help-26

[radiant marlin]

once you find that the rest is just calculation

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[dry oxide]

.reopen

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✅

[dry oxide]

Yea….

Can you give any suggestions for finding that angle? Cuz I don’t know calculus and don’t really wanna solve it by coordinate geo….

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[weak perch]

You don't need any of that

This probably I think it's mildly hard and I have no idea how to help you get to the solution editor straightout telling you the solution

Do you want a hint?

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[odd portal]

Hello

[odd portal]

At first I did row operations to get a row in each matrix with zeros, but I don't think that's correct. After I tried calculating what the determinant is, but I don't know how to show that X x Y is perpendicular to x and y with that information

If the determinant = 0, does that mean that X x Y is perpendicular to x and y?

[wooden matrix]

note that $x\cdot(x\cross y)$ is a scalar triple product, which is also equal to $det([x|x|y])$

[thorny flame]

Mosh

[wooden matrix]

and det([x|x|y])=0 by alternating

[odd portal]

yes, and the same thing applies to the other matrix

Thanks

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[grave juniper]

hello i am doing ap calc homework and for the question below I have given my question, the steps ive taken, the answer i got, and what it should be. thanks for the help

f(x) = 3x-2
lim Δx -> 0 [f(x+Δx)-f(x)]/Δx

to begin the solution i subbed in f(x) into the equation
lim Δx -> 0 [3x-2+Δx)-(3x-2)]/Δx
i then cancelled out the 3x and combined -2
lim Δx -> 0 [Δx+4)]/Δx
after inserting 0 for Δx i get the answer 4, yet the book says i should get 3

[grave juniper]

im wondering if i did the algebra of the equation wrong

[restive inlet]

you didn't simplify f(x+Δx)
properly

you didn't expand -(3x-2)
properly either

which looks like the exact same type of mistake

$a(b+c) \red{\neq} ab + c$

[thorny flame]

ℝamonov

[grave juniper]

it should be -3x+2

which would still cancel the 3x and the 2 too

[restive inlet]

you still need to

simplify f(x+Δx)
properly

[grave juniper]

how would I do that, this part of the lesson i didnt really understand

do i just put the f(x) in for x?

[restive inlet]

this is basic function notation

[grave juniper]

alright let me take a step back (3x-2+Δx) is not correct right for when I implement the function?

[restive inlet]

f(x) = 3x-2
to get f(x+Δx),
replace the x in 3x-2 with (x+Δx)
(inserting parentheses as needed)

[grave juniper]

((3x-2)+Δx)

right?

[restive inlet]

no

how are you getting 3x-2+Δx

[grave juniper]

im replacing x with 3x-2

[restive inlet]

that doesn't look like what you're doing

it looks like you're just adding Δx to 3x-2

im replacing x with 3x-2
you shouldn't be doing that either

[grave juniper]

in all honesty i dont know what else I would do i didn't understnad this part

[restive inlet]

starting with

f(x) = 3x-2
replace x with (x + Δx)
(and do nothing else)

[grave juniper]

3x+3Δx-2

[restive inlet]

not quite what i wanted, you went ahead and skipped steps but yes

[grave juniper]

ah sorry

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[quick lynx]

What digit will be at the end if n is a natural number

[pseudo bear]

@quick lynx What does the grey box mean?

[quick lynx]

nothing

it's error maybe

[pseudo bear]

OK, does natural number include zero?

[quick lynx]

yes

[pseudo bear]

OK, then 4n + 1 goes: 4(0) + 1, 4(1) + 1, 4(2) + 1, ⋯. This is 1, 5, 9, ⋯.

So, basically, you start with 3¹ and then repeatedly multiply it by 3⁴.

3¹, 3⁵, 3⁹, ⋯
3¹, 3¹ · 3⁴, 3¹ · 3⁴ · 3⁴, ⋯

[quick lynx]

so it's 3

last digit will be 3

[pseudo bear]

Yes, because 3⁴ has 1 in its ones place.

So, you have 3, 3 · 1, 3 · 1 · 1, ⋯.

[quick lynx]

yep

there is another one just like it but instead of +1 it's +2

[pseudo bear]

Well, it's the same sort of process.

You have 3², 3² · 3⁴, 3² · 3⁴ · 3⁴, ⋯, which is 3², 3² · 1, 3² · 1 · 1, ⋯.

[quick lynx]

so it's 9

[pseudo bear]

Yes.

[quick lynx]

thanks i guess it was easy

[pseudo bear]

Yes, you can see why it works either with modular arithmetic knowledge or with the old method for multiplying on paper.

When you multiply on paper, the digits in places other than the ones place can never affect the ones place.

So, you can look only at the ones place and forget the rest to see what the result's ones place will be.

[quick lynx]

okay i will ask if i will have questions thank you for help

[pseudo bear]

You're welcome.

[quick lynx]

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[fiery current]

help in graph theory

[fiery current]

want to double check if im doing things correctly

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[west hornet]

please only use one room at a time #help-5

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[fiery sigil]

So <K = 87 and <T = 42. But when I try to get value of k (the side), i get 26 while the textbook says its 36 can i get some help

[chrome glen]

Do you have your work

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[urban sequoia]

Hello, I would just like clarification on whether I'm doing this right or not.l

[urban sequoia]

I think the answer is all real numbers greater than 0?

[alpine dirge]

yea

this looks like surface area

[urban sequoia]

Yeah it's surface area of a cylinder. Thank you!

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[native light]

doing a truth table but this feels wrong

[native light]

my logic was that the left most NOT would invert all the others but now im getting everything as false which isnt great

because i have:

(p or not s) and (s then (t and not p)) but of course there are no situations where thats satisfied and it doesnt feel correct

i also tried solving by applying the leftmost NOT only at the end and I got this, which is a tiny bit better but not really

<@&286206848099549185>

[hot lynx]

u can reduce the given propositional logic to a shorter equivalent propositional logic

I tried reducing it, and I'm getting (p^~s)

[native light]

yeah i thought of doing that but i wasnt personally sure how to

also whats the ~? is that supposed to be NOT

[hot lynx]

negation of s

not s

[hot lynx]

do I send u the pic of the steps?

[native light]

sure

[hot lynx]

Let me know if u don't understand any step or my handwriting

[native light]

where does the T go?

is this one of the equivalence rules? im not too familiar with them

[hot lynx]

t^0 = 0

[hot lynx]

Yeah

U can do without reducing it also, bt then u need keep track of so many truth values and then combine them at the end

[native light]

alright thanks

so this is the answer right?

[hot lynx]

Yep

[native light]

ok thank you very much

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[native light]

just verifying my work again, like I said, I'm not too proficient at equivalance rules

[hot lynx]

Yep, all seems correct to me

[native light]

perfect, seems like i finally got the hang of it heh

thanks

again

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[viral flax]

I need to create 1 function with all those requirements. Can’t do it ): so can anyone please help me? <@&286206848099549185>

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@viral flax Has your question been resolved?

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@viral flax Has your question been resolved?

[pulsar falcon]

do 1 requirement at a time

[viral flax]

Did the first one. But then it becomes complicated since you have to meet the 2nd requirement while keeping the first met

[pulsar falcon]

so only change something at (5,10]?

that keeps the 1st condition met

[viral flax]

I don’t I understand

The function I got was X^2 - x. I believe

[pulsar falcon]

Average rate of change =5 for t in [1,5]. So if you just do your function so that is met then do something so average rate of change is =-20 for t in [1,10] by only changing something in (5,10]

[pulsar falcon]

you need a piece-wise function

[viral flax]

Yes

[pulsar falcon]

so why can't you do what I said?

[viral flax]

I don’t understand how you got the (5,10) interval

[pulsar falcon]

you just said idk how to do 2nd one while still keeping 1st condition, yes?

[viral flax]

Yes sir

[pulsar falcon]

I told you to only do something with your function in (5,10] then

that has nothing to do with 1st condition then

so 1st condition will still be met

(doing 3rd one before 2nd one might be smarter)

[viral flax]

Wait a minute, I think I understand what you meant by [5,10]

[pulsar falcon]

(Would recommend doing 4), 5), 1), 3) then 2) in that order tbh)

this avoids you having to redo/rethink your function

[viral flax]

I’ll try to do that

Alright. I’ll start with 4

Thanks btw

[pulsar falcon]

np

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[neon iron]

help pls

ohhh nvm

1 sec

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[visual merlin]

Ok, so I feel embarrased asking this but can anyone help me solve this problem? It is an economics problem but it requires math.
Here is the problem: In Country "Faraway," cigarettes are forbidden, so people trade cigarettes in a black market.
The cigarette demand is Qd = 12 − P , and the cigarette supply is Qs = 2P. Find the equilibrium price and quantity in the black market (the equilibrium price and quantity is where Qd: Quantity demanded and Qs: Quantity supplied, equal each other).
Can anyone help?

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@visual merlin Has your question been resolved?

[untold dirge]

let the equilibrium price be $P_e$

[thorny flame]

Ken G

[untold dirge]

then we know that at $P_e$, $Q_d = 12 - P_e$ and $Q_s = 2P_e$

[thorny flame]

Ken G

[untold dirge]

And also, $Q_d = Q_s$

[thorny flame]

Ken G

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[waxen jewel]

So I know how to do the substitution in general, but can someone explain how to isolate one of the variables to get the substitution staryed in the first place?

[gusty bane]

add one to both sides on the bottom equation

that's one example

you're isolating a variable like any other

[waxen jewel]

right, but then what order do I put them in

so like when I add one does it become x=1+4y or something else?

[gusty bane]

yes it becomes that

what else did you have in mind

[waxen jewel]

ahh, ok. I wasn't quite sure because there are some things where when you move it into a different order it becomes negative, etc.

[gusty bane]

wdym

[waxen jewel]

Like, I forget exactly rn, but in some things when a number is one one side of the equation/inequality, when you reorder it to the other side it becomes negative/positive depending on what it was before

i think i remember it mostly in graphs and stuff

[gusty bane]

in equalities, what you do to one side you do to the other

that's the rule

[waxen jewel]

oh, ok

i guess that just confused me, so i would end up getting some problems wrong because of it

but i think i get it more now

thank you

wait, one more thing: so if I just had x+y=11, for example, and I wanted to isolate x I would add y to both sides. So could I make that x=y-11?

or would it have to be x=11-y?

idk if it matters but just checking

[gusty bane]

the second one

it does matter

adding y to both sides would give you x+2y=11+y

[waxen jewel]

ohhh right

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[mortal zealot]

I don't follow how they got to the second step

[main mauve]

$\sum_{i=0}^{n-2}3^i = 3^0+3^1+3^2+3^3\dots +3^{n-4}+3^{n-3}+3^{n-2} \ 3^{n-1}+\sum_{i=0}^{n-2}3^i = 3^0+3^1+3^2+3^3\dots +3^{n-4}+3^{n-3}+3^{n-2}+3^{n-1} = \sum_{i=0}^{n-1}3^i$

[thorny flame]

quantum

[main mauve]

@mortal zealot

[mortal zealot]

ahhhh right because it's minus not plus

in the n-2

thank you

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[strange pendant]

i need help

[strange pendant]

a) Estimate the width of your index finger (in centimeters AND in inches).

b) Determine the number of times that you can place your index finger alongside the other to cover the width of a standard piece of letter paper.

c) Estimate the width of the paper by multiplying the estimated width of your finger with the value obtained in (b), in both centimeters AND inches.

[vital relic]

is this a real homework lol?

[strange pendant]

ya

[vital relic]

I cant rlly say much except to estimate?

if ur not sure for a), get a ruler

[strange pendant]

ok

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[strange pendant]

no I actual need help

[rich fossil]

what a homework

[strange pendant]

yes Its due in 20 min so I need the awsers now

[rich fossil]

dude

just

write numbers

it's not like they're gonna walk up to you

pull a ruler outta their bum

and measure your finger

lol wtf

just make an estimate

[strange pendant]

ok

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[neon iron]

To check if a system x(t) oscillates, is it enough to check if x(t) = 0 for some t?

[vital relic]

What is a system

[neon iron]

A function x(t)

[vital relic]

???

$x(t) = t^2$

[thorny flame]

Shuri2060

[vital relic]

Does not oscillate

but x(0) = 0

[maiden adder]

you may also have an oscillation while the system never intersects 0

[neon iron]

I'm talking about functions that can oscillate, like some general solution of a ode

[maiden adder]

such as $y = cos(x) + 2$

[thorny flame]

Khazali

[neon iron]

Hm, in this cases what do you think is a good way to check if a function oscillates?

[vital relic]

Definition of oscillate

$$x(t) = x(kn+t)$$

[thorny flame]

Shuri2060

[vital relic]

for all integers n

and some k not 0

[maiden adder]

does an oscillatory system have to be infinite? or can we include systems that decay over time

[neon iron]

Oh, I had forgot it existed, thank you!

[neon iron]

It can be either one, but I guess that answer is enough for now. I'll see if I can solve the problem if it, if not I come back

[maiden adder]

great

also consider the domain

[vital relic]

not sure how i'd define this version?

if you allow decay

[maiden adder]

if we define sin(x) to be oscillatory, then we must also specify its domain, since:

[vital relic]

https://i.imgur.com/esPLsYj.png

I assume u meant this for decay

[maiden adder]

indeed

[vital relic]

yh im not sure how u'd define that

[vital relic]

?

[maiden adder]

$f: [0, pi/4] -> R]
f(x) = sin(x)
$

does not oscillate

[vital relic]

sure.

[maiden adder]

oh sorry i don't know how to use the bot yet

[vital relic]

Only way I can think of for decay

is you have f(x)g(x)

f(x) decays, g(x) oscillates

But might not encompass all cases

[maiden adder]

i'm sure it's got some e^-k term

[leaden tusk]

damped sine wave

don't know the general equation but that's what it's called

[vital relic]

Sure if you want to consider specificly damping

But this counts as decay too?

[maiden adder]

there you go

i think it is something along these lines:

$f(x) * e^(-kx) : k is some real number$

messed that one up too

my bad

[thorny flame]

Khazali

[empty sail]

$f(x) * e^{-k} : \text{k is some real number}$

[thorny flame]

dldh06

[maiden adder]

yeah that's better, but i also forgot to multiply the k by x

$f(x) * e^{-kx} : \text{k is some real number}$

[thorny flame]

Khazali

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[prime hollow]

i did question 1, but i'm .03 off. and I am not fully sure how to do part b.

[vital relic]

show method?

[prime hollow]

i tried making a tree diagram

but im not 100% sure i did it right

[vital relic]

show us 🙏

[prime hollow]

the first two branches are bag 1 and bag 2. then off those two points i have red and blue with their probability

[vital relic]

6 red 10 blue
7 red 3 blue

[vital relic]

Ok and what do you get

what are you adding/multiplying

[prime hollow]

for part a, i got .42

but the answer is .45

i was multiplying numbers, then adding their products

[vital relic]

yes

can you tell me

what youre multiplying

and adding

like the numbers

[vital relic]

uhhh

I don't understand why you're putting all of that stuff on the bottom

$$\left(\frac{6}{16}\right)\left(\frac{7}{10}\right) + \left(\frac{10}{16}\right)\left(\frac{3}{10}\right)$$

[thorny flame]

Shuri2060

[vital relic]

The final answer should just be the top you've written

[prime hollow]

i was thinking that

cause they did match u

up

[vital relic]

Sorry, I'm a bit lost with where the bottom comes from

[prime hollow]

idek, i wrote it like a week ago

[vital relic]

,w 6/167/10+3/106/16+10/167/10+10/163/10

[thorny flame]

Results provided by WolframAlpha

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Requested by Shuri2060#9525

[vital relic]

@prime hollow Well the bottom is meant to be 1

You probably calculated/inputted it correctly

maybe ur misreading 0's for 6's

you write them similarly

[prime hollow]

i see

that makes sense

[vital relic]

The bottom is just the entire tree

you dont need to divide by it

[prime hollow]

figured so

why

[vital relic]

Because the sum of all possible events = 1

dividing by 1 isnt needed

We have 2 independent events

Lets call it A

and B

A' = not A

B' = not B

[prime hollow]

of course, makes sense

[vital relic]

P(AB+A'B+AB'+A'B')

I think u had that on the bottom

[prime hollow]

is the tree right?

[vital relic]

well im not sure about drawing a tree

for this

yeah i dont think thats the right tree

Your first branches should represent the 1st bag

right?

[prime hollow]

im not sure

[vital relic]

let me try draw it

[prime hollow]

i wasn't sure how to make it

[vital relic]

like this?

with the probabilities marked

[prime hollow]

i think that would make more sense in terms of order of events

[vital relic]

I should change my diagram

the R and B should go where the numbers are

and the probabilities should go where R and B are

Then you multiply along the tree to check the event

[vital relic]

I want to check 'Tails then Heads' I do 0.5 * 0.5

[vital relic]

I want to check 'Both same' I do 0.5 * 0.5 + 0.5 * 0.5

If that makes sense?

Thats probably better (random tree I found)

Then if I need 'same outcome' I do P(B, B) + P(W, W) = 9/64 + 25/64

[prime hollow]

hmmm, how do these relate to the original problem?

[vital relic]

you replace the probabilities

with the right ones

and 'B, W'

[prime hollow]

hmm

alright

i got it

I also had a question about another question that I've been trying

[vital relic]

?

[prime hollow]

ik its probably simple, but how do u do a?

[vital relic]

hmm

I would draw a tree first

Can you show me after u do that? or would u not be sure how

[prime hollow]

i did this

[vital relic]

looks good

So now go along the tree and figure out when u have an umbrella

[prime hollow]

i got it

finally

[vital relic]

👌

[prime hollow]

and then

for part b

i tried using the conditional probability formula

[vital relic]

You can. But you can also use your tree

Well tbh I would prefer formula (in the end u kinda have to use it)

[prime hollow]

i was confused about what you put in the denominator

[vital relic]

ok so

do you get what the conditional probability is after?

like in terms of what it means

[prime hollow]

i just know, its the the probability of A &B happening over the probability of what B is. Or it is also the probability of A given that B happens.

[vital relic]

Yeah so if u take your probability tree

,rotate

[vital relic]

Given you have an umbrella

means you consider only the branches where the umbrella exists

As 'the things that can happen'

So thats the 1st and 3rd branch

Then you are interested in the chance the 1st branch (rain) occurs in that 'universe'

$$\frac{\textnormal{1st branch}}{\textnormal{1st branch}+\textnormal{3rd branch}}$$

[thorny flame]

Shuri2060

[vital relic]

Kindof like this vaguely 🤔

[prime hollow]

makes sense, imma try it

[vital relic]

$$\frac{\mbb{P}(\textnormal{1st branch})}{\mbb{P}(\textnormal{1st branch})+\mbb{P}(\textnormal{3rd branch})}$$

I think the formula will tell you to do that

[thorny flame]

Shuri2060

[prime hollow]

it got a closest enough answer

thx

[vital relic]

is it not the same 🤔

[prime hollow]

it got .79 repeating

[vital relic]

oh ok

,w (2/34/5)/((2/34/5) + (1/3)(2/5))

[thorny flame]

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[prime hollow]

and the answr is .8

yeah

good enough

[vital relic]

I think u typed something in wrong

it should be 4/5

[prime hollow]

i got close enough

and i understand how to do it

[vital relic]

👌

[prime hollow]

can u help me with more probability questions?

[vital relic]

I can but as far as I can see the approach is similar for all

lets see

yes, again you can do a tree here

[prime hollow]

i tried

but

if u leave the umbrella at store one, then the probability for to leave it at the second store is 0. thus how does the tree work without only getting zeros?

[vital relic]

That branch does not exist

you cut off the tree there

because they cant leave the umbrella again

[prime hollow]

so what would the tree look like?

[vital relic]

@prime hollow sorry, bad writing. L is leave. U is not leave

[prime hollow]

i generally get it

or

maybe not

i dont get the top part

[vital relic]

its the same as a normal tree

that says 4/5

the bottom aprt is just missing

If you work along the branches the sum should still be 1.

1/5 + (4/5)(1/5) + (4/5)(4/5) = 1

[prime hollow]

i get that it all needs to add to one

but doen't that show its more likely to not loose the umbrella at the first store?

[vital relic]

huh?

what do u mean

the question says

probability he loses it is 1/5 right

[vital relic]

so yes?

the chance he doesn't lose the umbrella after 2 stores is 16/25

which is more than 0.5

So after 2 stores they probably still have it

[prime hollow]

oh, i just assumed he lost the umbrella

do u multiply U and L

?

idk

i still dont get it

[vital relic]

U and L arent numbers

are u asking about your question?

https://i.imgur.com/IiQ0VrG.png

Do you get what 1, 2, 3

represent

individually

@prime hollow

Like what is (1)

[prime hollow]

1

is he doesn't loose the umbrella at all

[vital relic]

yep

Remembers at shop 1 THEN Remembers at shop 2

(2)?

[prime hollow]

lost at 2nd store

got it

[vital relic]

yep

[topaz sinew]

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[neon iron]

hey

[prime hollow]

i am not a amazing at this kind of thing, but my guess would be c

[empty sail]

Don't give answers

Also, you need to wait at least 15 minutes before pinging helpers

[prime hollow]

sorry

[dry otter]

You can try and eliminate the wrong answers

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[empty sail]

It's because the purpose of the server is to teach people, not give out answers

[empty sail]

.close

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[empty sail]

Like I said, the purpose is to teach, not give answers. You want answers, go somewhere else

[wooden matrix]

if it's shitty, why are you still here?

[empty sail]

Well, if you recently joined, then you should have read #❓how-to-get-help

And one of the rules

[wooden matrix]

I highly doubt you depleted all the options for help.

@neon iron If you don't want to use the server's resources properly, just leave. No one is forcing you stay here and make a fool of yourself.

[grand cloud]

<@&268886789983436800>

[dark swan]

oops

fuck wrong ban

[empty sail]

lol

[wooden matrix]

lol

[grand cloud]

lol

[frosty imp]

🇷🇸

[topaz sinew]

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[polar kiln]

how will i work this out

i made it to y = mx + c

so i got y = 3x-20 over 5

what do i do next

is the answer 3x?

[wooden matrix]

you dont need 3 channels.

[polar kiln]

sorry man ... im always left hanging mid way through my help and no one helps me out so therefore im left unnoticed

[wooden matrix]

yeah, you still only should have 1 channel.

[polar kiln]

i dont know how to close them

[wooden matrix]

.close

[polar kiln]

.close

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[polar kiln]

is the gradient for this 3

i worked it out and got 3x as my inital gradient

[soft island]

get it into the form y=mx+b

[polar kiln]

alright

what do i do next after this step

@soft island

[gray basin]

Didn’t do it correctly.

Y can’t be negative.

[polar kiln]

oh

wait

i did it the other way round easier way

hows this

[gray basin]

Yes

[polar kiln]

what next

or is that it

[gray basin]

Now you can write the fraction separately but you don’t have to

Now identify the gradient

[polar kiln]

the gradient is 3 over 5 right?

[gray basin]

Yes

[polar kiln]

thanks

@gray basin thank you for staing till the end

i really appreciatew that

[gray basin]

👍

[polar kiln]

😃

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[mossy hazel]

I need help, can you help me?

[topaz sinew]

@mossy hazel Has your question been resolved?

[frozen perch]

i believe you can use residues

[topaz sinew]

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[topaz egret]

Hiya, could I get some extra clarification on what this is actually doing, slightly confused by what the map does.

[topaz egret]

My understanding is an action on a group takes elements of the group and maps each of them to a permutation in the set omega, is that correct?

[topaz sinew]

@topaz egret Has your question been resolved?

[vital relic]

I think the 2nd defn is more intuitive

The first has me 🤔

Ok I see (for the 1st), but not sure how to explain

[topaz sinew]

@topaz egret Has your question been resolved?

[vital relic]

Yes.

The set of permutations, of elements in omega

I guess it is easier to visualize if you think about how the actions combine.

[topaz sinew]

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[vapid lichen]

.reopen

[topaz sinew]

✅

[vapid lichen]

@topaz egret one should see motivating examples of group actions before the formal definition

a group action encodes a certain way in which the members of a group permute a set of states

an example is the group D_8 of the symmetries of a square acting on the set of configuations of the square (the rule is defined by how we expect every combination of flips/rotations to affect a square)

so the result of a flip acting on a square is what we expect, the same square but vertices flipped

now the product rule in the definition just formally states how we expect compositions of flips/rotations to affect a square

[topaz sinew]

@topaz egret Has your question been resolved?

[topaz sinew]

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[zinc thicket]

ik how to find tangent line but how to find horizontal?

[wooden matrix]

what's the slope of a horizontal line?

[zinc thicket]

parabola?

[wooden matrix]

??

[zinc thicket]

o wait

[wooden matrix]

parabola isn't a number

[zinc thicket]

sry i thought u asked about the shape of the equation

0

[wooden matrix]

yes, and where does a slope of 0 occur on a graph?

[zinc thicket]

at the highest or lowest point of parabola

[wooden matrix]

yes, the vertex

so find the vertex of that parabola

[zinc thicket]

o ok

5/2

?

[neon iron]

Uhh

Yes

[wooden matrix]

yes

[zinc thicket]

o and then i plug in that value to find y and then thats the tangent line?

[wooden matrix]

yes, y=that value would the be the eqn of the tangent line

but that's not what you're asked for

you're just asked for x=5/2

[zinc thicket]

ahhh ic

so only x=5/2

[wooden matrix]

yes

[zinc thicket]

ah ic tysm for the help!

.close

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[crisp girder]

hello my problem is as follows, Solve 5cos(7x)cos(3x)+5sin(7x)sin(3x)=0 for 0≤x≤2π. I started by seeing it was a sum identity and i wrote is as 5cos(7x-3x) = 0which is 5cos(4x) = 0. I was wondering if the next step would be to divide by 5 on both sides but Im not sure if you are allowed to do that

[acoustic tangle]

Of course you can divide by 5

[crisp girder]

and then i would be left with cos(4x)

=0

[acoustic tangle]

Yes

[crisp girder]

okay thank you very much

.close

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[dreamy iris]

Can someone help me solve part 2 using de moivre's theorum?

.close

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[solid depot]

The lengths of the sides of a triangle are 13, 14 and 15. A circle touches two shorter sides and
its center is on the longest side of the triangle. Calculate the radius of that circle.

[solid depot]

could somebody help?

[sour bloom]

have you drawn a diagram?

which lengths/angles can you calculate?

also 13-14-15 triangles are quite nice that they decompose into 2 right triangles

[solid depot]

i dont know how to calculate anything at the start

im stuck

[solid depot]

a diagram?

[sour bloom]

yeah, of course, sketch a diagram to help you visualise

[solid depot]

by diagram do you mean a sketch?

[sour bloom]

yeah that

[solid depot]

yeah

I did

what should I first do?

[sour bloom]

yeah that's a good diagram

do you know any angles or lengths?

[solid depot]

I wasnt given any additional angles or lengths

this is all I know

[sour bloom]

what can you say about the angles alpha and beta

[solid depot]

they are even?

but how do you know that?

[topaz sinew]

@solid depot Has your question been resolved?

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[atomic marsh]

Do you know how to set this problem up?

[placid drum]

yea

i made like two overlapping triangles

[atomic marsh]

Do you have some working out that you could show me? (Just to get an idea of what you need help with exactly)

[placid drum]

i think we have to do a system now

oops got it

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[atomic marsh]

Seems right!

Yep, just a substitution :) Glad you got it.

[placid drum]

THANKS

bro

.close

uh

[atomic marsh]

The command went through earlier, I just replied late X)

It'll close in a moment

[placid drum]

oh

ok ty

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[ornate rose]

Oh ok cool thank you very much

What about this one?
What makes that aaabba? Lol
This one got me I have no idea at all

[topaz sinew]

Please don't occupy multiple help channels.

[sour bloom]

please don't occupy multiple help channels for the same question

[ornate rose]

Sorry

I didnt know if I made a new

one

!close

[sour bloom]

.close

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[last yacht]

I’m confused on how I would factor with only x. Question is “solve by factoring x^2 - x - 42”

[wheat laurel]

find a and b in (x+a)(x+b). you have a+b=-1 and ab=-42.

(you can confirm that this method works by just expanding (x+a)(x+b))

oh hi yottachad

[solemn wigeon]

well what are the factors of 42?

howdy

[last yacht]

1, 2, 21,42

[solemn wigeon]

and what are the factors of 21?

[wheat laurel]

....that's not all

[last yacht]

Those are the ones off the top of my head

[wheat laurel]

there's more

[last yacht]

1, 2, 3, 6, 7, 14, 21, 42

Bro I’m dumb it’s 6 and 7😂

Ok found the answer lol

[wheat laurel]

yep!

[last yacht]

Thank you!

[wheat laurel]

(also it's actually 6 and -7, since the sum needs to be -1) (you probably already know this, but just in case....)

[last yacht]

Awesome, thanks!

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[tight plover]

Hello, could somebody please help me why what I did was wrong?

[tight plover]

The question asks to find f^-1(f(x))

For the equation given

Now it’s all completely wrong and makes no sense with the graph either but I don’t understand why especially since in a question before the solution was found by the exact same method?

[gleaming thunder]

little trick here for f^-1(f(x)) would be to use the symmetry of the parabola

[tight plover]

Yes that is what made me understand that what I did was totally wrong but I’d like to know why because I don’t only work with parabolas

I don’t understand really understand the method we were taught for determining f^-1(f(x)) in the second picture. Well I thought I did but I messed up totally for the first screenshot so I don’t understand

[gleaming thunder]

f^-1(f(a)) = {x \ f(x) = f(a)}

just need to solve for x given a certain y

[tight plover]

Sorry I don’t follow

[gleaming thunder]

set y = f(x). Suppose x' is such that f(x') = y. Use quadratic formula to find x'. This usually gives 2 values, which are the 2 elements of the set

[tight plover]

But this is what I did

In the first screenshot

[gleaming thunder]

you end up with a condition on y. That does not look like an x value

y is held constant. You have a formula for delta, use it to compute x' such that f(x') = y

[tight plover]

Okay I see

Still get them a bit confused

Um as for finding f(R)

So we determined f(R) in the second screenshot as a condition of y

But it is totally wrong for screenshot 1, why?

I am just trying to understand when we use the method on the y condition and why we use it

[gleaming thunder]

$f(\mathbb{R}) = {y \in \mathbb{R} | \exists x \in \mathbb{R}, y = f(x)}$

[thorny flame]

themateo713

[gleaming thunder]

Hence reasoning on y values is natural, you're deducing for which y values there is such an x, and determining that is finding f(R)

it is however not the only way, sometimes there is simpler. For instance, exp(R) = R+* because an exponential is strictly positive and can reach any positive number.
In the case of a parabola, you can look at its convexity and extremum to determine which interval of values it hits

that interval (for a parabola) being precisely [f(-b/a), sgn(a) inf)

[tight plover]

That makes sense

Although

I swear I did everything right and got all values between the range of -7 and 9 only

Even though the real answer is ]-infinity, -3]

I know this because the maximum of the parabola is -3 BUT why did I then find ]-7,9[

[gleaming thunder]

this goes wrong when you write delta =

that is absolutely not the discriminant

[tight plover]

I wrote >= 0

For the delta

Ah right I wrote delta equals

Please walk me through it I really don’t understand what you mean :( I’m sorry

[gleaming thunder]

it's just the wrong formula. You look at whether delta >= 0 to see whether there are solutions (so to find f(R)). When looking for the solution to f(x') = f(x), you look at the solution, not at whether delta >= 0, you already know delta >= 0 because x is solution

you want the full walkthrough for f^-1(f(x)) ?

[tight plover]

I think I understand how to do the f values

I am okay at determining f(x)=f(x’)

[tight plover]

But you say here that you look at f(R)

I looked at delta >= 0 to find f(R)

Except my f(R) with that equation is ]-7,9[

[gleaming thunder]

ok, it sounded like that sreenshot was for f^-1(f(x))

[tight plover]

It was

Well

They’re two questions sorry

They ask for f(R) too

[gleaming thunder]

don't try to do 2 questions at once

[tight plover]

“ direct image”

[gleaming thunder]

especially since you need to write them separately too

[tight plover]

😭😭😭 you’re right

So let’s pretend I was doing the first screenshot for f(R)

I did the condition on y

[gleaming thunder]

you tried

[tight plover]

I did try

😭😭😭

[gleaming thunder]

what's the formula for delta

[tight plover]

b^2 - 4ac

[gleaming thunder]

what did you write

b² - 4sqrt(c)

why a sqrt ?

[tight plover]

if b=-4 and a =-1 and c=-7-y

True

Wtf was I doing

[gleaming thunder]

it made me think you got confused between just delta and the full quadratic formula (where you did also not divide by 2)

[tight plover]

I was trying to follow example from second equation and I think I was trying too hard for that

To follow it

You’re right thank you a lot

I understand the difference now

Between fR and the f-1

[gleaming thunder]

if you struggle with that, do review the definitions

and also, I think we don't throw that around for no reason "faut maîtriser les maths de lycée avant d'essayer les maths du supérieur"

[tight plover]

I had maths for only 2 years before I went to uni so really not enough time to master everything

😭😭😭

Trying

[gleaming thunder]

tout un chapitre en première et des utilisations répétées pour l'oublier en L1 (je suppose), c'est quand même dommage

[tight plover]

Okay I’m trying to write in French but it keeps autocorrecting since I didn’t install French keyboard

It is dommage

[gleaming thunder]

ok lol

[tight plover]

I don’t know what l1 is

I’m on Swiss system

[gleaming thunder]

ah ok

license 1 = première année d'université (perso je suis en prépa)

[tight plover]

Yea i am l1 then

I heard the prépa system in France is better than first year of uni sometimes due to competition or something

Don’t really know

Impressive that you’re in prépa

Thank you for the help

[gleaming thunder]

the top students go to prépa (usually), the weaker students to uni. So you naturally have a higher level where students do in 2 years what they do in 3 (with a different focus). Personellenement je suis en MPSI à Louis Le Grand, donc on peut dire que je m'en sors très bien (si tu connais pas, c'est une des 3 meilleurs prépas de France sur le plan scientifique)

[tight plover]

That’s very cool, I didn’t know

Always cool to know how they do it in other cities. I have family in France so it’s definitely an option

[gleaming thunder]

you think of studying in France ?

[tight plover]

Maybe. If I fail my uni here 😭😭😭😭

I go to EPFL

[gleaming thunder]

oh

nice

[tight plover]

I’m not cut out for it

As you can see

But I’ll try

[gleaming thunder]

I do

[tight plover]

I’m trying

[gleaming thunder]

not to feel harsh but I do

[tight plover]

Yeaaa it’s tough

[gleaming thunder]

la rentrée est en février non ?

[tight plover]

September

Well second semester is February

[gleaming thunder]

what are you doing right now then ?

[tight plover]

Studying for my retake exams because they were last week but I tested positive with Covid lol

The retakes are in march so

I’m actually surprisingly passing the year so far which is shocking

[gleaming thunder]

covid friend. Me too so instead of doing my homework I'm here on discord doing math anyways

[tight plover]

But it’s the moyenne of two exams

[gleaming thunder]

*mean/average

[tight plover]

Well it’s good practice I guess

I’m pretty strong in geometry and also got really great grades in computer science

But analyse?

On oublie

[gleaming thunder]

https://cahier-de-prepa.fr/mpsi2-llg/docs?rep=16, si tu veux t'entrainer, voici le site de notre prof. Il met tous ses pdf en ligne. Comme ça tu peux prendre de l'avance si tu veux, ou travailler des exos de TD (il serait normal que tu peine sur du niveau 1, et passe 1h ou 2 sur du niveau 2)

calculer des dérivées c'est algorithmique

l'analyse entière est très algorithmique

[tight plover]

Thanks a lot that’s very nice of you

Yea I’m not very good at structure

But I’m trying

[gleaming thunder]

faut juste être bon en calcul

[tight plover]

Faut pas oublier les formules

[gleaming thunder]

et se souvenir des méthodes oui

[tight plover]

Bref il faut que je continue

Merci beaucoup

[gleaming thunder]

ok bonne chance

[tight plover]

C’est vraiment gentil

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[small cradle]

frere

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[small cradle]

4

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[neon iron]

Can anyone help pls

With part d

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[neon iron]

<@&286206848099549185> ?

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[wary solar]

anyone know what sign the answer should be at?

[wary solar]

for this inequality

would it be greater than or less than

[visual lodge]

@wary solaruse this formula when you get to the part when you're supposed to square the equation

https://cdn.discordapp.com/attachments/877051782184992848/933943558497529856/images.png

👍

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@wary solar Has your question been resolved?

[wary solar]

i dont know calculus man

im talking about the inequality sign

which way does it go

[visual lodge]

its the Pythagorean theorem (a^2 + b^2 = c^2) try that and i think u will be good 👍

replace a with x, b with y, and c being 20 or 1/20

[wary solar]

ok thanks

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[bronze shard]

hi can someone please help me

[bronze shard]

im not too sure on how to solve this

[wooden matrix]

what's f(pi)..?

just repeat what was said per the last time you asked this style of question.

[bronze shard]

why are finding f(pi)?

[wooden matrix]

cause you have g(f(pi))

so you need to know what f(pi) is

[bronze shard]

so would i set cosx=pi?

[wooden matrix]

no, f(pi) = cos(pi).

[bronze shard]

ohhhh

my bad i havent done trig in a while

thats -1

right?

[wooden matrix]

that had nothing to do with trig, it was just function notation

but yes, cos(pi)=-1

[bronze shard]

i mean yea function notation

[wooden matrix]

so g(f(pi))=g(-1).

[bronze shard]

ohh

and now i can

replace the x in 2^x+5 with -1

and solve

right?

[wooden matrix]

yes.

[bronze shard]

so i got 11/2

will i write my final answer as

g(f(pi))=11/2

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[rapid terrace]

Is my solution right?

[rapid terrace]

It is how I solved.

[empty sail]

So I got that same rank but not that RREF that you got

You messed up some algebra somewhere

[rapid terrace]

could you tell where the problem is?

[empty sail]

If you did C3 - C1, you don't get all zeros

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@rapid terrace Has your question been resolved?

[rapid terrace]

please share the solution

[empty sail]

You can use online RREF calculators

To check your work

[rapid terrace]

I don't understand the final answer.

[empty sail]

When you did operation 4, you canceled out the 1s, they never fully cancel and become 0

[rapid terrace]

Correct now?

[empty sail]

You should have stopped after operation 4

Do you recall the rules for RREF?

[rapid terrace]

yeah okay

is everything else correct?

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@rapid terrace Has your question been resolved?

[empty sail]

Yes, because the matrix after operation 4 is the same as the calculator you used

[rapid terrace]

Okay, thanks a lot 🙂

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[neon iron]

How to find the product of,
cos(x) cos(x/2) cos(x/4) ... cos(x/2ⁿ)?

[old ether]

So you're looking for a simplified version of $\prod_{k=0}^{n}\cos\left(\frac{x}{2^{k}}\right)$?

[thorny flame]

Chixen

[neon iron]

Basically I need to find the product of cos(x) cos(x/2) cos(x/4) .... Upto infinity. Is it possible to do?

[old ether]

Oh, so up to ∞, not n.

$\prod_{k=0}^{∞}\cos\left(\frac{x}{2^{k}}\right)$?

[thorny flame]

Chixen