help-36

[tired walrus]

??

in your case, 4^x + 4^x + 4^x + 4^x becomes 4 * 4^x

the thing you're adding four of is, well, 4^x

[gritty flume]

Thanks

[granite ivy]

Revise your laws of exponents

There are just 5 or 6 laws

[gritty flume]

But they don’t look what I’m doing

[jagged siren]

You need to restart at the basics of exponents to be able to do something like that as it incorporates many different laws.

[gritty flume]

But time

[jagged siren]

There's like less than 10 rules that are super basic

@gritty flume

[gritty flume]

I dii ok t understand the last one

I don’t

When do I know what to use

Some of the questions make me change big numbers into smaller odd numbers

[final saddle]

@gritty flume Has your question been resolved?

[jagged siren]

What question are you confused on

And show me ur workings already

[gritty flume]

I just want to know when to use each rule

[jagged siren]

When they give you the first bit you turn it into the second part??

[gritty flume]

Number 20 is what I can’t do

[jagged siren]

Show me what you have already

[gritty flume]

I haven’t done it

Idk the rule for this

[jagged siren]

Use the last rule i sent in reverse

[jagged siren]

You first need to simplify what you have underneath

And the way you did it is interesting

[gritty flume]

I’m having an issue

With the factorization

[jagged siren]

Where did the over 1 over 5!n come in

Hold on

[gritty flume]

Wdym

[jagged siren]

Was looking at hr first photo

[gritty flume]

Oh ok

[jagged siren]

I don't know how to explain this simply

Ill try

[gritty flume]

Ty

[jagged siren]

Where did the 5^2 go??

How did it before 5^0

According to Law 1 on the page. When you multiple the exponents add. However 5^n+2 does NOT equal 5^n × 5^0

As anything to the power of 0 equals 1 so..

[gritty flume]

Oh

[jagged siren]

You get me?

Try again, with this in mind now

[gritty flume]

I’m so sorry

It’s my bedtime now

I go to a boarding school

I’ll take a photo of this and reopen this tomorrow

At 5am for me

[jagged siren]

Ok babes... Goodluck...

[gritty flume]

Tysm

Ty

[jagged siren]

Hope it goes well!!

[final saddle]

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[final saddle]

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[crude narwhal]

i cant figure out how to model the equation

[ripe jewel]

how many comic books were given to her brother?

[crude narwhal]

15

[ripe jewel]

how many were not?

[crude narwhal]

we dont know

[ripe jewel]

use a variable

[crude narwhal]

15x

?

or we just let that be x

[ripe jewel]

there's more than one way to set it up, but

I would let x be the total number of books in the box

15 of them are given to the brother

how many are left? that's how I would do it

[crude narwhal]

x-15 ?

[ripe jewel]

yes

I would do it that way

[crude narwhal]

would there be more variables

[ripe jewel]

let's see if we can get away with no more

so she sold x-15 books

what was her profit per book?

[crude narwhal]

1.50

[ripe jewel]

so her total profit for all x-15 books was...

[crude narwhal]

x-15=1.50

?

[ripe jewel]

you know that cant be right because the LHS is books and the RHS is money

[crude narwhal]

true

[ripe jewel]

when you sell or buy things in bulk, you multiply the price by the item

2 dollars per donut? i bought 12 donuts?

12*2

[crude narwhal]

330/1.50

?

[ripe jewel]

she's earning 1.50 per book

i want to know how many books she sold

and each book is sold for its price

think of it with 1 unit

if a bookstore says that

it's 1 dollar per book

and you buy 1 book

it's $1 (1 book)

[crude narwhal]

yep

[ripe jewel]

multiplied

so, you take however many books she sold

and multiply it by the profit per book

it should be

(dollar amount)(book amount)

[crude narwhal]

1.50x

[ripe jewel]

.

[crude narwhal]

but we dont know how many books where sold

[ripe jewel]

yes you do

all except 15 of them

[crude narwhal]

im confused

can you show what the equation would be other than x-15

[ripe jewel]

that's not an equation at all

the other way would be to make a system of equations and you probably didn't learn that yet

[crude narwhal]

yea i havent heard of that

[ripe jewel]

and it would be harder

let's take a simpler example

[crude narwhal]

we get stuff and set it to 0

[ripe jewel]

you're selling mangos

you sell them for $3 each

if you sold 4 of them

what was your revenue?

[crude narwhal]

12

[ripe jewel]

what if you sold t of them?

[crude narwhal]

t3

[ripe jewel]

what if you started out with M of them, but you threw out 11 because they were rotten, and you sold the rest?

instead of 3M, it's...

[crude narwhal]

t3-11

[ripe jewel]

almost but not quite

3(M-11) you need parentheses

which you should realize because with

3M - 11

the first term is in dollars

and the second term is in mangos

[crude narwhal]

oh yes

and 11 mangoes gone

[ripe jewel]

she didn't sell the entire boxfull (unknown amount, use a variable)

she sold all except 15

[crude narwhal]

300(x-15)

[ripe jewel]

hmm nope

because we're multiplying by the profit per each one

just like with the mangos we multiplied by the price per single fruit

so the profit here is like the price in the previous example

I think it would help a lot to see more examples, if you can find problems like this

since the profit of $1.50 is written as for each one

and you're findinjg the profit of all of them

you multiply $1.50 times the number of comic books sold

which is a variable, and isn't the whole boxc

[crude narwhal]

1.50(x-15)

[ripe jewel]

yes!

now , we need to find what to set it equal to, and that's a bit tricky

[crude narwhal]

would it be 330

[ripe jewel]

no, because it says profit, and profit = revenue - cost

[crude narwhal]

true

[ripe jewel]

so, what is the total profit?

[crude narwhal]

30

[ripe jewel]

yup!

so the total amount of 1.50(x-15) is also equal to 30, and that should allow you to solve for x

[crude narwhal]

so is it 1.50(x-15)=30 which is 35

x=35

on the text book it says 75; $4.00

[final saddle]

@crude narwhal Has your question been resolved?

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@crude narwhal Has your question been resolved?

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[buoyant compass]

I need help

[buoyant compass]

How do I do this question

<@&286206848099549185>

The 42m is throwing me off

Do I do 42-25

I get 16 and that does not seem right because the base should be bigger

[glossy zephyr]

Thinking of it, no, you dont even need triangle similarity

This is just rule of sines

[hollow flare]

yes

multiple ways to do it

[glossy zephyr]

You can get the lengths of DE and AC as is with this

[slate grail]

Cossine rules underrated

[glossy zephyr]

We dont have enough info to make use of law of cosine

[buoyant compass]

Can you help plug the numbers in

[glossy zephyr]

Try to find all the angles you know

[buoyant compass]

I have them all 70,90,20

[hollow flare]

you only need one side to know all the elements of a triangle if you know the angles

and u have 2

[buoyant compass]

The 42 is throwing me off

[hollow flare]

don't do that, just fill in the triangles and then subtract the difference

[glossy zephyr]

Law of Sines assings a proportion between a triangle's side and its opposing angle.

for ABC, triangle

What is your known side and opposed angle?

[buoyant compass]

if the one side is 70 the other has to be 70 right?

then i just add together and subtract 180

[glossy zephyr]

oh, the angle

[buoyant compass]

yes

[glossy zephyr]

try to construct this equation for the left triangle

$\frac{A}{\sin\alpha}$

[soft zealot]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[glossy zephyr]

What is A and what is alpha?

[hollow flare]

rembember this word, socatoa
Socatoa
"S"in= "O"/h
"C"=a/h
"T"=o/a

[buoyant compass]

ik soh cah toa

a is 20 degrees

[glossy zephyr]

and A?

[buoyant compass]

i dont have that side

[glossy zephyr]

Remember, opposing sides

[buoyant compass]

so the top is 40 degrees

wait just start me from step 0 im sorry

[glossy zephyr]

Law of sines, we have to construct an equality that lets us solve for "L" in this diagram.

[buoyant compass]

so a is 20 degrees

[glossy zephyr]

yes, and A?

[buoyant compass]

sine law

[glossy zephyr]

$\frac A{\sin\alpha}$

[soft zealot]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[glossy zephyr]

you already got that "alpha" is 20 degrees

What is the length of its opposing side?

[buoyant compass]

25.447?

[glossy zephyr]

Whats your reasoning for that one?

Remember, we are at ABC

[hollow flare]

perhaps it's using the smallest triangle

use known data first

[buoyant compass]

but im pretty sure the 42 m is for both triangles

[hollow flare]

no

only angles

[glossy zephyr]

we are just looking at the left one

ignore the right triangle

turns out that they have nothing to do with eachother

[buoyant compass]

42 is only from C to B

[glossy zephyr]

Yes, so, how would the fraction look like?

[buoyant compass]

wait so if the bottom of the left triangle legnth is 42 cant i just soh cah toh it

[glossy zephyr]

yes, you can, but law of sines is quicker

Straight up tells you the length.

[hollow flare]

but solving the entire triangle will strengthen your understanding of how a triangle works

[buoyant compass]

122

.8

[hollow flare]

now find the short

same way

[buoyant compass]

its 70

[hollow flare]

then - them

[buoyant compass]

122-70

52

[hollow flare]

no

[blazing perch]

i dont want to barge in but i think op's instructor wants them to use both the triangles to arrive at the solution. this seems to be a 10th grade lesson on application of trig. i suppose law of sines hasn't been taught yet?

[buoyant compass]

they have but i dont undetrand law sins

[blazing perch]

alright

[glossy zephyr]

theres no known relation between the two

[hollow flare]

ideally, we would calculate the entry height, but the boy has difficulties with entry concepts

[buoyant compass]

wait i get it i thibk

[glossy zephyr]

Yeah, they are similar and thats it, but you dont have enough info to use that fact.

[hollow flare]

mastering angles will help you even with circles,try to understand that first

[blazing perch]

ohh

[buoyant compass]

129

-70

[hollow flare]

no

the 122 you got was fine, but you messed up your short wire

do the same process for it

[buoyant compass]

115

-70

[hollow flare]

where are you getting the 70 from?

[buoyant compass]

cause im not doing sin law and im subtracting the 70 from the other wire to get how much taller the other one is

what asnwer did you get?

i will still need to show my work so it wont help but it will put me on the rigfht track

iya know what im just gonna ask my teacher but i appreciate the help

[hollow flare]

the two triangles have the same angles,
and each triangle gives you a different side
Since in both you need the hypotenuse, you'll use sin(angle) = opposite because they are all opposite, and then you solve for the hypotenuse

in one, your best angle is 20º and its opposite is 42m; in the other, your best angle is 70º and its opposite is 70m

once you have both hypotenuses, subtract them, the difference is your answer

[buoyant compass]

Tbh I was just stressing out too much cause I had like 10 mins to hand in this assignment so I wasn’t thinking clearly so I already handed it in and this was the last question on the assignment

[hollow flare]

Trigonometry is beautiful (and incredibly useful), don't get discouraged

[buoyant compass]

Yes what I learned so far was act pretty fun

[final saddle]

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[gritty flume]

Can someone explain this too me pls

[gritty flume]

It’s different compared to the one Ik

And how do I plot this pls

[ivory vessel]

is this an exam?

[gritty flume]

Yeah

[ivory vessel]

help during an exam is not allowed

please close this channel

[dense coral]

it says 2021 though

[gritty flume]

It’s a past paper

How would I have my phone in an exam🥀

[ivory vessel]

oh sure

my bad

[proven radish]

input various values of x then plot it

[gritty flume]

No like how do I use this formula

The one Ik is diff

Ik two

[proven radish]

got this one right?

[gritty flume]

Yes

[proven radish]

ull simplify it further by opening up all terms and multiplying em i.e. multiply -2, (x-3) n (x+1)

then find the turning point as given in point 2 of the parabola form

i presume uk how to find ranges n domains?

[gritty flume]

I use the -b/2a right

For x

[proven radish]

yes

[gritty flume]

And suvmbstitute into the formula for y

[proven radish]

check check

im not checking as my class is starting in 5 mins

[gritty flume]

Gl

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[warm python]

Firstly here proving $(T*)^{-1}=(T^{-1})^*$ is enough, is it not

[soft zealot]

wai

[opaque ember]

yes

[warm python]

And could I have a hint pls

I think I should just unpack it using the defn

[opaque ember]

what def?

[warm python]

of T*

[opaque ember]

whats the def of inverse?

[warm python]

T(T^{-1})=I

[opaque ember]

yes. try working with def of inverse and properties of adjoint

[warm python]

will do

tq

.close

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[tall vector]

Did I simplify this correctly?

[tall vector]

I forgot the rules regarding multiply exponents

[vital crag]

,tex .exp rules

[soft zealot]

riemann

[tired walrus]

,w simplify (4x^3-1)(x^4+x+1) - (4x^3+1)(x^4-x+1)

[soft zealot]

Results provided by WolframAlpha

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

Requested by ann_dec

[tired walrus]

6x^4 - 2 is in fact the correct simplification @tall vector

you claim to have forgotten the rules but you seem to have messed them up an even number of times

[tall vector]

I think im starting to hallucinate numbers 🤯 thank you

.close

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[tired walrus]

??

[tall vector]

😴 😵 😵‍💫 🤕

[tired walrus]

?????

[final saddle]

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[jagged flare]

Let $(a_1,b_1,c_1), (a_2,b_2,c_2), \dots, (a_n, b_n, c_n)$ be all integer triples $(a,b,c)$ such that $a\geq 4, b\geq 5, c\geq 6$, and $a+b+c\leq 25$. If
$$S=\sum_{k=1}^{n}\binom{a_k}{4}\binom{b_k}{5}\binom{c_k}{6}$$
find the number of positive factors of $S$

[soft zealot]

ihave<skissue>

[robust mulch]

whoa

[jagged flare]

i got a hint that said that this was a combi question??

[robust mulch]

no way really

i couldnt tell

[jagged flare]

yeah wtf me neither

[robust mulch]

prehaps a good place to start is figuring out how many triples you have

[jagged flare]

i was thinking of changing that massive sum into like a combi model which would simplify to something much nicer and computable? but i have no idea how to do that

[robust mulch]

committee forming is always an idea, but it does take some time to figure out

dont get frustrated if you cant see the committee forming argument right away

[jagged flare]

uhh whats a committee forming

[robust mulch]

its making a story out of your combi problems

[jagged flare]

oh

[robust mulch]

did you figure this out btw?

[jagged flare]

oh wait sorry i missed that

so you can change it to w+4+x+5+y+6+z=25 <=> w+x+y+z=10 (nonneg int), by stars and bars its 13c3=286 if im not mistaken

[robust mulch]

seems right to me

just bash 286 cases

[jagged flare]

😭

[robust mulch]

you clearly have never tried an aime problem /s

[jagged flare]

eh rarely

[robust mulch]

ill brb

let me know if you have anything

[final saddle]

@jagged flare Has your question been resolved?

[jagged flare]

from 25 people in a committee, consider dividing them into 4 groups, A,B,C,D. from groups A,B,C, choose 4,5,6 people respectively to be specialists

i think this is equal to S? im not too sure, but if im not mistaken this is equal to 25c4×21c5×16c6

honestly ive only ever encountered a solution like this once, so maybe im completely wrong

[robust mulch]

try it with a smaller number first

idt 25 is special, so maybe try 16, see what happens

[final saddle]

@jagged flare Has your question been resolved?

[jagged flare]

ok if its 16, then the triples are (4,5,6) (4,5,7) (4,6,6) (5,5,6). S=1+7+6+5=19
from my committee idea... ok its not even close

fuck

[robust mulch]

if you had 17, you also add (4,5,8), (4,7,6), (6,5,6), (5,6,6), (5,5,7), (4,6,7)
which get you +28+21+15+30+35+42=+171
or in total 191

of i got my math right

i was hoping that would be 20C2

but its off by 1

[jagged flare]

huh wait

19+171 is 190 no?

[robust mulch]

omg im bad at math

my mind was stuck on 20

well i showed my hand

i think its (15+n+3)C(n)

-# could i tell you why? no

[jagged flare]

where its <=15+n i reckon?

[robust mulch]

what is <= 15+n

[jagged flare]

a+b+c<=15+n

[robust mulch]

whatever the constraints of the original problem are

it fits that

[jagged flare]

yeah ok fair

[robust mulch]

im playing it super fast and loose with the combi here

ok yeah

you have your specialists, and your bars

(15+n+3)C(n)=(18+n)C18

you line up your (in this case 25) members (so n=10)

then you set up 28 spots

you pick 18 of them to be specialist or bar

the first 4 of these are specialist, the next is bar, then 5, bar, 6, bar

the line fills in the spots where there are no bars

something like
..::::.|.::.:.::.|::.::.::|..

this accounts for your groups if you do it the more convoluted way

with the sum

[jagged flare]

oh wait so this committee argument was equivalent? rather i evaluated it incorrectly?

[frail moat]

Which is orginal que? @jagged flare

[jagged flare]

.

[frail moat]

Hmm.. 13c3 combinations

Meh there would be some method like a series pattern other wise this be too lengthy for combinations

[jagged flare]

ok i think i understand it now, thank you guys!

.solved

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[dense palm]

Guys is this correct i dont have anyone to help me

[dense palm]

I got this on chatgpt and when i finished it he said that i need 8 ones instead of 4

[boreal pilot]

I have no idea what that is

[dense palm]

Lol its a k-map

[boreal pilot]

yep
never heard of it

[abstract bramble]

Karnaugh maps are a way to help humans simplify boolean expressions and hopefully develop more efficient circuits

[dense palm]

Do you know if its correct or no

[abstract bramble]

no clue im not in that field

tried to teach myself kmaps before when ppl ask here but i still dont get it 💀

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@dense palm Has your question been resolved?

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[worthy radish]

How to find x out of these?

[worthy radish]

For example if it’s

(X-6)^2 = 0

Then it’s 6?

[trail pilot]

yes

[worthy radish]

But the third example?

How does it work

[worthy radish]

Is 6 final answer?

[velvet pivot]

Doesn’t it have to be a perfect square?

[calm ermine]

Third example is perfect square

[velvet pivot]

Such as $(x-4)^2=0$

[soft zealot]

🩷Aurora💜

[velvet pivot]

I’m sorry I thought you take the square root of 6

[calm ermine]

Nope

[velvet pivot]

6x6 is 36

[calm ermine]

U only take the squre root if it's of the form
x^2-a^2

[velvet pivot]

It becomes (x-6)(x+6)

[calm ermine]

No thats only true if its
x^2-6^2 not
(X-6)^2

[plucky rover]

Those are not equations, those are expressions

You can't exactly find x without an equation

Other than you know, circling it and saying you found it

[velvet pivot]

You’re right my bad!

[worthy radish]

Ik… but this is when solving = 0

So either I can use delta or these when solving?

[plucky rover]

In that case you would set each factor individually to zero

What the heck is delta

[calm ermine]

😭

[worthy radish]

😭

[plucky rover]

No seriously I'm not familiar with your notation and convention

[proper dagger]

discriminant

[calm ermine]

How does discriminant solve equations

[plucky rover]

Ah, we just use D

[plucky rover]

I mean short answer is quadratic formula

[calm ermine]

It just states the type

[calm ermine]

Oh well ig yea

[plucky rover]

You don't really need to

[calm ermine]

These are all factorable

[plucky rover]

You can factorize them and then set each individual factor to zero

[calm ermine]

So its a waste of time

[worthy radish]

How’d I use without the delta formula ?

I can’t cause c is 1

So I have abc

[plucky rover]

Multiply the entire thing by 2

[worthy radish]

Not ab

.close

[final saddle]

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[final saddle]

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[light topaz]

how do i integrate this

[light topaz]

on top if it was only cosx it would be easy substitution but i cant seem to figure it out

i think its integration by parts but cant exactly figure out what to do

[final saddle]

@light topaz Has your question been resolved?

[leaden moon]

I don't know much but you can try the identity $\sin^2x + \cos^2x = 1$

[soft zealot]

1 divided by 0 equals Infinity

[warped perch]

Just needa play with trigonometric identities

[light topaz]

?

which ones

[warped perch]

Like this one

[light topaz]

i already tried the cos(2x) one

tried that too

[severe verge]

Might want to rewrite your integral into $-\frac 13 (1-\frac{4}{3 \sin^2{x} +1})$

[soft zealot]

(De)Carbonized

[light topaz]

mhm

let me guess

uh

wait what that still doesnt help

WAIT WHAT HOW DID THAT HAPPEN

wait

[severe verge]

?

[light topaz]

I can write it as

$$\int_{0}^{\frac{\pi}{2}} 1-\frac{4sin^2x}{cos^2x + 4sin^2x} dx$$

[severe verge]

actually do you want to continue with this approach?

[soft zealot]

bagelguy3

[light topaz]

I want to continue with an approach that works

ur the one who recommended it

[severe verge]

alr

now the problem is we want to integrate $\frac{4}{3 \sin^2{x} +1}$

[soft zealot]

(De)Carbonized

[light topaz]

uhh

i dont get it?

what about the sin^2x

[severe verge]

which sin^2 x?

[light topaz]

the one on the numerator

[severe verge]

didn't I say we can rewrite the thing we want to integrate into this?

[light topaz]

ok but how do you rewrite into that

I get the denominator but how did you get rid of the cos^2x on the numerator

[severe verge]

just rewrite it as 1 - sin^2 x

[light topaz]

right...

I see

That still makes it

[zealous storm]

[light topaz]

$$\int_{0}^{\frac{\pi}{2}} 1-\frac{4sin^2x}{1 + 3sin^2x} dx$$

[soft zealot]

bagelguy3

[zealous storm]

I would do this problem by defining another integral like $$J := \int_0^{\frac \pi 2} \frac{\sin^2(x)}{\cos^2(x) + 4\sin^2(x)} \dd{x}$$

[soft zealot]

mmmm7

[zealous storm]

and calling the original integral something like I

[light topaz]

okay but this isnt the same as the original integral

[zealous storm]

no

look at the numerator

[north juniper]

I+4J= pi/2

[light topaz]

ik can see that but

[zealous storm]

yes I + 4J is trivial to compute

[light topaz]

interesting

[zealous storm]

and I + J

is also trivial

well a u-sub should do it

[light topaz]

whats stopping you

[zealous storm]

?

[light topaz]

from u subbing i t

cuz i tried and i cant

U sub doesnt work because theres an additional power on the numerator and that complicates things....greatly....

[severe verge]

I + J is 1/(3sin^2 x + 1)

[zealous storm]

u can ig

[severe verge]

divide both numerator and denominator by cos^2 x

then let u = tan x

[light topaz]

yes and u get

(1+u^2) on the denominator

i already tried that

[zealous storm]

re-express your thing as $\int_{0}^{\frac \pi 2} \frac{1}{1 + 4 \tan^2 (x)} \dd{x}$

[soft zealot]

mmmm7

[zealous storm]

dividing the numerator and denominator by cos^2(x)^

which is valid given the upper and lower "bound" of the integral

now u can u sub(?)

[light topaz]

$$\int_{0}^{\infty} \frac{1}{(4u^2+1)(u^2+1)} dx$$

[zealous storm]

that does look oay yeah

[light topaz]

what was the command for infinity

[zealous storm]

also it's \infty

[bleak granite]

\infty

[soft zealot]

bagelguy3

[north juniper]

ok thats kinda trivial

[light topaz]

now partial fractions?

[zealous storm]

yes

[light topaz]

E

I hate partial fractions

what about complex analysis

[zealous storm]

Lol

[severe verge]

thua

[light topaz]

think about it

[zealous storm]

turning away from basic algebra

to consult complex analysis

[severe verge]

I'm defeated

[light topaz]

I just have to find complex roots, find the residue and compute

I just use cauchy residue theorem

[warped perch]

Wait wait😭

[severe verge]

the "just" here sounds loose

[zealous storm]

[warped perch]

I'm close to answer... Wait a min

@light topaz

[light topaz]

k

[severe verge]

mf is trying to speedrun a !nosol

[zealous storm]

😭

[light topaz]

wait

i think i can do it

[soft zealot]

Results provided by WolframAlpha

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

Requested by mmmm70177

[north juniper]

wth is ts

[zealous storm]

residue theorem 😭

[north juniper]

ok that's far away from shit ive learned

[bleak granite]

just use partial fraction, why do all this?

[light topaz]

because

partial fr actions suck ass

[bleak granite]

it's literally just algebra

[zealous storm]

i like the I + 4J and I + J thing the most lol

[bleak granite]

[\frac{1}{(4u^2+1)(u^2+1)}=\frac{Au+B}{4u^2+1}+\frac{Cu+D}{u^2+1}]

[soft zealot]

ΠαϳαμαΜαμαΛλαμα

[severe verge]

I'll bow out, I'm totally defeated from this

[light topaz]

alr

thanks

i can take it from here

[light topaz]

doesnt work

[zealous storm]

it does

🤨

[north juniper]

i think you can compute I+J very easily

[light topaz]

actually

[zealous storm]

^

[light topaz]

i forgot

complex analysis

[north juniper]

sec^2/1+3tan^2

[light topaz]

interesting

multiply by

sec^2

[zealous storm]

wait it's sec^2(x)/ (1 + 4tan^2(x))

[light topaz]

ahh

no wait that doesnt work

its

[north juniper]

oh yeah that

[light topaz]

$$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+3sin^2x} dx$$

[soft zealot]

bagelguy3

[north juniper]

1/2arctan(2tan(x))

[zealous storm]

what

[light topaz]

I + J

[north juniper]

ok lets not make 1 on denominator

[light topaz]

ok

$$\int_{0}^{\frac{\pi}{2}} \frac{1}{cos^2x + 4sin^2x} dx$$

[soft zealot]

bagelguy3

[north juniper]

devide ts by cos^2

[light topaz]

holy crap

that is just insane

[zealous storm]

,, I + J = \int_0^{\frac \pi 2} \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x) + 4\sin^2(x)}

[soft zealot]

mmmm7

[light topaz]

it works....

[zealous storm]

(forgot dx^)

and yeah you can simplify the integrand

[light topaz]

how come I didnt think of that

[zealous storm]

and multiply and divide by cos^2(x) like the other person mentioned

[north juniper]

idk thats kinda hard

considering J

[light topaz]

mhm

anyway guys

i understood

thanks for all your help

[calm ermine]

If u want to evaluate from 0 to pi just sub in to the bottom eq and subtract

Im in a rush so my writting is messy af

Just use partial fractions

I did it in one page

[warped perch]

Well it took more space cuz I wrote with description

[calm ermine]

Fair

Its not really that hard of an integral tho

[warped perch]

Yea that's also possible

[calm ermine]

P sure partial fractions is the simplest

I wonder can u do this with weirstrass sub too?

[warped perch]

But... Final answer?

[calm ermine]

Just sub in

I cba

[calm ermine]

Probably not actually

[warped perch]

Just check it out, used a property of definite integrals

[calm ermine]

Ud end up with a function of degree 4 which is just not worth it

[warped perch]

Nope

[calm ermine]

O

Ill try later ig

I mean ud end up squaring cos which is (1-t^2/1+t^2)^2

Ud deffo get some function of degree 4 i think

Probably not i haven't done it yet

Maybe simplify or sum

[warped perch]

Did you even check my solution?🥀

[calm ermine]

Oh gimme a sec

[zealous storm]

it's wrong

also

[final saddle]

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.

[warped perch]

Pls elaborate

[zealous storm]

J is not what u defined it to be earlier

so I = J is a true statement following from a faulty premise

[calm ermine]

The answer is pi/6

Using this

[warped perch]

Ohh.... Wait

[calm ermine]

@light topaz

[warped perch]

Thanks for the correction

[calm ermine]

@zealous storm am i correct

[zealous storm]

yes

[calm ermine]

😊 yay

[warped perch]

🥹

[final saddle]

@light topaz Has your question been resolved?

[final saddle]

Channel closed

Closed by @light topaz

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[light topaz]

bruh

aasnt this shit

been closed

I did it with complex analysis

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[sharp wraith]

i initially tried heron's formula but i think it would be too long, i tried then drawing a new diagram, considering the labelled areas A and B. i think when area A = area B (which i think occurs when p or x = 1, then the area of the triangle would be the same when x = 0, would this be useful in determining the third point that minimises the triangle?

[thorny otter]

How much calculus do you know?

[sharp wraith]

i think enough for this q if it requires

[north juniper]

find a way to minimize h

[thorny otter]

Yeah this is what I was about to suggest

[sharp wraith]

oh thats so smart thanks

yh i think i can continue appreciate both ur help

.close

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[rustic wedge]

a menial way to do this: use the determinant formula for area of a triangle

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[spring haven]

<@&268886789983436800>

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[shell condor]

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[pure cliff]

how the fuck do i do this plz help.

idk how to do iii

[zealous storm]

what would you get if you subtract the area of triangle ABC from the area of sector ABC

[pure cliff]

is the area of the triangle abc 18

[zealous storm]

do you know what type of triangle ABC is?

[pure cliff]

equilateral

[zealous storm]

yeah and what's the formula

for the area of an equilateral triangle

[pure cliff]

1/2 base x perpendicular height

?

[zealous storm]

that works

u can use that if you want

or there's also a direct formula to compute the area of an equilateral triangle

whatever works

[pure cliff]

ok

So

What noiw

[zealous storm]

wdym what now

[zealous storm]

it's not 18^

[pure cliff]

Oh

idk

idk what to do im loosingh my mind

[abstract bramble]

calculate the area of the triangle??

[abstract bramble]

ok if you cut it in half, it makes two 30-60-90 triangles

that means one leg is 3, and the other leg (the height) is 3sqrt3

[pure cliff]

ok

[abstract bramble]

so it's 3 * 3sqrt3 / 2 * 2

[pure cliff]

9 root 3

[zealous storm]

yes

now find the area of sector ABC

[pure cliff]

12pi

so i minus

maybe

[zealous storm]

how'd you get this?

[pure cliff]

a=(1/2)(r)^2(theta)

[zealous storm]

okay and what did you use

as theta?

[pure cliff]

60

[zealous storm]

no

[pure cliff]

60/360

[zealous storm]

that formula requires that theta is in radians

i mean if u want

intuitively

it's just 1/6th the area of a circle with radius 6

[pure cliff]

is 60/360 not right

[zealous storm]

that's one part of the required calculation yeah

[zealous storm]

^

[pure cliff]

i just subbed iun

[pure cliff]

to this

[zealous storm]

no

[pure cliff]

and got 6pi

[zealous storm]

forget that formula

[pure cliff]

ok

[zealous storm]

theta needs to be in radians

which i'm guessing you're not comfortable with

[pure cliff]

idk what that really means

Ya

[zealous storm]

if n is 60 then that means it's 1/6 of the total circle area

because n = 360 would cover the whole circle, right?

so find the area of the circle

and divide by 6

[pure cliff]

Im lost

what circle

[pure cliff]

oh

howq do i find the area of a circle

nv m

[zealous storm]

😭

[pure cliff]

Im cooked

My teacher just introduced this to us today and gave us the hardest questions to start

Ok

Area is 36pi

6Pi

[zealous storm]

u can use that to think about the area of a circle

but okay anyway

[zealous storm]

yes

[pure cliff]

ok

[zealous storm]

entire green bit is 6pi

remove the area of the triangle

and u get this strip

[pure cliff]

Ooo

So 12pi

• area of triangle

[zealous storm]

wdym 12 pi

😭

u just computed the area of the sector

and it wasn't 12pi

[pure cliff]

ya but there is 2 sectors

[zealous storm]

let's focus on one of those first

[pure cliff]

yes

[zealous storm]

if u can find the area of one of those

then at the very end

u can multiply by 2

[pure cliff]

wait wtf is 6pi then

is it not his green par

[zealous storm]

yes

6pi - area of triangle = one of the two shaded regions

the two shaded regions are symmetric

so two of the shaded regions is just double the difference we computed prior

[zealous storm]

^ just find this first

[pure cliff]

3.261098653

[zealous storm]

😭

give exact answers

not some random number

[pure cliff]

what

I did 6pi-area of triagnle

6pi - 9root3

[zealous storm]

i mean write it as 9 root(3) instead of

[soft zealot]

Result:

15.58845726812

[pure cliff]

what

[zealous storm]

....

when you're dealing with irrational numbers

like 5 sqrt(5) or whatever

it's better to write 5 sqrt(5)

instead of

[soft zealot]

Result:

11.180339887499

[zealous storm]

anyway

area of one of those things = 6pi - 9 sqrt(3)

so double that since u have two of those shaded things

[pure cliff]

thanks

Im done

.close

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[warm ether]

<@&268886789983436800>

oop, too slow on my part it would seem

.close

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[sonic spindle]

Hi, I got this question in an exam (done and dusted) and got that h=18.97, can someone solve this, and explain in simplistic terms how they got h (if possible)

[late rose]

,rotate

[sonic spindle]

Sorry mate

[late rose]

is 24 the length of the entire bottom leg of the (presumably) right-angled triangle?

[sonic spindle]

It is right angled, I apologise for not stating that

Consider the length of the bottom side of the LEFT triangle as 24, consider the length of the bottom side of the RIGHT triangle as x

[final saddle]

@sonic spindle Has your question been resolved?

[late rose]

so the entire bottom leg of the right-angled triangle is 24 + x?

[sonic spindle]

Yes

[late rose]

okay

consider tan(34) and tan(20) as the ratios of sides.

can you do that?

@sonic spindle

[sonic spindle]

What ? Tan 20 is h/(24+x)

[late rose]

okay

and tan 34?

[sonic spindle]

h/x ? What did you think it would be?

[late rose]

good. i was just confirming you knew how to do that.

notice that we can express h in terms of x in two different ways. do so, and equate them together.

[sonic spindle]

It's this question, I didn't get this as the answer though, and neither did my mates, https://brainly.com/question/29570015

Sorry, I only just found it online

[late rose]

no worries, you provided the question in enough detail to solve

[late rose]

^

[sonic spindle]

Yeah, I remember using

tan20=h/(24+x)
And
Tan34=h/x
And substituting h or X from one equation into another

[late rose]

exactly

actually, now that i think about it, it may be easier to find x in terms of h and substitute that into the other equation

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[late rose]

.reopen

[final saddle]

✅ Original question: #help-36 message

@sonic spindle Has your question been resolved?

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[bleak granite]

<@&268886789983436800>

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[sour kindle]

Struggling to show that (g1 g2)*f(x) = g1 * (g2 *f(x))

But i might just be confused with the notation for the group action in this case.

[signal vector]

use the fact that (ab)^{-1} = b^{-1}a^{-1} in a group

show us your working and we may be able to say if youre confused with the notation

[rare girder]

so you want [((g_1g_2)\cdot f)(x)=(g_1\cdot(g_2\cdot f))(x)] for all $x\in X$

[soft zealot]

Flip

[rare girder]

grrr brackets

[sour kindle]

I might have made a dumb error idk

[rare girder]

this is a little (uninteresting) error: you have an x on the right, no x on the left, but it's fixed if it read (g_2 circ f)(x) = f(g_2^-1 . x)

the same applies on the next line

[sour kindle]

Oh yeah my bad but thats not the reason I can’t get it work right

[rare girder]

it might help if you change your notation for functions slightly

$(g\bullet f)(x)=f(g^{-1}\cdot x)$ is a definition for the function $g\bullet f$, but when you're manipulating such functions, it might help to explicitly state input-output relationships, like [g\bullet f:x\mapsto f(g^{-1}\cdot x)]

because then you can write stuff (for yourself) like [g_1\bullet(g_2\bullet f):x\mapsto f(g_2^{-1}\cdot x)\mapsto f(g_1^{-1}\cdot(g_2^{-1}\cdot x))]

if that makes sense??

it kinda helps me

oh I'm silly and missing literally every f imaginable

I'll put them in

[soft zealot]

Flip

[sour kindle]

yeah sure

[soft zealot]

Flip

[sour kindle]

wait

[rare girder]

alternatively you can read it left to right: [g_1\bullet(g_2\bullet f):x\mapsto (g_2\bullet f)(g_1^{-1}\cdot x)=f(g_2^{-1}\cdot(g_1^{-1}\cdot x)]

[sour kindle]

yeah that makes sense but dont we end up with the wrong expression form each

[soft zealot]

Flip

[rare girder]

no, right? it's socks and shoes

what you have written at the end seems good to me

[sour kindle]

huh

[rare girder]

although to end with the punchline I would've written these two in the opposite order

[sour kindle]

wait im so lost my two expressions dont match tho

[rare girder]

but they do

[sour kindle]

haha

[rare girder]

$(ab)^{-1}=b^{-1}a^{-1}$ is what I've heard called socks and shoes

[soft zealot]

Flip

[rare girder]

in the morning you put on your socks first (b), then your shoes (a)

how do you undo this? first you take off your shoes (a^-1), then your socks (b^-1)

[sour kindle]

this is my issue tho

i might be crazy

[rare girder]

if it helps, remember that you really want to show that $g_1\bullet(g_2\bullet f)=(g_1g_2)\bullet f$

[soft zealot]

Flip

[rare girder]

the right side is favorable because it equals f((g_1g_2)^-1 . x) = g_1g_2 o f(x), and fortunately it's what you ended up with

[sour kindle]

i swear these arent equal dude

i know (ab)^-1 = b^-1 a^-1

its only equal is g1 = g2 right

unless abelian

[rare girder]

or if g1 and g2 commute, yes, what you have written here is generally correct

however, we have never asserted this

[sour kindle]

we dont just want the case of g1 = g2 to work or when G is abelian because we dont know that G is abelian.

so i dont understand how these are equal

[rare girder]

maybe it'll be a little clearer if you write out $((g_1g_2)\bullet f)(x)$ explicitly

[soft zealot]

Flip

[rare girder]

let me reread

[sour kindle]

if im wrong right now im quite sorry, let me quickly retry with better notation hopefully

[rare girder]

no I understand your question now, I apologize

[sour kindle]

all good

i think the issue comes from how we treat the group actions on x inside the function, i feel like it needs to act differently

[rare girder]

no, so the problem is we're both misunderstanding something I think

[sour kindle]

hmm okay

[rare girder]

without me even thinking, I wrote two things that are blatantly different but assumed they were equal

[rare girder]

this is somehow wrong

[grim nebula]

what the hell is happening here

[rare girder]

this is correct

[sour kindle]

haha idk man

[rare girder]

I'm just bad at precomposition

[grim nebula]

(g1 . (g2 . f))(x) is f(g2^-1 g1^-1 x)

you computed it wrong

when you do g1 acting on g2 f you gotta shove g1^-1 onto the input

not onto the left

(g2 f)(x) = f(g2^-1 x)

[sour kindle]

why does g^-1 go inbetween them?

[grim nebula]

because as a function, g2 f is f(g2^-1 (-))

[sour kindle]

like (g^-1 x) is just another element of x like k say so shouldnt it stilll go onthe left

[grim nebula]

what?

[sour kindle]

like the action of g^-1 * x is just another elemtn ini the set X right?

[grim nebula]

the action of g is defined as (gf)(x) = f(g^-1 x)

put it another way

[rare girder]

as an element yes, but the action G on X asserts that every little g corresponds to a function g* : X -> X, x -> g.x

[grim nebula]

if $L_g : x \mapsto gx$ is the left multiplication map, $gf = f \circ L_{g^{-1}}$

[rare girder]

it's better in this context to think of the elements of G as themselves functions on X

[grim nebula]

,, g_1 (g_2 f) = (g_2 f) \circ L_{g_1^{-1}}

[grim nebula]

you gotta shove x multiplied by g1^-1 into g2 f

[rare girder]

$g_1\bullet(g_2\bullet f)=(g_2\bullet f)\circ L_{g_1^{-1}}=(f\circ L_{g_2^{-1}})\circ L_{g_1^{-1}}$

[soft zealot]

Flip

[sour kindle]

i havent seen the L_g notation before but whats the difference between the full circles and open cirlces in this case then

[rare girder]

L_g is made up

I just took it

[grim nebula]

i'm using \circ for function composition

L_g is a function

f is a function

you compose them

[rare girder]

open circles is function composition, bullet is the action G on the set of functions F

often function composition is written as juxtaposition, like the composition of g after f is just "gf"

but here we might want group actions to be juxtaposition

[rare girder]

hopefully this is easier to read

[sour kindle]

okay i think ill sit on this for a while an try to understand

[rare girder]

sorry for the confusion

[sour kindle]

all good

.close

[final saddle]

Channel closed

Closed by @sour kindle

Use .reopen if this was a mistake.

[final saddle]

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[cedar obsidian]

To find eqn of SHM how can I find phi if I have found theta = 30°

[tired walrus]

30° ≠ pi/3. do you mean pi/3 or pi/6?

kindly mark theta on your diagram @cedar obsidian

[cedar obsidian]

I'll simply write 30° for now

[cedar obsidian]

There ^