help-0

[wheat isle]

ohhhhhhhh

[rustic ruin]

and q/1 is the product

you only see the coefficients

[wheat isle]

ohhh okk

x^2 - p/1 + q/1?

[lone heart]

@wheat isle Has your question been resolved?

[lone heart]

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[vague mirage]

Kos 😅

Bro you nailed it

[jaunty crane]

no

need to prove

[vague mirage]

Cos?

[jaunty crane]

cosec x actually

oh the answer?

its cos

mb

its in malay

[vague mirage]

Ok

[jaunty crane]

lol

any clues

[vague mirage]

Just give me 2 min

[jaunty crane]

ok

[vague mirage]

First expand (cosec x - cot x)^2

[jaunty crane]

ok

[vague mirage]

${({cosec\theta-cot\theta})^2}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

done

[vague mirage]

Then write them in the place of sin and cos

Like cot = cos/sin

A d cosec = 1/sin

[jaunty crane]

oh ya

[vague mirage]

Yupp

[jaunty crane]

wait

[vague mirage]

Then show me what you got

${cosec^2\theta}+{sec^2\theta}-{2cosec\theta sec\theta}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

this?

[vague mirage]

Bro do it like this

Use the formula

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

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

oh

[vague mirage]

Yupp

[jaunty crane]

mb

wait

[vague mirage]

Yes

[jaunty crane]

aight done

cosec x2 - cot x2 + 2cosec x sec x

right

[ocean seal]

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

[vague mirage]

You will get this

[jaunty crane]

wait why + then -

[vague mirage]

Bro did you forgot the formula 😭

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

[jaunty crane]

oh

[vague mirage]

Yupp 2ab

[jaunty crane]

right

[vague mirage]

Hm

[vague mirage]

Copy this

[jaunty crane]

okk

done

[vague mirage]

Now what will you get after simplifying this

[jaunty crane]

wait

[vague mirage]

Yes

[vague mirage]

$\frac{1+cos^2\theta}{sin^2\theta}-\frac{2cos^2\theta}{sin^2\theta}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

howww

[vague mirage]

[vague mirage]

Vro 2.cosecx.cotx

Not secx

[jaunty crane]

oh mb

wait

[vague mirage]

Yes

[vague mirage]

Again this is right

[jaunty crane]

aight done

[vague mirage]

$\frac{1+cos^2\theta-2cos^2\theta}{sin^2\theta}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[vague mirage]

You will get this

[jaunty crane]

yes

[vague mirage]

What's ${sin^2\theta}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[vague mirage]

In ${cos function}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

1-cos2x?

[vague mirage]

Yess

[jaunty crane]

oh just plug in?

[vague mirage]

Yupp

Plug it in

And show me what you got

[jaunty crane]

sub what?

[vague mirage]

Good

[vague mirage]

I mean to say that Factorise
1+cos^2 x - 2cos x

${1}+{cos^2\theta}-{2}×{cos\theta}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

what?

[vague mirage]

Factor this

(a-b)^2

Use this formula

Imagine 1=1^2

[jaunty crane]

oh the ax2+bx+c?

[vague mirage]

No not the quadratic

[jaunty crane]

oh wai t

[vague mirage]

$({1}-{cos\theta})^2$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

ok done

what should i do now

[vague mirage]

Now use (a-b)(a-b)=(a-b)^2 here

[jaunty crane]

what???

[vague mirage]

$\frac{(1-cos\theta)(1-cos\theta)}{1^2-cos^2\theta}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[vague mirage]

Got it?

[jaunty crane]

oh yes

[vague mirage]

${(1-cos\theta)^2}$=$(1-cos\theta)(1-cos\theta)}$

[ocean seal]

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

[jaunty crane]

wait why is the 1 below with power

[vague mirage]

What's a^2-b^2

${a^2}-{b^2}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[jaunty crane]

oh 1

like the c2=a2-b2?

pytha

[vague mirage]

Vro it's algebra

Not right angled triangle

[jaunty crane]

right

so the formula is actually sin2a+cos2a=1'2?

[vague mirage]

😭

$\frac{(1-cos\theta)(1-cos\theta)}{(1-cos\theta)(1+cos\theta)}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[vague mirage]

😭😭😭😭😭

[jaunty crane]

ok whatttt

im so dumb

[vague mirage]

Can you solve it or do you need help in division

[jaunty crane]

this?

[vague mirage]

Yes

[jaunty crane]

just cut cut?

[vague mirage]

Yes

[jaunty crane]

i just dont understand the part

below part

[vague mirage]

Below part you mean the denominator

[jaunty crane]

ya

nvm i got did lol

[vague mirage]

I am not gonna wish Happy Birthday to my sis

[vague mirage]

Ok 1-cos^2

And 1^2-cos^2

[jaunty crane]

lol what?

shit i prob spent ur 30 mins doing my hw

🙌

mb bro

[vague mirage]

1 hrs 💀

[jaunty crane]

oh man

even longer

[vague mirage]

And I'm 14 yrs old

[jaunty crane]

a genius 14 yo

[vague mirage]

Is there any difference in between 1^2 and 1

[jaunty crane]

nope

1'2=1 yes?

[vague mirage]

Yes

So without writing 1-cos^2 we could write 1^2-cos^2

[jaunty crane]

ohhh

[vague mirage]

And we know
a^2 - b^2 = (a-b)(a+b) a=1 b=cos ok don't get confused

[jaunty crane]

oh alright yes

[vague mirage]

That's it

[jaunty crane]

oh understood

[vague mirage]

Yupp

[jaunty crane]

thx bro for spending ur 1 hr to help me 💀

appreciated

[vague mirage]

Yeah that's what she said

[jaunty crane]

💀 💀

[vague mirage]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 = 1

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[jaunty crane]

🫰

bye

.close

[lone heart]

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[vague mirage]

Bye

[lone heart]

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[alpine sable]

Hello, I need some help setting up an equation for a question

[analog falcon]

yo

whats the question

[vague mirage]

Yupp what's the equation

[analog falcon]

also, https://dontasktoask.com for future

[alpine sable]

1a) I know what the answer should be just by thinking about it, but I have no idea how to set up

[vague mirage]

Need to find the derivative

3x^2 -12x+12

[alpine sable]

It’s under 5.4 exercises not the review questions Mb

[analog falcon]

you can let one number is x, then second will be 20-x
and subsequently sum of their squares will be $x^2+(20-x)^2$. You need to minimise this sum of squares

[ocean seal]

n11

[alpine sable]

thank you so much

[analog falcon]

hmm how do we graph

oh

yeah x² will cancel out ig

[vague mirage]

,w plot 3x^2 -12x+12=y

[ocean seal]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[vague mirage]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 = 1

[ocean seal]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[alpine sable]

.close

[lone heart]

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[analog falcon]

oh yeah no x² won't cancel out

[lone heart]

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[strange fractal]

help pls!

[strange fractal]

I di this but it doesnt work

[lone heart]

@strange fractal Has your question been resolved?

[lone heart]

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[vale wigeon]

!15m

[lone heart]

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.

[vale wigeon]

you need to post your question BEFORE pinging helpers.

[empty cedar]

Just think about it for a second.

[vale wigeon]

this looks like a test, and it says "graded" at the top.

[empty cedar]

it does say 1-4 1-5 quiz

[sly mantle]

.close

[lone heart]

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[golden canyon]

what is E?

[formal nebula]

Integer part

[vague coral]

floor function

[formal nebula]

Yup

[vale wigeon]

the second equality seems easier to prove than the first if you have access to some number theory

namely if you are able to show that no perfect square is congruent to 2 mod 4 then you will have shown that floor(sqrt(4n+2)) cannot be greater than floor(sqrt(4n+1))

[formal nebula]

Wat about the first

[lone heart]

@formal nebula Has your question been resolved?

[formal nebula]

Not yet

[formal nebula]

.close

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[rough yoke]

any other methods to find the last coordinate in a square (like an actual square) when given 3 coordinates other than B-A=C-D?

[lone heart]

@rough yoke Has your question been resolved?

[lone heart]

@rough yoke Has your question been resolved?

[lone heart]

@rough yoke Has your question been resolved?

[lone heart]

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[languid cape]

.close

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[weak parcel]

<@&286206848099549185>

[lone heart]

@weak parcel Has your question been resolved?

[weak parcel]

@wind cloak yo u there bro?

.close

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[alpine sable]

How do I find all the roots of

[alpine sable]

X^3+3x^2-25x+21

[unkempt flame]

you solve the cubic equation

do you know how to solve cubic equations?

[alpine sable]

No

[ionic vine]

Rational root theorem

[alpine sable]

Uhhh

[vague mirage]

Do you know what is zeros of polynomial or rational root

[alpine sable]

Yes

[vague mirage]

Then find it

[unkempt flame]

you take the last element

the one without x

and factor it

in your case it's 21

so 21 can be 21 * 1

or -21 * -1

so it has factors -21, 21, 1 and -1

[vague mirage]

3,-3

7,-7

[unkempt flame]

as well as

yeah

those 4

so yeah

all in all

8 factors

1, 3, 7, 21, -1, -3, -7, -21

the rational root theorem says that if a cubic polynomial has a rational root it has to be one of these numbers

so you plug in each one until you get the expression to equal to 0

[vague mirage]

Try 1 first @alpine sable

${x^3}+{3x^2}-{25x}+{21}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[unkempt flame]

pluging in 1 gives us 1^3+3*1-25+21 = 1 + 3 - 25 + 21 = 0

[vague mirage]

So x-1

[unkempt flame]

which means that 1 is a root of our polynomial

[vague mirage]

Is a factor

And from now on there are 2 methods

But in this method

You have to divide

$\frac{x^3+3x^2-25x+21}{x-1}$

[ocean seal]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[vague mirage]

By long division method

Or synthetic division

[alpine sable]

Oh so you divide it to find the roots ?

[vague mirage]

We divide to factor the cubic polynomial

We have already founded a root

Which is x-1

Now we will find more 2 cause a cubic polynomial have 3 roots

And the roots means the factors

[alpine sable]

The result is x^2+4x-21

So I just factor that?

[pine kettle]

can you

looks factorable

[alpine sable]

Yeah I got -7 and 3 for roots

.close

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[warm temple]

How do you do this

[warm temple]

I set n*7^n >= 10000

But how do you actually find n manually

[vale wigeon]

trial and error

this isnt the kind of inequality that can be solved manually

7^4 = 2401 though so one would think n=4 falls short just barely

[warm temple]

Alright thanks

Thought there would be more to this than just trial and error

.close

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[lone heart]

@pearl relic Has your question been resolved?

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[south jasper]

This is what I did

,rccs

[fierce prairie]

,rotate

[small silo]

The simplification youve done between the second and third lines are wrong

[south jasper]

what?

isnt that how you do it

[small silo]

No, take what you did with the x^2 terms:

X^2 .... = ... -X^2

Should become

2X^2 ... = ...

[south jasper]

what i did distributive propet

[small silo]

That part is correct, I meant the step right after that

[south jasper]

the -32 6x-16?

[small silo]

Yeah, it should be 2x^2 - 14x - 16 = 0 at that step

[south jasper]

x^2 plus x^2 is 2x^2?

[small silo]

Yes

[south jasper]

ok

now what

[small silo]

Solve that equation. It will give you the points the x-coordinates they intersect at.

[south jasper]

ye how do you find y tho

and how do you solve that equation

[small silo]

Once you have the x values, substitute them into either of the equations.

Because the two curves intersect, it means they have the same x ans y coordinates at those points. So either curve works for finding the y value.

Make sense?

[small silo]

Its a quadratic equation. Do you know the quadratic formula?

[south jasper]

no

is it like x= a+ or - sqr root b / x

[ocean seal]

custard

[small silo]

I meant the squareroot to cover that whole bracket. Is that familiar?

[lone heart]

@south jasper Has your question been resolved?

[lone heart]

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[ornate condor]

..

[sour dove]

bruh

[ornate condor]

use a calculator

[high rapids]

<@&268886789983436800>

We are not stupid

[night geyser]

.close

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[ornate condor]

[lone heart]

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[alpine sable]

Hi guys good evening. How can I test if a vectorial space Is linear?

[alpine sable]

For example:

Here the text ask: which of below application within vectorial space are linear?
A. true
B. False
C. True
D. True

I know that the two properties for a Linear transformation are
T(u+v)=T(u)+T(v)
and
T(alphau)=alphaT(u)
But i don't if i have to apply them and, if yes, how I have to apply

[ionic osprey]

You just put u+v into the definition of the map T that you are given and see if the result equals T(u) + T(v). If that's true then you do the same for alpha*u

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

can somone explain

shouldn't it be the oppoisite

ik its a stupid question but i cant wrap my head around it

[hard patio]

Do you know what those signs means?

[marsh rapids]

It's like $<$ and $\le$

[alpine sable]

yes

[ocean seal]

themateo713

[alpine sable]

just give me a minute

[marsh rapids]

One is strict and one isn't

The only difference is whether it includes the case where the two are equal

[high rapids]

Yeah for 1v) it can be both A=B and A<B

[alpine sable]

thank you guys @high rapids @marsh rapids @hard patio

ill figure it out

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[alpine sable]

Hello @ionic osprey thank you fro answer, but what do you mean with "Definition of the map"?

[lone heart]

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[frozen compass]

Hi! I’m a little confused on how to approach this problem

[frozen compass]

Haven’t really learnt work and rate relationships

[high rapids]

$$1/(x+20) + 1/x = 1/24$$ I think

[ocean seal]

Diesel

[frozen compass]

Oh interesting

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[livid torrent]

I wish I has nabbed this channel. Zero is my lucky number. Ach. Anyways, best wishes and regards to whoever gets my favorite help channel next. Lol.

[lone heart]

Please don't occupy multiple help channels.

[livid torrent]

I'm sorry I did NOT know it would do that

I am just a friendly individual fuck >___>

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[gaunt harness]

ok this is really just a simple question but why is the derivative of 4e^t itself

[gaunt harness]

like I just need to know why cus it just dont make fun sense

[lucid barn]

hello, have you see Maclaurin series ? it can help understand it

[gaunt harness]

no I havent

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[prime dragon]

Hi

[prime dragon]

Could someone help me with this question 🙋‍♂️

[placid zinc]

https://unacademy.com/content/ncr-formula/

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[austere compass]

Is $\mathcal{L} := {x \in \mathbb{R} \ | \ x< -1 \lor -1 < x < \frac{5}{3} \lor x > 3}$ correct notation wise?

[ocean seal]

Levens

[austere compass]

Wait I have a different question now, ignore the one above

How do I find the solution set of $|x+3a| - |x-a| = 2a$, where $a \in \mathbb{R}$

[ocean seal]

Levens

[austere compass]

<@&286206848099549185>

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[wind cloak]

💀 sorry

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[proud viper]

Not sure where to even begin here

[heavy cloak]

You have a isocele triangle

So sv and tv must be equal

[proud viper]

So how would I go about solving for X?

[heavy cloak]

Lenght of sv = lenght of tv

2x + 6 = 3x - 6

[proud viper]

So X = 12?

[heavy cloak]

I guess

[proud viper]

Thanks

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[midnight flicker]

hi

[lone heart]

Please don't occupy multiple help channels.

[midnight flicker]

I need some help with part f of a question

.clos3

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[wind bloom]

how do i know which values to plug in?

[wind bloom]

like why did my teacher use these specific values

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[lusty knoll]

Hi

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[brittle sun]

Hey fellas,
I'm trying to solve the following equation
3 <= x^2 < 4
I'm not very fluent on the technique of how to prove this properly. Meaning, I can come up with the answers just fine, but I don't know how to explain it. Any help with the proof would be much appreciated

[golden canyon]

You should probably split it up into 2 cases: 1 where x is positive, 1 where x is negative

then you can take the sqrt, if x is negative sqrt(x^2) = -x

[brittle sun]

Oh! That makes sense

[brittle sun]

So quite literally go:
"If x >= 0: ..."
"If x < 0: ...", correct?

[golden canyon]

yes

[brittle sun]

Seems pretty obvious in hindsight haha

Cheers mate, that'd be all

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[desert tusk]

I have no idea where to start even with the hint

[vale wigeon]

cosine law

and for b, consider the area of ABC in two different ways

[desert tusk]

Do I use the cosine law like this then?

I think my problem is how I am going to use the info given (sin angle BAC = 3/5) to use that in cosine law.

Is there any hints you can give?

[vale wigeon]

"cos(3/5)" bad

your angle is not 3/5 radians...

[desert tusk]

I forgot to mention

my angle is supposed to be in degrees

not radians

since we don’t use radians in my syllabus yet

[vale wigeon]

yes but 3/5 is still not its measure in any unit.

it is its sine.

and writing "cos(3/5)" to refer to "the cosine of the angle whose sine is 3/5" is wrong.

BC^2 = 12^2 + 7^2 - 2 * 12 * 7 * cos(BAC).

[desert tusk]

so do I have to find angle BAC itself?

[vale wigeon]

no

[desert tusk]

because the question specified I can’t do so

[vale wigeon]

you do not need to find BAC itself

but you can and should find cos(BAC) from sin(BAC)

knowing BAC is obtuse

[desert tusk]

so it’s more than 90 degrees and less than 180 dehrees

[vale wigeon]

and knowing a particular trigonometric law, the most basic of them all

[desert tusk]

it doesn’t involve the tangent law, right?

where tan theta = sin theta / cos theta

or cosine theta = adjacent / hypotenuse

[vale wigeon]

no you are overthinking it

what is the most basic trigonometric fact

the most basic trig identity

[desert tusk]

it involves right angled triangles only?

[vale wigeon]

cos^2(x) + sin^2(x) = 1.

this is what i am talking about.

[desert tusk]

I swear to god, that is never in my previous textbook

[vale wigeon]

then your textbook is shit and should be burned

[desert tusk]

Let me double check one more time

Because I seriously don’t know that

[vale wigeon]

shame!!!

[desert tusk]

yeah no it’s not here

This is what it all had

in my previous syllabus

this education system is fucked up

[desert tusk]

but iirc, this is part of something called trigonometry identity?

[vale wigeon]

trigonometric identity

yes

[desert tusk]

Is there a lot of assimilate about it

[vale wigeon]

what

[desert tusk]

i mean, is there a lot to learn about it

Or it’s very short and simple

just wondering

[vale wigeon]

i think you may have misused the word "assimilate"

there are a number of trigonometric identities worth knowing

i would not call it short and simple

[desert tusk]

well in the BrE it said to learn something and make it stay in your mind so you can use it daily

but never mind

[vale wigeon]

"in the BrE"?

what

[desert tusk]

British Enlighs

Or AmE, American English

[vale wigeon]

are you looking at some dictionary or something

[desert tusk]

that’s one of the English words’ definition I’ve recorded in a small notebook

form Cambridge dictionary

[vale wigeon]

hm.

in my experience this is kind of uncommon

[desert tusk]

I found this dictionary pretty good at explaining things than Google’s shitty explanation

[vale wigeon]

"assimilate" as i've seen it used usually means "become similar (to something)"

[desert tusk]

well we have our own different ways to learn English

but whatever, I’ll see more about trig

Trig identities

now what the hell is a cosecant or secant, not even in my textbook too

never mind

the inverse of cosine

[vale wigeon]

reciprocal

tbh you can always replace sec and csc with 1/cos and 1/sin respectively and nothing will change

[desert tusk]

what’s with these funny names I have to learn now

so cosine and cosecant are the same thing I suppose

What’s the difference between inverse and reciprocal?

[vale wigeon]

no they are not

[desert tusk]

I meant arcsine for cosine

[vale wigeon]

and still no

not even if you meant arccosine

arccos(x) is the angle whose cosine is x

sec(x) is 1/cos(x)

[desert tusk]

so they are inverse functions to each other?

[desert tusk]

and secant is the reciprocal of cosine?

[vale wigeon]

yes

[desert tusk]

I think I’m very close to the answer

Using the trig identity you mentioned, I got cos angle BAC = 4/5, which is wrong because the angle BAC is obtuse, so it has to be a bigger value, right?

[vale wigeon]

no

in fact quite the opposite

for angles between 0 and 180° cos is a DECREASING function, and in fact the cosine of an obtuse angle is NEGATIVE

[desert tusk]

So if it’s an obtuse angle, I’d have to change the 4/5 to negative?

[vale wigeon]

it is precisely BECAUSE angle BAC is obtuse that, when solving for cos(BAC) and getting that it is either 4/5 or -4/5, you TAKE the negative solution.

you do not CHANGE anything.

[desert tusk]

oh great I forgot to put the plus and minus sign when square root a number

FINALLY

i found the answer

i should be able to do the rest now

thanks

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[ember glacier]

Hi

[languid bolt]

Hi

[ember glacier]

How did this person determine that 2.) was in quadrant 4

[languid bolt]

are you asking how -pi/6 is in quadrant 4 or how did he get there?

[ember glacier]

How he got there

Both

[languid bolt]

ok so

tan is negative only in the quadrant 2 and 4

[ember glacier]

ya

[languid bolt]

quantumbee can probably explain better, ima just leave

wait he's gone now

[ember glacier]

What if quantum bee leave

[languid bolt]

bruh

[ember glacier]

🐝 ⬅️🪹

Yooo

[languid bolt]

they just solved for theta

[ember glacier]

Ya but how did they determine it was in quadrant 4 instead of quadrant 2

Oh my b ur typing

[kindred anchor]

$\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)=\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{-\pi}{6}$

[ocean seal]

QuantumBee

[kindred anchor]

$\frac{-\pi}{6}$ lies in 4th quadrant

[ocean seal]

QuantumBee

[ember glacier]

Why does tan^-1(-radical3/3) equal tan^-1(-1/radical3)?

[kindred anchor]

What when you multiply with √3 so you can get 3?

[ember glacier]

Oh you multiply both the num and dem by 3?

[languid bolt]

,,\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}\sqrt{3}} = \frac{1}{\sqrt{3}}

[ocean seal]

help

[ember glacier]

When do you know when to do that

[languid bolt]

like this

practice i guess

[ember glacier]

Oh but I mean specifically why did we multiply both the num and dem by radical 3

Cancel out I mean*

[languid bolt]

it's not really necessary... it just helps when evaluating trig functions

you can do it without cancelling it out if you have calculator

[ember glacier]

Proffesor doesn’t allow calculators ☹️

But actually I got some progress out of this

Thanks for help

[languid bolt]

you're welcome

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[alpine sable]

If the traffic is still heavy at that moment, we will not go to the cinema this evening at 4.00pm. It is 4.00pm now and the traffic is still heavy, therefore we will not go to the cinema
In the 100m competition, the existing record can only be broken by either RunnerA or RunnerB. RunnerA and RunnerB did not win the race, thus the record was not broken.
November has always been the wettest month in US from 2000 to 2021, except for 2016 and 2018 where unexpectedly January and March were respectively the wettest. Therefore, it is logical to assume that November 2022 will also be the wettest month for the year 2022.
All tigers are protected species as defined by WWF. The tiger and the West African lion are tigers. Therefore the tiger and West African lion are protected species.

Which is deductive or inductive argument and it is valid, sound, strong or cogent?

[quiet gyro]

hi,there

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@alpine sable Has your question been resolved?

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[alpine sable]

.reopen

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✅

[alpine sable]

If the traffic is still heavy at that moment, we will not go to the cinema this evening at 4.00pm. It is 4.00pm now and the traffic is still heavy, therefore we will not go to the cinema
In the 100m competition, the existing record can only be broken by either RunnerA or RunnerB. RunnerA and RunnerB did not win the race, thus the record was not broken.
November has always been the wettest month in US from 2000 to 2021, except for 2016 and 2018 where unexpectedly January and March were respectively the wettest. Therefore, it is logical to assume that November 2022 will also be the wettest month for the year 2022.
All tigers are protected species as defined by WWF. The tiger and the West African lion are tigers. Therefore the tiger and West African lion are protected species.

Which is deductive or inductive argument and it is valid, sound, strong or cogent?

[empty cedar]

do you know the difference between deductive and inductive argument

[alpine sable]

nope

[empty cedar]

same

basically deductive reasoning goes like this

Apples are red. My shorts are red. So therefore, shorts are apples.

inductive reasoning is like this

Fruits are apples. All fruits are red. So therefore, apples are red.

[alpine sable]

Hmm idk correct or not

[empty cedar]

this my next topic so i dont know shit yet

but soon in around 7 weeks i will

so just wait around 2 months for a reply

[alpine sable]

brub XD

[empty cedar]

go onto google theres a lot of examples there

i mean its quite obvious that you cant just predict the weather based on past events

[alpine sable]

Do you think this topic is difficult?

[empty cedar]

idk give me 7-8 weeks

[civic delta]

hi

hi

hi

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[alpine sable]

8x = 2.4x

[alpine sable]

um

what is...the question?

i wasn't asking a question

there was a question here which got deleted, the guy came in discussion and asked for help so it got deleted

that was a hint, on completing the square

so why are you posting this again?

oh i see

i understand now

no no, the question was still there when i said that

this isn't your channel, sorry for the confusion

np

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[vale remnant]

Hello!

[vale remnant]

I have messed up here

But I cannot find where I went wrong

[earnest mortar]

(5y)² is 25y² not 5y²

(See second line)

[vale remnant]

Oh my

How did I

Tysm for your help, it’s appreciated!

.close

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[rigid isle]

answer pls

[mortal trellis]

what have you tried

[rigid isle]

nothing

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[boreal verge]

that looks like hell

i guess the only way to approach this would either be with ln x^y = ylnx and take it in front of the fifst integral since its differentiating in terms of x

then just do the 2nd one with the result of the first

or, if you know the formula, merge the 2 integrals into 1

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[chilly dove]

can anybody help me on this?

[lone heart]

@chilly dove Has your question been resolved?

[chilly dove]

<@&286206848099549185>

[prime badge]

you divide it by 2

it becomes the sum of dimensions

[chilly dove]

yea

[prime badge]

then you choose whatever

any number

[chilly dove]

i dont really understand tho

[prime badge]

i guess you can't choose any number, it's only safe to choose something with x or y

e.g. x

or 2x+2.5y

[chilly dove]

so 2x+2.5y could work?

[prime badge]

that would be one of the dimensions, you set the other one so it adds up to half

oh the other is x in that case, coincidentally

[chilly dove]

2x+2.5y and 4x+2.5y works right

[prime badge]

nope

[chilly dove]

i dont get it then

[prime badge]

ok

[chilly dove]

.close

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[chilly dove]

.reopen

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✅

[chilly dove]

can anybody help me on this? i dont really understand

[lone heart]

@chilly dove Has your question been resolved?

[lone heart]

@chilly dove Has your question been resolved?

[alpine sable]

I am not exactly sure what the question is asking for but I think u need to divide by 2

after u divide by 2 u will get the length of the parallel sides

3x + 2.5y

then u can write it as 3x * 2.5y or 2.5y * 3x

not sure tho

[chilly dove]

oh ok

sorry for the late response

[chilly dove]

could i just write it as 7.5xy? again sorry for the late response 😅

[copper rock]

So actually what sLoct answered isn’t quite right.
Perimeter of a rectangle is computed like this:
P = 2W + 2H where W and H are the width and height
so 6x + 5y = 2W + 2H
or 3x +2.5y = W+ H
now there’s an infinite number of solutions here, the most obvious one is when W = 3 x and H = 2.5y (which is what sLoct mentioned)
You could also have for instance W = x + 1.5y and H = 2x + y

[chilly dove]

ohhh okay thank you very much for explaining to me well, your the one by far to help me the most on this question 🙂

[chilly dove]

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[chilly dove]

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✅

[chilly dove]

How can I find an algebraic equation for A?

[tacit arch]

21 = 3 * 7 and 14 = 2 * 7

[chilly dove]

my bad i forgot to put the B part for this question and i meant to ask B, how can i do B?

[tacit arch]

then use 3n = 3 * n and 2n = 2 * n

[chilly dove]

wdym by 3n = 3 * n tho

i dont really get it

[tacit arch]

do you understand this?

[tacit arch]

here n is the number in the top left corner

[chilly dove]

oh ok

[chilly dove]

sorry for the late response but what can i do for c?

[tacit arch]

The first row is then m-1, m, and m+1

[lone heart]

@chilly dove Has your question been resolved?

[floral sail]

@chilly dove

@tacit arch

is this what you were going for?

[floral sail]

it'll be a shame if you didn't need the answer anymore

[chilly dove]

nah ill be sure to read that, tysm for ur time for writing that

i really appreciate it

[floral sail]

well, tell me if it solves your problem! 🙂

[chilly dove]

alright

i have to go to bed now tho ill talk to you tmr

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@chilly dove Has your question been resolved?

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[neon smelt]

i tried solving it myself but my answer was way different from the answer key,

[vague coral]

and what did you get

[neon smelt]

this is the answer key

[vague coral]

but you arent done yet...

you can simplify this

[neon smelt]

for the first term in the bracket how do simplify the denominator

[vague coral]

$\left(\frac{-k}{1+\frac{L}{x}}\right) \left(\frac{-L}{x^2}\right) = \frac{kL}{x^2 + xL} = \frac{kL}{x(x+L)}$

[ocean seal]

Ｈｅｒｅｌｓ

[vague coral]

dont you say you dont know how to multiply two fractions

[neon smelt]

dont i have to make it not a fraction first

for the L/x

[vague coral]

what ?

[neon smelt]

so i can just multiply L/x by x^2?

[vague coral]

$x^2 \left(1+\frac{L}{x} \right) = x^2 + Lx$

simple

[ocean seal]

Ｈｅｒｅｌｓ

[vague coral]

what do you want to do ?

[neon smelt]

i wanted to like make the L/x into either only L or only X

[vague coral]

wdym

you are overthinking this simple calculation

[neon smelt]

like since its a fraction in a fraction

yeah maaaybe

i got it now tho

[vague coral]

not maybe, you are

[vague coral]

thats fine

[neon smelt]

yeah i got it

thxxx

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[vague coral]

np

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[wraith sun]

any hint to solve this question¿?

[lament glen]

multiply the first equation by -3 so the LHS on both are equall

[wraith sun]

yeh but after that eq 2 become zero

0=k

[lament glen]

?

no you get k - 3h = 0

[wraith sun]

Ohhh

[wraith sun]

Then?

[lament glen]

wait not k - 3h = 0

k + 3h = 0

mb

then you move 3h to the other side

k = -3h

[wraith sun]

yeh
but what is the values of h and k

[lament glen]

nothing

h can be anything, and k = -3h

[wraith sun]

ahh alright!!!

Tysm 😊

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[strong bane]

Hi

[strong bane]

C(M,x) = (Ox, Txy, Txz)
But also :
C(M,x) = [O(M)].x
I don't understand how to read a matrix, do you read by row or column?
Because for me it is :
[O(M)].x = (Ox, Tyx, Tzx)
Not (Ox, Txy, Txz)

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[grave trellis]

hello

[grave trellis]

A Meteorite is moving at height of 5.0 x 10^3 km above earth’s surface. What is the value of the field strength g at this height?

i've tried using the f = GMm/(r^2) formula

but the only variable i have is the r

[marsh rapids]

g = G M_earth / r^2 at any point

both G and M_earth should be known

[marsh rapids]

because the mass of the other object cancels out in the computations

since a = f/m

[grave trellis]

hmm so: g = (6.67 * 10^-11 )* (5.972 × 10^24 kg)/(5000 km)^2?

the value of M being the mass of the earth from google

[marsh rapids]

units matter

so no km

[grave trellis]

meters?

[marsh rapids]

yes

[grave trellis]

g = (6.67 * 10^-11 )* (5.972 × 10^24 kg)/(5000 000m)^2?

[marsh rapids]

yes

I'd expect something like 2.4 m/s² at that height off the top of my head

,w 6.67E-11 * 6E24 / (11.371E6)²

[grave trellis]

yeah the answer is about there

[marsh rapids]

,w radius of the earth

[ocean seal]

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[marsh rapids]

about 2.4 yeah

makes sense. I rounded the wrong way round

so it should be a bit more than g/4

since we did a bit less than double r

[grave trellis]

so i should add the radius of the earth as well

?

[marsh rapids]

to get the distance to the center of the earth yes

[grave trellis]

mmmm i see

yuppp i got the answer

thank you so much <3

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@candid comet Has your question been resolved?

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[calm perch]

Hey guys im having troubles to get the result, and the f'(x) to ? can u guys help me?

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[placid spire]

How to find f'(1)?

this looks complicated

first I have to find f'(x) obviously

[boreal verge]

whats når

is it if

[placid spire]

when

[boreal verge]

oh oki

[placid spire]

find f'(1) when

I have the solution but i dont understand it

They first use the chain rule, idk why

but then they use fundamental theorem next

why do they take the x-x^2 and put that as u

is that the chain rule???

idk

[boreal verge]

tbh idk either

i guess they just evaluate -cos²t as if it wasnt out of the integral and just the normal function for -0.5 and u

idk why the x-x² is there tho

[placid spire]

They chose x-x^2 as the core

and then they multiplied with the core derivated

(x-x^2)'

chain rule

[boreal verge]

im not qualified enough to really help

oh

[placid spire]

what rule is this

[velvet pelican]

leibniz integral rule

or like differentiating an integral

[boreal verge]

yeah i had no idea that existed wow

[placid spire]

how come I havent learned this

[velvet pelican]

i mean the concept is that the antiderivative of f(t) is F(t), then the integral is F(v(x)) - F(u(x)).
Differentiating this gives f(v(x))v'(x) - f(u(x))u'(x)

[placid spire]

is this the fundamental theorem?

[velvet pelican]

umm i havent heard of this being called that

[placid spire]

I got this in my formula papers

[placid spire]

it says u(x) and v(x)

but in my example we have -0.5

Theres no x

[velvet pelican]

u(x) = -0.5 then whats du/dx?

[placid spire]

0?

[velvet pelican]

yea

so that term vanishes

[placid spire]

but in my case

what happens with -cos^2

[velvet pelican]

f = -cos^2

[placid spire]

whats my derivative tho

f(v(x)dv/dx - 0?

and v(x) is x-x^2

f(x-x^2) dv/dx?

-cos^2 * (x-x^2) derivated?

[velvet pelican]

-cos^2(x-x^2) * d/dx (x-x^2)

yea

[placid spire]

but its actually -cos^2 * (x-x^2)dv/dx - (-cos^2 * (-0.5))du/dx?

but the right side becomes 0?

[velvet pelican]

du/dx = 0

so yea

that term becomes 0

[placid spire]

but what does f(u(x)) mean?

when u(x) is 0.5

[velvet pelican]

f(0.5)

[placid spire]

so that means -cos^2 * 0.5?

[velvet pelican]

-cos^2(0.5) yea

since your function is -cos^2(x) or whatever

[placid spire]

yea

and the derivative of that is 0

hm okay

[velvet pelican]

yes

[placid spire]

I feel like I need to calculate the derivative of that

using chain rule

but its possible to see it right away?

[velvet pelican]

derivative of what

[placid spire]

-cos^2(0.5)

[hearty valve]

Hi can i ask question here?

I got a problem on Riemann Sum

[velvet pelican]

its constant tho

so its derivative is 0

[velvet pelican]

use an empty channel

[placid spire]

oh ye

do you understand what he did here tho?

using chain rule

[velvet pelican]

yea they used u = x-x^2

then f'(x) = df/du * du/dx

by the chain rule

[placid spire]

hm

its not the same strategy u showed?

[velvet pelican]

it is tho

just another representation of it

[placid spire]

so whenever I see an integral like that

I can just find the derivative by taking the number at top and call it u?

like use the chain rule like that

[velvet pelican]

yea

[placid spire]

no matter what the limits are?

[velvet pelican]

like

[placid spire]

or does the bottom limit have to be a constant without x?

[velvet pelican]

i'd rather use this

[placid spire]

like -0.5

[velvet pelican]

in this case yea

otherwise you'd have another term

just like the rule i showed you

[placid spire]

hm okay

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[abstract shuttle]

Het