]

i can't figure out the scale factor

i really don't know

[minor belfry]

thats okay

so

we know that the small triangle is length 4

right?

and the big triangle is length 10

[dense tinsel]

yes

[minor belfry]

so thats what we compare

to go from 4 -> 10

[dense tinsel]

oH WAIT

[minor belfry]

what do u do

[dense tinsel]

BOTH SIDES ARE ALSO 6 AND 4

[minor belfry]

nono, 6 is the trick number

[dense tinsel]

• 2.5?

[minor belfry]

do not use 6

[minor belfry]

yea, same as 5/2

which is 10/4 simplified

[dense tinsel]

THAT'S THE SCALE FACTOR THAT'S WHAT THE BOOK GOT

[minor belfry]

yes

[dense tinsel]

LETSSE GO

[minor belfry]

okay now

we have a length

and we want to scale an area

i.e.

cm -> cm^2

so what do we do to the scaling factor

{cm -> cm^2}

[dense tinsel]

2.5^2?

wait no

sorry

[minor belfry]

yes

ur right

absolutely correct

[dense tinsel]

2.5^2?

6.25

[minor belfry]

(5/2)^2

yes

[dense tinsel]

so 5/2 is the scale factor

and do we do with 6.25

[minor belfry]

5/2 is how u go from length on small triangle to big triangle LENGTH

and AREA is LENGTH SQUARED

so u do 5/2 squared

i.e. 6.25

[dense tinsel]

stop

stop

my brain hurts

gimme a sec

[minor belfry]

its alright

take ur time

[dense tinsel]

ok do we multiply 6.25 by 16

or divide

[minor belfry]

its tricky to get working in ur head

[minor belfry]

yes

multiply

[dense tinsel]

OH MY GOD

MY WORLD

IS CRASHING DOWN ON. ME

[minor belfry]

did it work

okay im gonna say something, and i want you to really think about it

if i have a length on a small object

and am scaling it up by a factor of 3, then i just multiply the length

if im comparing lengths, then want to find an area

[dense tinsel]

NO WAIT

I GOT ALL OF IT

[minor belfry]

then i need to multiply by the factor used by length squared

since area is length squared

[minor belfry]

bet

[dense tinsel]

so

SO

6.25 * 16

=

100

100 - 16 = 84

WHICH ANSWERS C. II

[minor belfry]

yes

[dense tinsel]

ALRIGHT

LETS GO

TWO MORE

REAL QUICK

[minor belfry]

bruh alright

[dense tinsel]

ill send pic

dude im also tired asf

[minor belfry]

alright

its all good aha

[dense tinsel]

all fucking righty lets see what we're working with

would V = 100

[minor belfry]

yes

[dense tinsel]

nope

nope

V = 270 and i have no clue why

[minor belfry]

no

u were right

wait no

u were close

not 100

close

wait no

im dumb lol

[dense tinsel]

by close you mean 170 units off but ok

[minor belfry]

its hard to read sideways

what unit is length, what unit is volum

[dense tinsel]

dude

fuck

wait

[minor belfry]

try comma rotate

[minor belfry]

,rotate

yea there

[dense tinsel]

there we go

[minor belfry]

this

[dense tinsel]

lemme draw it real quick

[minor belfry]

draw what?

[dense tinsel]

length is cm, volume is cm^3

im drawing the diagram

[minor belfry]

how do u go from cm to cm^3

[dense tinsel]

in my notes

[minor belfry]

u cube it right?

[dense tinsel]

you cube

[minor belfry]

bet

[minor belfry]

perfect