help-13

[ember heath]

or sin^3 x / cos^2 x

[dire geode]

yes can you see this now

[ember heath]

yeah..do i divide denominator by sin^3x?

[glass sky]

why will you do that tho?

you already have the sin cube in numerator just go w this u sub no?

[ember heath]

so that i can get some cos^2 x/ sin^3 x terms which i can write as 1/du

[glass sky]

why do you need 1/du in the integral?

we are trying to replace dx with du na?

[ember heath]

idk bro i was just trying to get u and du somewhere 😭

[ember heath]

but what about the cos ^2 x

[glass sky]

you can multiply it in numerator and denominator , and see where it gets you ( or you can recognise something regarding this in the denominator)

[ember heath]

can i take it out from the bracket

in denominator

[glass sky]

you can 🤷‍♀️give it a shot

do you have options for this na?

[ember heath]

nope

[glass sky]

or is this subjective

[ember heath]

just a worksheet my teacher gave

[ember heath]

sounds like u dont want me to do that

what do i regocnize

[glass sky]

one of your thought process could be like
ah usually options helps a lot , but its fine , you can think of this in this way , how will i integrate tan inverse which is in the denominator , only if i could sub arc tan as t

[glass sky]

i do want you to do that , go with your gut while integrating

[ember heath]

hey its working out

that bracket becomes u^2+1

[glass sky]

yep.

[glass sky]

this stratement makes more sense if you know the answer , i just realised

[glass sky]

but yeah exactly can you see it now

[ember heath]

but i still have kind of a mess
du/(u^2+1)arc tan u

[glass sky]

and.. whats the derivative of arctan u

[ember heath]

1/1+u^2

[glass sky]

bingo?

t = arctan u
dt = 1/(1+u^2) du

[ember heath]

do i make another sub?

[glass sky]

yeah you can technically

[ember heath]

oh thats cool

[glass sky]

or you could make one sub only which was t = arctan(secx + cosx) directly

but yeah the former is safer aproach

[ember heath]

i cant even check the answer

[glass sky]

its ok what did you get

[ember heath]

but is it
ln|arc tan (sec x+cos x)|

+C

[glass sky]

yes

good1

[ember heath]

oh ok thanks a lot

[glass sky]

rieman was the mvp on the u sub

[ember heath]

yeah @dire geode could u tell me why u took that for the u sub

[glass sky]

i was goingf to suggest t = arctan (secx + cosx ) but that method made it more easier

[ember heath]

and also im kinda new to integration..u have any tips?

[glass sky]

uhh maybe do the ncert integration thoroughly ( monthly for revision) , it builded my intuition on this
so just 7th chapter

[dire geode]

cos^4 + 3cos^2 + 1 looked like it related to (cos + 1/cos)^2

[ember heath]

oh ok

[dire geode]

but also another usual thing to do when you have a complicated composition like
arctan(f(x)) then just try u=f(x)

[ember heath]

ohh alright

good to know

thanks guys

.close

[cedar kiln]

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[cloud sigil]

can anyone help?

[oblique flare]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[solar juniper]

cant you just write out the components of each vector

and then like calculate it directly

[cloud sigil]

this is how the question goes.
no given components

[blissful glade]

assume vector A is structured as a vector of (a,b,c) and so on for B and C

[dire geode]

You make them up as variables

[stiff saffron]

hello I Want to learn multi variable in calculus can guys tell me a plan

[cloud sigil]

.close

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[velvet gust]

Hellk

[velvet gust]

Help

[lone dune]

go ahead

[velvet gust]

Idg

See, my tutor doesn5 k ow how to teach and he has us homework and we havent go this far in log

Idg the question

I've tried using the addition law and subtraction law eve power law

I'm not even sure

Pls help

He give us simple examples but hard classwork and homework so that's y idg

[sullen cipher]

Alright so do u know that if two logarithm functions with same base r in addition form ,the values r multiplied

[velvet gust]

Yep

[sullen cipher]

Waiit a sec

[velvet gust]

I understand the subtraction law and power law

[sullen cipher]

Alr

So same for two logarithm functions being in subtraction form,they xan be written in divided form

[sullen cipher]

Alr

Can u work on 2 no then

Basing on the subtraction law

[velvet gust]

Ok

Idg no.1,4 and 5 but I will try

[sullen cipher]

Actually what's the Q no 1

It's a bit messy

[velvet gust]

Its at the bottom

Srry

For me no.3 is 100

[sullen cipher]

Oh alr

[sullen cipher]

Can u tell how u got that

[velvet gust]

Is it correct

Well

The 2 at the back of log5

Is the power/index

[sullen cipher]

2?

[velvet gust]

And 5 is the no

Yea check

No.3

[sullen cipher]

Yea got u

[velvet gust]

Thxs

But no.4 and 5 r tricky

I know that there in base 10 tho

[sullen cipher]

Wait

I meamnt i got it that it's 3 no

[velvet gust]

What's happened

[sullen cipher]

But i dont get how u got 100

5² gives u 25

And u have log30 in added form and log9 is subtracted form

So u r supposed to convert log 25 and log30 into multiplied form and then log 9 in divided form

[velvet gust]

Wdym

yea

[sullen cipher]

Like tell me what's log25+log30

[velvet gust]

That's 25 × 30

[sullen cipher]

And log(25×30) -log9?

[velvet gust]

Where did u find log 30

Oh bro it's 36

[sullen cipher]

Oops

Alr 36 then

Now log100 makes sense

Sorry

But dont forget the log

[velvet gust]

Ok

[sullen cipher]

Let's do 4

[velvet gust]

Sure

That's what Im doing

Im i getting it

Imma resend

[sullen cipher]

Oh no ,u cant do that

Log125-log5=log(125/5)

Not log125/log25

Lemme give u a hint
Use the power rule here

[velvet gust]

Ohn, thats how he gave us

[velvet gust]

Alr

[sullen cipher]

U mean That's how he taught u?
Or like that was the question he gave u

[velvet gust]

That's how he gave us the question

[sullen cipher]

The question is alright

[velvet gust]

Yep

[sullen cipher]

So did u figure it out

[velvet gust]

Let me show u one of the question that involved that

I'm not sure it

[sullen cipher]

Do u know what's 5³

Ping me once u r back

[velvet gust]

Ere

[velvet gust]

thats 125

[sullen cipher]

,rccw

[velvet gust]

@velvet gust brb

[velvet gust]

Brb

[sullen cipher]

Exactly

So u can write 125 as 5³ and proceed

[sullen cipher]

Donr put the equals here

Put them at front

And whatver was done here seems correct

[sullen cipher]

But what u did here isnt right

[sullen cipher]

Read this part once more

[velvet gust]

No, this is the correction to the classwork

[sullen cipher]

Which is a correction to which one

[velvet gust]

Is this correct for no.4,

[velvet gust]

This is one is correction

The rest r homework

[sullen cipher]

No

It isnt

[velvet gust]

Ohn

Where did I make error

[sullen cipher]

Like i said log125 -log 2
5=log125/5

[velvet gust]

No bro, it's 5 not 25

[sullen cipher]

But u xant right log 125-log 5=log125/log5

[velvet gust]

Sorry if my hand writing is some how

[sullen cipher]

Nah dw
Use power rule

5³=125

[velvet gust]

Yea. Ik

Alr

Thats what i did

[sullen cipher]

But that's what u did,so it's wrong

[velvet gust]

I true 125 to 5^3 and 5 just like that

[velvet gust]

Ohn

From the part where i did 5^3 is wrong right

But how

[sullen cipher]

Remember log5³=3log5

[velvet gust]

Yea

[sullen cipher]

U can take the power to coefficient

[sullen cipher]

So log5 cancels out

[velvet gust]

Explain again

I turn log ^5 10 to 1

Ok

[sullen cipher]

Log125/log5=log5³/log5=3log5/log5

This is wrong

[velvet gust]

So I would just cancel out

[sullen cipher]

So whatever u do after that will automatically be wrong

[velvet gust]

Ok

[sullen cipher]

,rccw

[sullen cipher]

Check this

[velvet gust]

Is it correct

[sullen cipher]

Yes

this is right

[velvet gust]

Thxs

[sullen cipher]

Try 5 too

[velvet gust]

What

[sullen cipher]

Same way

[velvet gust]

WAIT

ok ill try 5

[sullen cipher]

What did u not understand

[velvet gust]

This is subtraction law for us

But first check this

[sullen cipher]

Wow i m blind

[velvet gust]

I'll resend

[sullen cipher]

Nah i got u

My eyes r too powerful

Jk

[sullen cipher]

Yes i wrote the same thing

Js pointed out the mistake u did

Log(a/b)=loga/logb

[velvet gust]

Yea right

[velvet gust]

What

There the same thing

[sullen cipher]

No they r not

If u r allowed to use calc u can calculate

Log(5/2) is not equal to log5/log 2

[velvet gust]

?

Ok the fraction 5/2 which is the base in log

Is not equals to log5/log2

Thxs vic

[sullen cipher]

Yup

Let's try 5

We follow the same procedure herr

[velvet gust]

Brb

[cedar kiln]

@velvet gust Has your question been resolved?

[sullen cipher]

Ping me once ur back

[cedar kiln]

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[tranquil oracle]

if you want to LaTeX, dollar signs around the math

[wraith dagger]

brandon

[tranquil oracle]

ah okay the "stones on chessboard" just makes the problem a bit more complex than necessary

Basically once you pick d, the chessboard is determined by whether $x^3-y^2$ is divisible by d

[wraith dagger]

Element118

[tranquil oracle]

essentially, it's asking for the number of solutions to $x^3-y^2\equiv0\pmod{d}$. You can compute locally (for each prime power) and combine.

[wraith dagger]

Element118

[tranquil oracle]

The prime factorisation would help because you can combine local solutions for each prime power into a global solution for d (remember chinese remainder theorem?).

[wraith dagger]

brandon

[tranquil oracle]

yeah you'll have to consider the prime power factors

might need to start with primes and do hensel lifting

primes is really easy once you know the structure of multiplication mod p

[wraith dagger]

brandon

[tranquil oracle]

to count solutions you use CRT

[cedar kiln]

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[knotty cosmos]

could i get help with this question? im having a little trouble
23. The subscription prices to a club for men and women are in a ration 4:3. Two men and 5 women pay a total sum of $460. What is the subscription fee for each man?

[knotty cosmos]

OH nvm i got the answer mb

😭

.close

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[idle marten]

I need help verifying I don't understand how to start this.

[silk gust]

what's 2pi?

in degrees ig

[idle marten]

I finally moved on from simplifying

lol

im on verifying

[silk gust]

:D

[vital jay]

Do you know many trig identities about sin and cos?

[idle marten]

this is an example they gave

I know some

[vital jay]

Well the quotient identity from that example would probably be useful

[silk gust]

@vital jay ||correct me if im wrong, but arent the angles inside the tangents the same value?||

which is why they're equal

[flint cape]

yes*

*no

[idle marten]

since it says verify

does that mean

sometimes

theyre not correct?

[vital jay]

Kinda? Depends on the approach you're allowed

[iron wren]

Whats the question

[idle marten]

.

[vital jay]

Nah by verify they mean show that it holds in general

[iron wren]

Any angle +2pi is the same output through trig functions

[flint cape]

"verify that I have two apples on me" doesn't mean "there is a probability that I might currently NOT have two apples on me"

It means I need you to check

[vital jay]

Like |x| = x isn't true because even though |5| = 5, |-5| =/= -5

Show that it's always true for all values of x

[flint cape]

Why is it that any maths concept I keep going back to apples to explain them

I might consider using the addition formula for tangent:

[vital jay]

Which is what they do here, by showing that tan(x)csc(x) = sec(x) regardless of what x is

[idle marten]

so how would I verify this

[flint cape]

5th one here:

wait no

[vital jay]

Or 6th

[flint cape]

6th one

That's a subtraction

[vital jay]

Yeah lol

[silk gust]

alternatively, the arguments inside the parenthesees are equal so

[idle marten]

these?

[vital jay]

Yup!

[flint cape]

True-ish; but if I have to describe e.g. turns on a pump, I'd need to be careful

[vital jay]

So long as you know what tan(2pi) equals

[idle marten]

oh

2pi is

360

degrees

so uhh

1/0?

isnt that undefined

wait

no

[vital jay]

Actually it's just 0

[flint cape]

Because turning something 5 whole turns isn't the same as not turning it at all, even if the end result is that the thing's orientation is the same, for instance - so while the intuition works, it's not really sufficient for a verification/proof

[vital jay]

If it were pi/2 or 3pi/2 it would be undefined

[idle marten]

oh 1,0

so it would be 0/1

[vital jay]

But pi and 2pi are both 0

[silk gust]

yeah fair

[idle marten]

so with tan2pi equaling 0

what do we do

[flint cape]

Well if we used one of these formulas?

[vital jay]

Well plug tan(2pi-x) into the bottom right formula here!

[idle marten]

I would use

the difference formula

right?

[flint cape]

ye

[vital jay]

Yup

[idle marten]

uhh

if it says

tan2pi-x

would I say

because b is -x

the numerator

[vital jay]

Close, remember we're using the difference formula

[vital jay]

So b should be x, and the numerator would be tan2pi - tanx

[idle marten]

yeah i was thinking tan(-x)

is equal to -tanx

[vital jay]

Oh wait

Actually

[idle marten]

so since theres two -

would it be +

[flint cape]

If you're using this formula, you're setting a = 2pi and b = x

[idle marten]

ooh

[vital jay]

Using the sum identity with b = -x would probably be better

[idle marten]

so since its a different formula the b value is changed to positive

?

[flint cape]

If you're using the addition formula (which is only what I suggested because you have a tan(-x) here),

you'd set a = 2pi and b = -x

[vital jay]

Since we want tan(-x) at the end

[idle marten]

okay

[flint cape]

-# damn my intuition is a gangsta

[idle marten]

so this?

mb for the messy handwriting

Im using a mouse

to write this

[vital jay]

All good

[vital jay]

So sorry about the confusion, but you might actually want to use the sum identity from here

Using a = 2pi and b = -x

[idle marten]

oh

[vital jay]

Since the original problem wanted us to show tan(2pi-x) = tan(-x)

[idle marten]

uhh im lost when do u use difference or sum

[vital jay]

So think about doing tan(2pi + (-x) )

[flint cape]

(spoilers - they're actually the same formula; just the difference formula comes from the fact that tan(-theta) = - tan(theta), so we can use the addition formula to get the difference formula)

[vital jay]

It's really contextual. Like, both of these identities can be used here, but they'll just find different (but equivalent!) results

[flat mica]

you can also make a geometric argument based on the unit circle

and the tangent line

[silk gust]

i think sum would be most straightforward here

[flint cape]

So here, because the RHS has (-x) inside the function, we might want to leave this be - so if we let one of the a and b be -x, we could keep that in the function

[idle marten]

oh so I can use either one its just in some scenarios one is better to use?

[flat mica]

sure, but, just pointing out theres an alternative

[silk gust]

yeah, ofc :)

[flint cape]

yeeeeee

[vital jay]

Looks good to me!

[idle marten]

okay so from here what do I do

[flint cape]

tan(2pi) equals?

[vital jay]

Well remember what we said about tan(2pi)?

[idle marten]

0

[vital jay]

Plug it in!

[flint cape]

So then this equals?

[idle marten]

tan(-x)/1-tan(-x)?

[flint cape]

almost

[idle marten]

do we keep the 0

or get rid of it

[vital jay]

Hmm, the issue is in your denomenator

[idle marten]

oh

[flint cape]

First, for the "this is equal to", replace all the tan(2pi) 's with 0's

(ugh, I hate pluralising expressions)

[vital jay]

You did 1 - tan(2pi)tan(-x) = 1 - 0*tan(-x) yeah?

[idle marten]

no i just completely got rid of 0

[flint cape]

But.... 0 apples doesn't mean apples

WE'rE BaCK tO AppLeS BoYs

[vital jay]

So you gotta remember adding 0 is the same as doing nothing, but multiplying by 0 has effects

[idle marten]

yes

[vital jay]

And we have tan(2pi) * tan(-x)

So you can't just drop the 0 here

[idle marten]

so

[flint cape]

-# I swear if at some point I acc write a syllabus that's just filled with apple metaphors, I will NOT be surprised

[idle marten]

would this be

the thing

[flint cape]

yeee

[vital jay]

Yup that looks good!

[idle marten]

what do I do with the 0 now

[vital jay]

Well what's 0 * 5?

[idle marten]

0

[vital jay]

Or 0 * 100

[idle marten]

so 0(tan(-x))

is 0

[vital jay]

Or 0 * 1000000000000000000000000000000000000000000000000000000000000000000000000

[idle marten]

?

0

[vital jay]

Yep!

[idle marten]

so

tan(-x)/1?

[silk gust]

yep!

[vital jay]

Yeah!

[silk gust]

and what is anything/1?

[idle marten]

just that

[silk gust]

yes

[flint cape]

-# the amount of shits I give about how big that number is

[idle marten]

I could say -tan(x) right?

[flint cape]

but yeeeeee

Well - we've already got tan(-x)

And the RHS of the thing we're asked to check IS tan(-x)

[idle marten]

oh I forgot the right side

yea.

[silk gust]

@flint cape if you have some number of apples and you divide them among one person, how much would that person get

[flint cape]

While that IS equal to -tan(x), it's not really needed lol

[idle marten]

yeah

so I verified that its the same

[vital jay]

So now you just need to write the whole thing up similar to that example from before!

[flint cape]

"And now, on Waes's A(pple) Level Maths, ..."

[limber dawn]

this is TRUE!

🎉

[idle marten]

for sum and difference formulas

do i use them when theres something like cos (90-x)

if theres two numbers in a parantheis

[silk gust]

yes

[idle marten]

ooh

[flint cape]

yee

[idle marten]

okay

[silk gust]

they are applicable literally anywhere lol

[idle marten]

thank you

[vital jay]

Just keep in mind those work with + and -

[surreal cave]

also they're good to know bcz most trig identities (excluding the Pythagorean ones) are derived from them

[flint cape]

You've got two angles, you're adding/subtracting them, and you got a trig function of that sum/difference

That's what these formulas do

[vital jay]

Sum/difference formulas don't work with cos(a*b)

[idle marten]

how would i know when it doesnt work

[vital jay]

If there's addition or subtraction they work

[silk gust]

when you see + or -, they work. if you dont, they dont work

[vital jay]

But multiplication and division are different

[idle marten]

ah

[surreal cave]

honestly I don't even bother memorizing them either, I just rederive them anytime I need them

[idle marten]

yeah I havent been memorizing them

ive been just looking at the formula

should i memorize them?

[silk gust]

i literally have been engraving them into my brain, too lazy to derive them

[vital jay]

Maybe if you need to for a test

[surreal cave]

literally on any test I save a corner to draw out this diagram each time

[idle marten]

theres no test for this unit

[vital jay]

But otherwise, just looking them up is easier so long as you understand how to use them

[surreal cave]

I always forget where each angle goes in what sin or cos

[silk gust]

either memorize them, or derive them like a nerd pajamamama llama lol

[idle marten]

does anyone know what you learn in precalculas

[silk gust]

one or the other, either works

[idle marten]

bc Im taking ap precalc

soon

[silk gust]

ik you learn trig

[vital jay]

Polynomials?

[silk gust]

yeah i think that too

composition of functions, that stuff

[idle marten]

so just more alg 2 stuff?

[vital jay]

I remember finding a lot of quadratic roots

[idle marten]

I learned all that in alg 2

is it just

[surreal cave]

conic sections

[idle marten]

more harder versions

[vital jay]

Probably?

[surreal cave]

complex numbers and de movires formula

I did some vector stuff

[idle marten]

someone in a higher grade in my school said that ap precalc is just about graphing the functions

[vital jay]

Probably trying to refamiliarize you with ratios and stuff

[idle marten]

idk if thats it tho

[surreal cave]

yeah you're gonna do a lot of graphing

[vital jay]

That sounds like it could be true

[surreal cave]

idk about the ap curriculum - as its new - but I foudn that to be constant in most precalc ones

[idle marten]

mm

[vital jay]

There's a lot of physical intuition for calculus that comes from graphs, so making sure you're good with graphs sounds like a good thing for precalc to focus on

[idle marten]

okay well thank you for helping

.close

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

[cedar kiln]

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[scenic jolt]

anyone can give me a starting point to how to solve this question

[scenic jolt]

thanks

[muted bear]

draw the radii to the two corners of the pavement

[scenic jolt]

ok one sec

let me redraw

hopefully thats what you meant

my initial reaction is to solve area of the circle and area of the triangle

then just subtract the area of triangle from the area of circle to get the area of remaiing segment

the shaded area

is that not the correct steps?

[fossil dawn]

area of sector, you mean?

[scenic jolt]

area of the entire circle

[fossil dawn]

why the entire circle?

[scenic jolt]

then i could just subtract the area of the triangle

to get the remaining segment area

[fossil dawn]

but if you do that, you have the area of the rest of the circle still inside

[scenic jolt]

yea ok nevermind

im so lost with this question

[fossil dawn]

you'd be erroneously including the red area

[tough finch]

you'd have to get the area of the SECTOR and subtract the area of the triangle to get the area of the segment

[fossil dawn]

i mean, you know what the green area is, right...?

[scenic jolt]

secto

[fossil dawn]

exactly

just take the sector

you don't need the entire circle

[scenic jolt]

ok step 1: get sector area

[fossil dawn]

step 1: get sector angle

for that, you need trig

[scenic jolt]

so should be easy

base is 5.4/2 = 2.7

[fossil dawn]

good

[scenic jolt]

then we use sine law?

[fossil dawn]

but remember that the angle you get from this preliminary calculation might not be the full sector angle

[scenic jolt]

to get the missing angle?

[fossil dawn]

be aware about which angle your current calculation will get you

if you are already aware, you can proceed

[scenic jolt]

this is where im at

[fossil dawn]

ok so far

[scenic jolt]

so i should find the 3rd side

then i can use sine law to find missing angle?

[fossil dawn]

you don't need to find the third side

you have two sides and one angle. just use the sine law directly

[scenic jolt]

ok

so angle A should be 35.9 deg

now i should be able to solve for the sector area

[fossil dawn]

as long as you remember that A is not yet the sector angle

then all good

[scenic jolt]

its not ?

ok so its 360-35.9

360-60 = 300deg

which is the sector angle

[fossil dawn]

nono

[scenic jolt]

i mean 360 - 36

so 24deg?

[fossil dawn]

your work found the green angle. but your sector is green plus red

[scenic jolt]

ohhh

thats why you wanted me to draw a second radius line

[fossil dawn]

to complete the sector, yes

this red angle is easy

[scenic jolt]

its the same angle

[fossil dawn]

correct

[scenic jolt]

36*2

[fossil dawn]

why?

[scenic jolt]

cause they are mirrored triangles

[fossil dawn]

more specifically?

[scenic jolt]

cause its a right angle

[fossil dawn]

specifically, it's because the two triangles are similar

[scenic jolt]

so when you flip a triangle that is 90 deg, its mirrored

isoceles

[fossil dawn]

in fact, not only are they similar - they share two sides and one angle that's in between those two sides.
they are outright congruent

[scenic jolt]

i think its isoceles

yea congruent

[fossil dawn]

that has no bearing on why the two angles are equal

but now that you have the full sector angle, do whatever next you need to

[scenic jolt]

k

that has to be right

or else im going to bed

[fossil dawn]

you're not done

yet

[scenic jolt]

but the area is correct right for sector?

[fossil dawn]

looks good except 35.9 x 2 != 72

[scenic jolt]

i rounded

close enough

[fossil dawn]

i encourage you to never round until you reach the final answer

otherwise, you risk compounding the rounding error

[scenic jolt]

ok

so whats step 2?

[fossil dawn]

as you said, the area of the middle triangle

[fossil dawn]

the diagram is a bit misleading down there

[scenic jolt]

isnt the shaded area the area of the sector?

[fossil dawn]

this is the correct area

you shaded only half the sector

[scenic jolt]

oh shoot!

yea i keep forgetting theres 2 sides

[fossil dawn]

mhm!

[scenic jolt]

k...gonna work on triangle area

[fossil dawn]

aight gl!

[glad cradle]

why did you plug 5.4 in the sine law? I thought it's supposed to be 2.7 since we're only considering half of it?

[scenic jolt]

this mght take a whike

lol

5 mins max

[fossil dawn]

oh wait i didn't even realize

[scenic jolt]

what?

i made mistake?

[fossil dawn]

@scenic jolt mistake alert

here

the triangle you are considering has half of the chord, not the entire chord

[scenic jolt]

oh should be 2.7

ok let rework that

[fossil dawn]

my bad, i didn't catch that one at first sight

[scenic jolt]

its getting a bit messy. hopefully ou can still understand

this has to be correct

[fossil dawn]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[scenic jolt]

i already know answer is 1.46...but need to know why

[fossil dawn]

i'm curious. if you decided to find the height of the right triangle, why not just use it to calculate the height without the sine?

[glad cradle]

why are we multiplying by sine here?

[fossil dawn]

since you are only considering the right triangle here

[oblique sundial]

I think you shouldn't approximate the result of B..

[scenic jolt]

why did i solve the height?

[fossil dawn]

no

if you solved the height, you can just use the most basic area formula

why the need for sine

[scenic jolt]

oh you mean

bh/2?

[fossil dawn]

yeah

then you can double it to get the whole triangle

[scenic jolt]

like i said.... there are like 5 different formulas to solve for area triangle

i get so confused sometimes

[fossil dawn]

yeah but why not go with the simplest hahah

[oblique sundial]

yep

[fossil dawn]

you have b and h

[scenic jolt]

cause i lose track which is the simplest

[fossil dawn]

fair, that happens

[scenic jolt]

and all of them formulas dont seem so simple anymore

[glad cradle]

it's not about picking the simplest, it's about choosing one that fits which information you have

[scenic jolt]

so is the area correct?

[glad cradle]

in this case you've got b and h so go for it

[scenic jolt]

but i guess i should multoply by 2

cause the triangle is mirrored

there

[fossil dawn]

did you round 8.7948 to 8.7?

[scenic jolt]

i just left it at 8.7

too many decimal digits

[fossil dawn]

if you wanna round, at least round it to the nearest tenth

[scenic jolt]

it was confusing me

oh ok

[fossil dawn]

but again, i suggest you don't round and just use the full amount

you can bash this into a calc anyway

[scenic jolt]

i know but sometimes there are like 10 decimal numbers

hard to keep track

yea not sure how to store the number in my calc

[fossil dawn]

in that case you could leave it in radical form

[scenic jolt]

i know theres a way but cant be bothered to figure out at the moment

[fossil dawn]

but try not to round until you're done

[scenic jolt]

ok ok

so is area triangle good?

[fossil dawn]

the more you round in intermediate steps, the more the rounding error in your final answer

[scenic jolt]

yea makes sense

[fossil dawn]

looks ok

except for the rounding error

[scenic jolt]

its ok for now...

so now i got area of sector and area of triangle

step 3

easy

sector area - sector triangle

should be final anser

final answer is a bit off but should be close enough

1.46

so to summarize steps

1. solve area of sector
2. solve area of triangle (plus mirrored side)
3. area of sector minus area of triangle = segment

makes sense now why we mirrored the triangle

its so that it covers both sides of the segment

[fossil dawn]

final answer is not exact most likely due to rounding error

[scenic jolt]

i think thats what at least i like to believe

yea....

[fossil dawn]

sector*

[scenic jolt]

yea..

i think theyre interchangebale terms

i got 2 more then im calling it a night....can i show you next question....all i want to know is general steps

and i will figure out the rest

[fossil dawn]

this is quite involved

[scenic jolt]

how many steps do you see

[fossil dawn]

did you shade that part of the diagram yourself

[scenic jolt]

no its like that

when they say total area of the resulting hole, they want area of the two circles?

[glad cradle]

try utilizing the fact their center is 10.5 cm apart

[scenic jolt]

my initial tohughts is

find area of the two circles

then subtract that middle section

and i already know we gotta do that same mirroring thing with the triangles like what we did in the last quesiton

somrething like that

[scenic jolt]

5cm

10/2

but i just want to know general steps

[glad cradle]

just like before, find the sector

find the triangle

[scenic jolt]

i cna figure out specific caluclatoins hopefully on my own

[fossil dawn]

i think once you've gotten the blue line, it's exactly the same

[glad cradle]

subtract the sector with triangle

[scenic jolt]

isnt the blue line 5cm?

because the distance from both centeres is 10

Step 1: find sector area
Step 2: find triangle area

what do the ymean Do not include two of the hatched area.

i only see one hatched area

is the second half supposed to be hatched as well?

im just going to assume it is

[fossil dawn]

basically the overlap

don't include two of it

[scenic jolt]

yea

ok im gonna work on

ill close this for now...if i get stuck i will create a new help...thanks again everyone. you're the beset. 🫶

[glad cradle]

btw the formula for the area of a triangle is bh/2 and no additional multiplication with sine

[scenic jolt]

i know

but i get confused sometimes on which formula is best to use

causes theres like so many of them

but ill try to keep that in mind for future questions

.done

[fossil dawn]

.close sir

[scenic jolt]

.close

[cedar kiln]

Channel closed

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

Please don't occupy multiple help channels.

[scenic jolt]

i basically calculated area of circle of one of the circles

calculate sector area of one of the circles

then subtracted sector area from circle area

to get area of remaining minus sector area

then just multiplied by 2 to get area of both circles minus their sectors

if anyone cares to review my work would be much appreciated

caues something is def wrong

[unreal pivot]

I think, based on what you said, your calculation is wrong

[scenic jolt]

which part

[unreal pivot]

You should find the total area drilled out

Not the area without overlaping part or the sector which overlaps

It should be circle1 + circle2 - overlapping sector

[scenic jolt]

so isnt that just the area of the two circles then?

[unreal pivot]

It isn't, since in circle1 + circle2 you summed the overlapping area twice

[scenic jolt]

hmm

ok so i need to find the area of the shaded bit in the centre

mult by 2

and then basically subtract from the area of the 2 circles

basicallt what you said

[unreal pivot]

Yes

[scenic jolt]

can you give me 3 mins

i should be able to come up with an answer

half the work should be already done

[unreal pivot]

Sure thing, take your time 😃

[scenic jolt]

🫶

somethings not adding up

[unreal pivot]

Let me calculate shaded sector area myself to check

[scenic jolt]

my work is all over the place sorry

basically

step 1: solve sector area for one of the circles

step 2: solve the triangle area for one of the circles

step 3: solve for shaded area (middle section) by sector area - triangle area

[unreal pivot]

What angle did you get between?

[scenic jolt]

this?

,rotate

[unreal pivot]

Okay 97.7209° is the whole angle or?

[scenic jolt]

yea the full angle for the sector

[unreal pivot]

I got the circle sector be 49,25 approximately (the thing shaped like a cake piece)

[scenic jolt]

but thats only for half right?

,rotate

[unreal pivot]

What angle is 97.7209°?

[scenic jolt]

let me draw it

its the full angle

[unreal pivot]

Okay the whole thing is 97

[scenic jolt]

ya

[scenic jolt]

yea exactlyt

[unreal pivot]

I got the red part to be 49,25 in area

And for triangle i will calculate now

[scenic jolt]

same sector area

, rotate

[unreal pivot]

28,62 for triangle

[unreal pivot]

So then shaded area should be 20,63

[scenic jolt]

same

,rotate

[unreal pivot]

321,65 the final answer?

[scenic jolt]

yea thats what i got

but its wrong

[unreal pivot]

How do you know it's wrong?

[scenic jolt]

cause its not part of the MC options

[unreal pivot]

Okay but what if the question setter made a mistake?

[scenic jolt]

i mean...its possible but i think more likely its a mistake on my part

what did you get?

[unreal pivot]

Same as you

But please

Check it with your colleagues too

Maybe i overlooked something

[scenic jolt]

thats the correct answer

we're close but i wouldnt say close enough

so i dont think its a rounding error

[unreal pivot]

We got 321,25?

It's a mistake on question setter's part, they mixed 1 and 7 on that right hand side keyboard

[scenic jolt]

ok i got this class tomorrow

ill verify with instructor

thanks anyway

🫶

[unreal pivot]

Np, check with your other colleagues too, they probably found the same mistake

[cedar kiln]

@scenic jolt Has your question been resolved?

[cedar kiln]

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[formal depot]

this was translated so tell me if you dont understand something, can anyone solve C and D or show me how to do it

[cedar kiln]

@formal depot Has your question been resolved?

[abstract breach]

you might want to go on the physics server

[formal depot]

.close

[cedar kiln]

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[cedar kiln]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[ornate echo]

Hey

[ornate echo]

How do I calculate this limit?

This is one of those indetetminate cases where it's 0/0

So, I should have to evaluate the lateral limits right

[sullen cipher]

Yeah u r on right track

[heady sapphire]

use lopitals

rule

[sullen cipher]

Plugging in x as -3

[heady sapphire]

it works

[sullen cipher]

Gives u an answer

[ornate echo]

Hm

[upper ruin]

No no, it's completely avoidable

[heady sapphire]

yeah it is

[ornate echo]

But then it's 3-(-3)

[heady sapphire]

yeah

so just add them

[ornate echo]

Which is 3+3=6

[sullen cipher]

Yep that's ur answer

[ornate echo]

Well

[heady sapphire]

yeah

thats the answe

r

[ornate echo]

But

[upper ruin]

Your function has simplified to -x + 3, right? So, what's the limit when x → -3?

[ornate echo]

In the solutions it says it's 0

Haha

[heady sapphire]

solutions might be wrong idk

[upper ruin]

Ohh that's a typo then, for sure

[heady sapphire]

i got this on mathematica

[ornate echo]

Really?

[upper ruin]

Yes

[ornate echo]

Ok thank god haha

[heady sapphire]

yeah definitely

[sullen cipher]

Yep unless the question had a mistake

[heady sapphire]

they probably got -3+3

instead of -(-3) +3

[ornate echo]

Cause I did this↑

But then checked solutions and ut wasn't correct

So I got confused hahah

[heady sapphire]

yeah you are correct tho so good job

[ornate echo]

Ok

is this a valid reasoning though?:

Cause like

In $\frac{(-x+3)(x+3)}{x+3}$

I can't just plug in -3

Since that would be dividing by 0

[wraith dagger]

stoicindiehacker369

[ornate echo]

So↑

[sullen cipher]

U r allowed to do simplification so that u can get something defined

[ornate echo]

Oh wait, yeah I can do that

It was just messing with my head

Cause

-3 doesn't belong to the function domain

But I forgot that doesn't really matter😅

Because the limit doesn't need to belong to the domain of the function

Ok

Thanks!

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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Remember:
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[true sapphire]

hello ,srry bro didnt see

[true sapphire]

.close9someoen was typing)

[cedar kiln]

Channel closed

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[flint cape]

dang, someone who actually understands Discord 🫡

[cedar kiln]

Available help channel!

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Remember:
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[marsh phoenix]

I need help

[hallow pelican]

!da2a

[cedar kiln]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[marsh phoenix]

It's uploadin

How do I solve ts

[hallow pelican]

why did you have to film it like a video like that

[marsh phoenix]

Idk

Anyway

[hallow pelican]

watch it makes me want to throw up(i have slight motion sickness)

[marsh phoenix]

Ho2 to solve

[hallow pelican]

what is the question

[marsh phoenix]

2nd pic

[hallow pelican]

,rcw

[marsh phoenix]

Crazy

[hallow pelican]

true

so, what have you tried

[marsh phoenix]

I m trying

[hallow pelican]

how big is the n+1-th triangle compared to the nth one?

[marsh phoenix]

Ye

[hallow pelican]

huh?

look at the first and second triangle, and then the second and third. only look at the edges, not the stuff within.

[marsh phoenix]

Lemme try then imma come back again ig

.close

[cedar kiln]

Channel closed

Closed by @marsh phoenix

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[cedar kiln]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[stray crypt]

hey, they want me to find a min k in which p(x) + k will be >=0

so i did p'(x) and found that x=3 gives a local minimum

[supple grove]

ok

so find p(3)

[stray crypt]

yeh but how can i prove that x=3 is global minimum

[blazing zephyr]

it is not a global minimum

[old dirge]

p(x) is a cubic equation

[livid hound]

question looks sus, what's the exact wording?

[blazing zephyr]

the lim as it goes to -inf is -inf and as it goes to inf is inf

[old dirge]

theres no global minimum

[hallow pelican]

very bad question

[stray crypt]

yall right im sorry

they say p is in the interval [0,4]

sry!

[old dirge]

oh

[hallow pelican]

oh lol

[stray crypt]

yep

[old dirge]

then you must check all possible minimums

[stray crypt]

u mean i need to do p4 p0 and p3 and p1

[hallow pelican]

nah you should first find out all monotonic intervals and find out values at certain points

[stray crypt]

cuz x=1 and x=3 are which f'(x)=0

[old dirge]

yes

and the minimum is the answer

[hallow pelican]

im going to control my colon 3 instincts

[stray crypt]

but i showed that p3 is a local minimum

it isnt enough ik

[stray crypt]

is there a quicker way or is it the safest one?

[old dirge]

i believe it's the only way

cause all are possible

[stray crypt]

so youre sayin this aint necessary at all

for me to show that p3 is the global minimum

[old dirge]

yeah

you don't need to show it's a global minimum

[stray crypt]

but i need some way to find a k in which p(x) + k >=0 in the interval [0,4]

global minimum will help isnt it right?

[old dirge]

um

you need the minimum from the interval to be greater or equal to 0

[stray crypt]

yes, p(3)

[old dirge]

well then you must solve p(3)+k>=0

[stray crypt]

p(3)<=p(x) for every x

p(3)=-4

-4<=p(x) for every x

so then -4 + 4 <= p(x)+k (in which k=4) will make p(x)+k>=-4+4=0 @old dirge

for every x

[stray crypt]

i can say this only when ik that p3 is the global minimum

am i good ^^ ?

[stray crypt]

in the interval [0,4]

[old dirge]

global minimum in the interval [0,4]

but otherwise it's correct