help-39

[faint crypt]

I dont know how to get they answer

Can anynody help

<@&286206848099549185>

[lofty lion]

Did you try to extend the two parallel lines in order to find equal angles and deduce the bearing ?

Moreover, if we write B(A,C) the bearing of A from C, you can notice that $B(A,C) - B(C,A) = \pm 180^\circ$

[jolly parrot]

RedxClaw

[pearl pond]

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[cursive wraith]

!status

[pearl pond]

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

[stoic imp]

I multiplied and divided by the conjugate

but dunno how to proceed

I mean we can cancel some stuff out

but after that what?

,, \lim_{n \to \infty} \frac{\left(\frac{3^n}{5n} + 2\right)}{\left(\sqrt{\frac{9^n}{n^2} + \frac{3^n}{5n} + 2} + \frac{3^n}{n}\right)}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,w lim n to infinity (2n)/(3^n)

[jolly parrot]

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[stoic imp]

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[random lintel]

how to do number 6

[tame ravine]

use x=-b/2a

to find the x coord

then plug it back in to find the y coord

[random lintel]

?

[tame ravine]

you forget to change sign for 10x-5

• 10 time -5

y should be -38

[random lintel]

ok i see

[tame ravine]

so the vertex should have been (-5,-38)

[random lintel]

what is vertex form mean thou gh

[tame ravine]

you mean y=a(x-h)^2+k?

[random lintel]

is that the answer

[tame ravine]

yes

[random lintel]

w

thanks big dog

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[cursive timber]

Hi I need help ^_^ I'm in an undergraduate formal logic course right now.

Are the two statements "G and J will not both do it" and, "G and J both will not do it" logically equivalent?

I want to express both statements as ~(G & J). Looking for some feedback on this because I feel like statement 1 implies that both will not do it, leaving room for one to do it. Statement two says that both will certainly not do it.

Ty

[sharp quest]

It’s a little vague to me, linguistically, but I’d take the first statement as an xor statement (exclusive or) and the second statement as a neither/nor (negation of or)

[marsh silo]

^

same

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[next chasm]

Can someone explain how to manipulate lim --> 0 sin(x+h) - sinx / h to get it into the form where the trig identity can apply.

[light helm]

they told you the trig identity to apply below

[autumn fossil]

you have sin(x+h) - sin(x)
and the identity is
sin(A) - sin(B)
so just apply it

[light helm]

which lets you express your difference:
sin(x+h)- sin(x)
as a product

[next chasm]

ok

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[eager arrow]

This is from MVC, I just want to know why is this the case? I can't intuit it.

[vivid grove]

Is |x| differentiable at x =0?

[eager arrow]

I remember it's not because it has slope 1 and slope -1 on both sides of x = 0

[vivid grove]

Similarly the same thing is going on in this question

[eager arrow]

Oh the case where the length of the vector is 0 but it has "velocity"?

[steep niche]

imagine uniform circular motion

|r(t)| is just length of position which is constant, so left side is 0

but right side, it is magnitude of velocity which is not zero

[eager arrow]

Oh yeah

[steep niche]

in other words, left side is the rate of change of magnitude of position while the right side is magnitude of rate of change of position

[eager arrow]

Ok oka I see it. Thank you very much!

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[pseudo mirage]

can someone help me find the bearing?

i'm still kinda blurry on this topic

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[quasi turtle]

So my question is this

I get a(1 + i) + b(1 - i)

Which breaks down to two equations

a + b = 0
(A - b)i = 0

I'm guessing with the real part, I drop the i

And a =b=0

But I can't understand what to do over c

[rustic gate]

what does this mean?

[quasi turtle]

That a and b don't have an Imaginary part over the real field

[rustic gate]

they don't, but why does that mean you can conclude that a = b = 0?

[quasi turtle]

Because you're left with the system

A + b =0
a - b =0

A = b

[rustic gate]

basically (a + b) + (a - b)i = 0 for a, b real means that you can conclude both a + b = 0 and a - b = 0

[quasi turtle]

Yeah and then get a =b=0

[rustic gate]

okay

what about over C

[quasi turtle]

That's My question

[rustic gate]

you just need to show linear dependence

are 1 + i and 1 - i multiples of each other?

[quasi turtle]

Oh yahweh I didn't see that

That's pretty simple

[quasi turtle]

Thanks for the painful obvious 🙂

Haha have a good day 🙂

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[rustic gate]

👍

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[quick flame]

Let ABC be an acute-angled scalene triangle with incircle ω and circumcircle Γ.
Suppose ω touches line BC at D and the tangent to Γ at A meets line BC at T. Two
circles passing through A and T tangent to ω meet line AD again at X and Y .
Prove that BXCY is a trapezium.

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[weary dragon]

Can someone tell me what I’m doing wrong? Pls ping me on reply

[pearl pond]

Please don't occupy multiple help channels.

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[hollow pike]

help?

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[obsidian grail]

Whats the best website that tells you current currency interest rate AND so I can efficiently change and calculate currency value (e.g. how many 5 euro is in dollar).

[jolly parrot]

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[pearl pond]

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[obsidian grail]

Does it work with Icelandic kronas and polish zloty?

[late prism]

,w convert 5 isk to pln

[jolly parrot]

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[mild grotto]

so for this querstion i got the asnwer A

so does that mean that if i have log(c)/log(b) thats = to ln(c)/ln(b)

[charred ocean]

Look up the change of base formula, but generally yes

[mild grotto]

cool cool, thank you

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[pastel widget]

Q9 please

,rotate

[pastel widget]

.rotate

ive only ever seen transformations

like

x' = -x-3

not x' = -y-3

for the first part i just plugged in the values

x' = -(-5)-3
= 8

y' = -3+2
=-1

(8,-1)

but idk if its right and idk how to describe those transformations

<@&286206848099549185>

[brave mesa]

lemme look

[latent quail]

Have you done a?

[pastel widget]

yes idk if its right

[pastel widget]

.

[latent quail]

No problem with it

[pastel widget]

ok thanks how do i do b?

im lost

like

is it reflection in y axis or x axis

[latent quail]

Have you learnt how to describe a function in matrix way?

[pastel widget]

i dont recognise the name so i guess not

i know that if i have like

[latent quail]

Name doesn’t matter

[pastel widget]

x' = 3x then thats a dilation by factor 3 in the y axis

no i mean idk what matrix way means

ive never heard of it

[latent quail]

Like this

[pastel widget]

why 3x+4y?

oh

on the right

thats interesting

is there any other way to part b?

because i didnt learn this way

@latent quail

nevermind I have solved it

thanks

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[desert solar]

Ho can soes the cramer method works

[desert solar]

how can does*

ti calcolate the metrix

help !!!!

It Is urgent

[tall flint]

why is it urgent?

[desert solar]

i spenning 2 hours o understand this

@tall flint

[tall flint]

what's your native language?

[desert solar]

France

Bro why you are ask so much question Just help me

Plsssss

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[hearty junco]

hi can someone tell me how to do this question without using cosine rule?

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[prime bane]

Can Somone Explain me how to solve these

[prime bane]

do square root on the other side

x+4 = 1

ya

[jolly parrot]

faiyrose

[prime bane]

the answer is supposed to be only x = -5

my answe sheet is showing x = 5 or x = -3

i dont get it how its -3

ohhh

its +-

sry i missed that point

how will it be -5?

[jolly parrot]

faiyrose

[prime bane]

ok i got it

[jolly parrot]

faiyrose

[prime bane]

tysm

ima close this

.close

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[stoic siren]

can someone try to explain this to me

i have not tried to understand it

nor have i been taught it in class yet

this looks easy to apply though im not gonna lie

[proven vale]

What don't you understand

They literally just applied the theorem

[stoic siren]

yeah ive just realised

theres nothing to ask for help with

😭 seems really easy

thank you!

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[versed lake]

guys im completely lose on how to do this question, can someone walk me thru it?

[versed lake]

<@&286206848099549185>

[snow moss]

What do you need to find?

[versed lake]

angle d

@snow moss

plz anyone..

[dusky wagon]

Hint : draw a radius to the tangent line

[versed lake]

and then?

ik it creates a right angle but

what else?

[dusky wagon]

Notice that this is a cyclic quadrilateral, and that the angle of two radii at the center is double the angle of two lines at the circumference that subtend the same arc

That should be enough for you to try and solve this problem on your own

But let me know if you need more help

[versed lake]

what does subtend mean-

[dusky wagon]

It basically means that the two radii touch the ends of the lines from the circumference

[versed lake]

is it 30?

FRICK WAIT NO

[dusky wagon]

Lmao

[versed lake]

its 50 :)!!!

[dusky wagon]

Yea there you go

[versed lake]

tyty

[dusky wagon]

Np

[versed lake]

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[scarlet surge]

are they isomorphic

?

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[storm harness]

in this case why do i need to multiply -2^1005i with 1 instead of i * 1, in this example, is it a situational thing or?

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[keen rain]

Any random variable X is dependent of itself right

[keen rain]

Unless it's constant

Since P(X=x,X=x)=P(X=x)=P(X=x)*P(X=x) if it's constant

But if it isn't this equality doesn't hold

[limber ravine]

Yes

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[formal marten]

how do you solve sin5theta = -1/5 what will be the theta values

[latent quail]

You can’t solve it without the help of calculator

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[finite pebble]

is there a parametric to cartesian conversion tool like symbolab or any python library doesn't matter

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[eternal oracle]

And so, it's me again with my brilliant questions. So I can't elevate expressions that are equal to 0 to the power of (-1)?

[sharp smelt]

no

[eternal oracle]

No what?

[midnight haven]

Why did you take an inverse ?

3rd to 4th step

[eternal oracle]

This is the degree (-1). I thought the expression would remain equal to itself.

[midnight haven]

Ig he meant "no you can't "

[midnight haven]

Do you mean a =a^-1?

[eternal oracle]

Oh, I assumed that this expression is x.

[autumn topaz]

well x does not equal x^{-1} in general

[eternal oracle]

x=a
1/x=1/a.
Some like that.

[midnight haven]

Oh

[eternal oracle]

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[pearl pond]

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[primal sorrel]

Does anyone here know Vector calculas??

[digital cargo]

several people do

[sharp vigil]

!da2a

[pearl pond]

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/

[primal sorrel]

Is the solution correct?

[ocean hornet]

think you made a mistake when computing the 2nd and 3rd components

[digital cargo]

your work is difficult to read; what is the derivative of x^2 + xyz wrt z?

[primal sorrel]

Zyx^2

I mean xyx^2

[digital cargo]

is that the case?

[primal sorrel]

Yes

[ocean hornet]

how

you're differentiating with respect to z right?

[primal sorrel]

Yes the third value is differentiated wrt z

[ocean hornet]

are you saying that $\frac{\partial }{\partial z} (x^2+xyz) = xyx^2$?

[jolly parrot]

Bair

[primal sorrel]

Yes

[ocean hornet]

bruh

how are you getting that

[primal sorrel]

I differentiated z =1 and left the other alone

Is that not the case?

[ocean hornet]

im very confused, we can first apply linearity and differentiate x^2 and xyz separately, x^2 is constant with respect to z, so it's 0 when differentiated right?

then xyz is just linear, so when differentiated it's xy

[primal sorrel]

I will do that question again as u stated

I was incorrect indeed

[pearl pond]

@primal sorrel Has your question been resolved?

[primal sorrel]

@ocean hornet is it correct now? I think i messed this question up. Cux i was applying chain rule to solve the gradient of a scalar in a previous question

Anyone?

[ocean hornet]

Yeah I think looks good

[primal sorrel]

Thank u

[pearl pond]

@primal sorrel Has your question been resolved?

[pearl pond]

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[stoic imp]

2. Let ( f : \mathbb{R} \rightarrow \mathbb{R} ) be a differentiable, non-identically zero function that satisfies
   [ f^2(x) = \int_0^x f(t) \frac{\cos(t)}{3 - \sin(t)} , dt. ]
   Calculate ( f(0) ) and ( f(x) ).

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

can wqe use ftc2

?

[merry carbon]

$f^2(x)$ representing $f(f(x))$, right?

[jolly parrot]

@merry carbon

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

pressumibly

[stoic imp]

im sure it doesnt mean second derivative at least

because that would be (2)

so its

g(x) = x^2
f o g(x) = g(f(x))

function composition

[merry carbon]

Yea, it's most likely not (though often that notation is used to mean (f(x))^2 instead, which is "interesting" )

[stoic imp]

😼

[merry carbon]

I think in either case, I’d use FTC sure

[stoic imp]

ftc1, ftc2, ftc3

[merry carbon]

(Actually on thinking about it, more likely to be (f(x))^2 they mean)

[merry carbon]

“The one which has the integral you see in it” (I never remember which order they’re in tbh)

The first one from here

[midnight haven]

wha

[vague drift]

You have immediately f^2(0) equal to 0 innit ?

[stoic imp]

its fundamental theo calcukus

[stoic imp]

?

[vague drift]

I just don't know if it is f(f(x)) or f(x)^2

[stoic imp]

f(x)^2

literally.

[vague drift]

So, f(0)^2 is 0

[stoic imp]

?

[vague drift]

'cause you will integrate from 0 to 0

[stoic imp]

non-identically zero function

[vague drift]

Bro, if you integrate from 0 to 0

It will obviously be 0

By definition

[stoic imp]

ok.

now what

[vague drift]

Now you have a square equal to 0

Mayyyyy we think and admit that a real who squared is equal to 0 CAN BE 0

😂

[stoic imp]

😂

yes.

[vague drift]

So, let's focus on the other point

Which I think is actually fundamental theorem

[stoic imp]

ok

[vague drift]

Bro

I'm literally helping you, why pinging 😭

[stoic imp]

how to solve?

canyou help me

[vague drift]

I used fundamental theorem

[stoic imp]

?

[vague drift]

I derivate on each part

Of the equation

[vague drift]

Yes, I just derivate the equality

By using the fundamental theorem

[stoic imp]

,, f^2(x) = \int_{0}^{x} f(t) \frac{cos(t)}{3-\sin(t)} dt

[vague drift]

Oh

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

Yes

It is what I wrote

Then derivate on the left

[stoic imp]

now what

[vague drift]

And on the right

[stoic imp]

,, \frac{d}{dx} (f^{2}(x)) = 2f(x) \cdot f'(x)

[vague drift]

Yes

It is what I wrote on the left

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

And now use the same idea for the left, but with your theorem

[stoic imp]

,, 2f(x) \cdot f'(x) = f(t) \frac{\cos(t)}{3 - \sin(t)} dt

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

yes but do not use t, is is x on the both sides

[stoic imp]

$\text{Let F a primitive to} f(t) \cdot \frac{cos(t)}{3-\sin(t)}$

[vague drift]

so you will have the integral equal to F(x)-F(0)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

if you derivate, you will find again f(x)

which leads to my equation

[stoic imp]

,, 2f(x) \cdot f'(x) = f(x) \frac{\cos(x)}{3 - \sin(x)} dx

[stoic imp]

how is that change of variable possible

[vague drift]

I just explained it

[vague drift]

.

[vague drift]

.

[stoic imp]

okay

now what

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

In your opinion ?

[stoic imp]

is this a diff eq?

[vague drift]

do you think it is ?

[stoic imp]

,, f'(x) = \frac{\cos(x)}{2(3 - \sin(x))}

[vague drift]

do you find it similar to what you've seen in class ?

you got it, now what in your opinion ?

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

we need to integrate

?

[vague drift]

what is the initial question ?

bro

[stoic imp]

finding the integral

[vague drift]

use your brain the goal isn't to use a bot

you can find it easily i guess

[stoic imp]

how

[vague drift]

don't you have a thing like :

$$
\cfrac{u'(x)}{u(x)}
$$

[jolly parrot]

dragonrard

[vague drift]

sin and cos you know

[stoic imp]

oh, we can u sub you mean

[vague drift]

?

[stoic imp]

u = 3 -sinx

[vague drift]

may be yes 🙂

[stoic imp]

u' = -cosx

[vague drift]

yes !

any question ?

[stoic imp]

yes

,, \int \frac{\cos(x)}{3-\sin(x)}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

u = 3 -sinx

u' = -cosx

du/dx = -cosx

dx = 1/-cosx

[vague drift]

oh bro don't go in this way

a primitive of u'/u is ln(|u|)

go faster ^^

[stoic imp]

?

cos(x) = u'

-sin(x) = u?

but d/dx -sin(x) = -cos(x)

[vague drift]

,, \int \cfrac{u'}{u} = ln(u)

[jolly parrot]

dragonrard

[stoic imp]

-cos(x)

[vague drift]

it's really close to our result don't you think ?

if you have a problem with the sign MAYYYYY you just add a - on the beggining

[stoic imp]

,, -\ln(3- \sin(x))

[vague drift]

try to derivate, do you find the result ?

is it even defined ?

[stoic imp]

sin is bounded $-1 \leq sin(x) \leq 1$

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

so in your ln everithing is negative. not defined

[stoic imp]

ye.

should i just do u sub instead

[vague drift]

no, set the minus before the ln

it is still not defined

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

now it is

don't forget the 1/2 before

and there you go

[stoic imp]

what is the answer

[vague drift]

,, f(x)= -1/2\ln(3- \sin(x))

[jolly parrot]

dragonrard

[stoic imp]

,, f(x)= -\frac{1}{2}\ln(3- \sin(x))

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[vague drift]

i think

[stoic imp]

ok easy

[vague drift]

lmao

[stoic imp]

.close

[pearl pond]

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[stoic imp]

thanks

[vague drift]

You're welcome

[stoic imp]

?

can you help me with new problem?

[vague drift]

It's too late in my country

May be tomorrow

[stoic imp]

ok

[pearl pond]

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[drowsy flare]

What did i do wrong

<@&286206848099549185>

,rccw

[pearl pond]

@drowsy flare Has your question been resolved?

[drowsy flare]

<@&286206848099549185>

[pearl pond]

@drowsy flare Has your question been resolved?

[drowsy flare]

<@&286206848099549185>

[pearl pond]

@drowsy flare Has your question been resolved?

[pearl pond]

@drowsy flare Has your question been resolved?

[pearl pond]

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[pearl pond]

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[dapper orchid]

Did I do this right

[marble bluff]

The 20% bit yea

(because that comes from the question)

for the other numbers, nope

why would 20% of the total, at 1500, be larger than 30% of the total (which you've said is 480)?

[dapper orchid]

Thought I could use the first number to calculate the rest

Wait would it be divided

[pearl pond]

@dapper orchid Has your question been resolved?

[pearl pond]

@dapper orchid Has your question been resolved?

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[molten garnet]

does anyone know how to approach this rotational motion physics problem?

[molten garnet]

i tried doing 250 = mgh

and then solving for h from there, but it says it's wrong

[pearl pond]

@molten garnet Has your question been resolved?

[molten garnet]

anyone?

.close

[pearl pond]

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[pearl pond]

@cunning kayak Has your question been resolved?

[pearl pond]

@cunning kayak Has your question been resolved?

[cunning kayak]

<@&286206848099549185>

[pearl pond]

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[cedar oxide]

in this triple integral, can I change the order of the integral into dy dx dz and move around the bounds

cause someone told me I cant but i thought I could

[sharp vigil]

you can change the order of integration but you would need to change the bounds to match (bounds can only depend on the outer integration variables)

[cedar oxide]

uhhhhhhh

this is the original problem right, and hypothetically Lets say I wanted to do dy dx dz

[midnight cosmos]

hey im so sorry to interupt but i have been sick in my school for ages and i have no clue about the upcoming exam can you please helkp me with some specific topics

[sharp vigil]

please find your own channel

instructions in #❓how-to-get-help

[cedar oxide]

for the problem I put in

[cedar oxide]

.

hypothetically lets say I wanted to do dy dx dz

which should still be mathematically logical

doing dy first, I can say that the bounds of y go from 0 to x

then how would I find the bounds of dx

...

[midnight cosmos]

conceptz can u pls help me

for the topics i stated

[cedar oxide]

i dont think so man im sorry

maybe find someone else?

or your own chanel :/

[midnight cosmos]

okay no probs

[cedar oxide]

.

[sharp vigil]

not quite

[cedar oxide]

oh

[cedar oxide]

sorry im having a hard time with this concept (no pun intended)

so what would be the bounds of dy

[sharp vigil]

we are imagining allowing y go from smaller to larger, and finding which surfaces it passes through along the way

[cedar oxide]

right, in that case wouldnt y go from 0

but it would end at where y = 0 intersects with z = 4-y^2

right

[sharp vigil]

the same bounds need to apply to every line of increasing y

[cedar oxide]

right

[cedar oxide]

if we quantify it by that

y is decreasing

cuz if i draw it out

it would look like

sorry that took so long, I was tryna find out why it didnt work

[cedar oxide]

therefore the bounds would go from 0 to x no?

but i gave that answer and it was wrong

[sharp vigil]

remember that the bounds may depend on x and z

[cedar oxide]

i mean

[sharp vigil]

but also notice that in increasing y, we enter the surface at y = x

[cedar oxide]

right

from my observations, its either bounded by y = x OR y = sqrt(-z+4)

so then would it be bounded between 0 and y = sqrt(-z+4)

[cedar oxide]

i think that would be it right

<@&286206848099549185>

[cedar oxide]

ok so apparently the integration is y = x to y = sqrt(-z+4)

.close

[pearl pond]

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[sharp vigil]

maybe you can better visualize with this: https://www.desmos.com/3d/4xmhrcej7v

[pearl pond]

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[worn hedge]

How do I do this?

I got the r value to be -0.3084 but for the second calculation, I keep getting the wrong answer

Never mind I figured it out

I just did the calculation wrong

.close

[pearl pond]

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[tepid panther]

The average monthly temperatures in Chicago can be modeled by a sinusoidal curve. The minimum average temperature is 27.2 (degrees) and occurs in January while the maximum average temperature in August is 73.7 (degrees). a. Let January correspond to x=o, Write an equation in the form of y = acos(bx) + d that would represent the situation. Show your work.

[tepid panther]

Okay so i'm preparing for my Trigonometry final and forgot how to do this.

I know that January is the minimum and x=0

so naturally, this would be sinusodial.

But the problem asks us to write the equation in cosine

[vocal lark]

that makes sense

because cos(x) = 1 when x=0

so you can put the amplitude in front to stretch it to the temperature it needs to be at t=0

the amplitude is half the distance between the min and max temperature

so A = (73.7 - 27.2)/2

what is that @tepid panther

[tepid panther]

so just to recap

to find the amp. you did max-min./2

I understand that

and cos(x) = 1 when x = 0

so that's why we're using cosine

[vocal lark]

yeah becasue the temperature isn't 0 when we begin. this will save us from needing to phase-shift the graph

[tepid panther]

okay

a = 23.25

[vocal lark]

ok

[tepid panther]

what is our b in this scenario?

like what does it mean

[vocal lark]

well hang on a sec

first we should shift the graph up or down so the transverse axis is where we need it

[tepid panther]

tranverse axis?

[vocal lark]

it's actually the same amount I think

this is a less complicated case

[vocal lark]

midline of the graph

[tepid panther]

oh

isn't the midline max-min/2?

which is the amp.?

[vocal lark]

one sec

oh

the problem is that I told you to subtract

I know better

sorry

we are averaging

(min + max)/2

[tepid panther]

okay so amp. would be max-min/2 and mid. would be max+min/2

alright

glad we cleared up the confusion

midline is 50.45 and amplitude is 23.25

[vocal lark]

f(t) = -23.25cos(bt) + 50.45

[vocal lark]

good job

b tells us how to scale the graph horizontally

recall first that the period of standard sine and cosine functions is 2pi

for cosine, we begin at the max when t=0 and reach the min when t = pi (half the period)

oh

and we want our function to be negative in front

[tepid panther]

why?

[vocal lark]

because -23.25cos(0) + 50.45 = 27.2
and -23.25cos(pi) + 50.45 = 73.7

we needed to start at the min and reach the max in half a period

now, the time between january and august is 7 months

so the full period is 14 months (sort of...temperature graphs, if you have seen them, are not perfect sine or cosine waves)

thus, b = 2pi/14 = pi/7

[tepid panther]

i think I understand

[tepid panther]

I understand this

[vocal lark]

well understand that it couldn't actually be 14 months or it wouldn't reset next year

but let's make sure this works

since I've already goofed twice at least

[tepid panther]

right

[vocal lark]

f(t) = -23.25cos[(pi/7)t] + 50.45

We can test this works in Desmos

success

so, recap:
We wanted cosine instead of sine because when t=0, we have a nonzero value. This saves us from having to calculate a phase shift.
f(t) = a*cos(bt) + k

first we want to find where the graph is centered (vertical shift aka transverse axis, "k"). That's the average of the highest and lowest point.

the amplitude "a" is the distance from that centerline to the highest or lowest point. Choose the sign that makes the graph start where it needs to

These problems give us the extreme points, which are ALWAYS a distance apart of half the period. So, the period of the graph is twice whatever that distance is

that means b = pi/(the distance between the extrema)

or more generally, 2pi/(twice that distance)

[tepid panther]

I pretty much understand everything but why the amplitude is negative

can you re-explain?

[vocal lark]

yeah

so from the midline, it's the same distance to the max as the min

[tepid panther]

right

[vocal lark]

when t=0, the graph is supposed to be at the min

[tepid panther]

right

making it a sin graph

[vocal lark]

so we need to SUBTRACT that distance from the midline

[vocal lark]

no, this does not make it a sine graph

check it out:

cos(t) + 1

let's say when t=0, we are at a max of 2, and when t = 2pi, we are supposed to be at a min of 0.

[tepid panther]

why would we be at the min if the period is 2pi?

[vocal lark]

hang on I flipped it to be simpler

more like the standard cosine function

[tepid panther]

oh okay

[vocal lark]

this is a standard cosine function with a vertical shift of 1

lol, I had to fix it again

[tepid panther]

this is like your tenth mistake 😆

[vocal lark]

I know

I'm cursed 💀

but I've been doing this longer than you'd think

ok the constant number we added, the 1, means the center of the graph is at y=1

the standard cosine function begins at a max of 1 and goes down to -1

(at the halfway point, when t=pi)

so actually, if we change nothing, we will start at the max of 2 and end at a min of 0

because cos(0) = +1, and cos(pi) = -1

+1 + 1 = 2; +1 -1 = 0

do you understand that?

now say we had to start at the min and reach the max after.

then we just have to multiply cos(t) by -1

and that will do it

the bold part is really the most important part, and that's all I'm really thinking about when deciding if I need the cosine term to be positive or negative

"from the midline, do I need to start above or below that line when t=0"?

above -> +
below -> -

@tepid panther does that help?

[tepid panther]

hold on

so if the function is at its minimum when t=0 from the midline, we'll subtract the amplitude and when the function is at its maximum when t=0 from the midline, we'll add the amplitude?

[vocal lark]

exactly that.

[tepid panther]

I don't necessarily understand the theory behind it but I understand that i need to do that.

Okay.

[vocal lark]

it's because cosine oscillates between max and min (or min and max if it's negative)

(as does sine)

we MUST shift our graph up to the midline

and then just figure out if we start with the max or the min

this case was the min

it helps to draw a picture.

[tepid panther]

yeah

[vocal lark]

@tepid panther

[tepid panther]

so the way I would have done it:

[vocal lark]

here is a graph of average temperatures over a year:

[tepid panther]

wait continue your thing

[vocal lark]

well, I'm basically done with that

but I was saying earlier the period obviously can't be 14 months, since there's only 12 months in a year

this is the typical shape of average temperatures we measure

[tepid panther]

right

[vocal lark]

it's concave-down as we approach the peak (dumps water), then concave-up (holds water) notice the curve is sloped differently on either side of a peak or valley

(fun-fact)

did you have a related question?

[tepid panther]

where's the fun fact?

[vocal lark]

I might have skipped it

temperature rises faster than it cools over the year

👆

[tepid panther]

cool

[tepid panther]

so the way i do it

[tepid panther]

I would subtract 23.25 from the maximum to match the sinusodial function

that's the way I usually look at it

[tepid panther]

but this makes more sense to me

[vocal lark]

I am glad you can understand a problem like this multiple ways!

[tepid panther]

okay

so that was part a.

[vocal lark]

lol

what else do they want you to find

[tepid panther]

b. Determine what the amplitude means in the context of the problem

c. Use your equation to predict the average temperature in October ( x = 9)

[vocal lark]

I'm curious what you think it means?

think about what we used it to do

it took us from the minimum to the maximum

so |a| is the magnitude of the most dramatic temperature swings from the (blank) temperature;
a is negative because (blank)

[tepid panther]

b is negative because...

because..

[vocal lark]

start with the first part

the absolute value (unsigned size) of a represents the most dramatic temperature swings from the (blank) temperature

[tepid panther]

b?

isn't that the period?

[vocal lark]

a.

sorry

I am cursed, remember?

[tepid panther]

okay, the absolute value of a represents the difference in temperture from the average between both the max/min (midline) to either the maximum or minimum

[vocal lark]

good

and it is negative because

[tepid panther]

the question asks for a cosine function and the function is at its minimum when t=0 is from the midline, so we'll subtract the amplitude to match the functions

[vocal lark]

mm...kind of

[tepid panther]

yeah i didn't really hit the nail on the head

[vocal lark]

simpler: just say we had to start with the minimum when x=0 whereas the standard cosine function begins at the maximum, so we negated it

[tepid panther]

i like that and actually understand it

[vocal lark]

oh good

life tip: don't drink diet soda

and especially don't drink 2 liters every other day for 10 years

what else we got

oh, ok

calculate f(9)

and as a bonus, do you think this will be an under-estimate or an over-estimate of the actual temperature in October? 🧠

[vocal lark]

I was right that time

[tepid panther]

buddy threw in life advice 😆

[vocal lark]

keep your short-term memory

[tepid panther]

def an underestimate

climate is getting hotter

[vocal lark]

this is true

[vocal lark]

but this is not

recall that our model has a period of 14 months

but there are only 12 in a year

[tepid panther]

right

[vocal lark]

we need the temperature to be back at the minimum when t=12

[tepid panther]

right

[vocal lark]

but that won't happen until later, when t=14 (hypothetically)

[tepid panther]

right

[vocal lark]

so the temperature is higher than it should actually be when t=9

sine and cosine functions don't perfectly model temperature patterns

(although a combination of sines or cosines could be artibtrarily close)

[tepid panther]

so just plug in 9 for t here to get the temp. for october?

I think the way our teacher does it is to have us use a graphing calculator

pretty hard to do w/o a table of values

[vocal lark]

huh?

yeah

of course

why do you need to graph it?

[tepid panther]

I was thinking of a previous problem where we're given just a chart of values and we need to find an equation

sorr

[tepid panther]

but was my thinking right here?

[vocal lark]

yeah

[vocal lark]

Reminds me of this hilarious line from Meet the Mathematicians:
"You can model almost any smooth-continuous function with sines and cosines"
"I have smooth-continuous curves, Greg. Can you model me?"

[tepid panther]

😆

Thank you for the help, Mr.Melvin

[vocal lark]

np

[tepid panther]

greatly appreciated

I have more questions but its getting late.. can I shoot you a message tomorrow?

[vocal lark]

yeah ok

I can't answer until later in the day though

[tepid panther]

thats fine

[vocal lark]

ok

good night

[tepid panther]

good night

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[lofty finch]

pls help

I dnot get what going on

why would i do m1 x m2

<@&286206848099549185>

[pearl pond]

@lofty finch Has your question been resolved?

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[brazen swift]

hi, i wanted to ask if i find a parameter in the angle and like thats my answer, do i write 180-(said parameter) for the value (incase its an obtuse triangle) or do i just leave it at that since the reason for doing 180-Shiftsin is because the calculator only returns one answer for an angle which i know the values of

sorry if this is hard to read

[uneven smelt]

whats the question?

[queen lodge]

I don't think you necessarily have to do it, but your calculator might require you to do it to get the right answer

remember to check if you're calculator is using radians instead of degrees.

[brazen swift]

no question just theoretical

want to make sure incase something like that is in my exam

[brazen swift]

nono you didnt get me i meant like
sin(a) = x
a = (coefficient of x) * x
that returns me the angle of a single angled triangle
so in case of an obtuse triangle is it 180-a ?

[queen lodge]

the 180-a when you show how you solved it. the answer should just be an angle

[brazen swift]

ok ty

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[void grail]

what does it mean that we assume elements of logic?

[void grail]

what are the basic building blocks of logic, and what are those made of?

and can implication p -> q be further abstracted?

how to think about implication with linear algebra?

what is the point of the last two rows of the conditional? do we use conditional and implication interchangeably?
F T T
F F T

it was very interesting to see that moving from the philosophical implication that includes "conditionality" and "causality" to one that only includes "conditionality" - seems to be a step towards abstraction

are there resources that explain this? 🙂

[queen lodge]

In computer science we have something called boolean operators

AND, OR, and NOT

[pearl pond]

@void grail Has your question been resolved?

[void grail]

what are those operators made of?

[queen lodge]

Not made of anything to the core they are abstract, but the can be made real using transistors.

https://www.youtube.com/watch?v=JQBRzsPhw2w

[void grail]

thanks

not made of anything?

[queen lodge]

Logic gates can be made out of mechanical systems it can be made of transistors and the flow of electrons. But is still an abstraction https://en.wikipedia.org/wiki/Boolean_algebra

[pearl pond]

@void grail Has your question been resolved?

[pearl pond]

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[teal hazel]

how would i do b?

[trim willow]

Apply the formula you have been taught for compounds interest

[teal hazel]

yes but for this one the money is also been added anually

and compunded monthly

[trim willow]

And the formula you have been given that shouldn't be a problem

[teal hazel]

well i got 8642

im nit sure if that is correct tho

[trim willow]

Idk either since you just gave an answer

[teal hazel]

well i did

a1= 300

a2= 300 (1.003)

a12= 300(1.003)^11+300

a252= 300(1+1.003^13+1.003^23....+1.003^251)

and used sum

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[severe flame]

i dont understand the transition between the 4th step to the 5th

how does 3x+6/4 turn into that

[bitter epoch]

(3x+6)/4 = (3x)/4 + (6)/4 = (3/4)x + 3/2

[severe flame]

what

[bitter epoch]

So we can split the numerator into two fractions since it's addition

[severe flame]

by how do we change it

and why

how do u do that

[bitter epoch]

Suppose we have a whole fraction, that being 1/1 = 3/3, right?

[severe flame]

yes

[bitter epoch]

3 thirds is equal to one

We can also split the 3/3

Into 1/3 + 2/3

[severe flame]

yea sure

[bitter epoch]

Which is still equal to one

[severe flame]

yes

[bitter epoch]

We do the same thing here

[severe flame]

but why

and how

[bitter epoch]

The reason we do it is because we want to isolate x

We want some number "a" times x by itself, plus another number "b"

[severe flame]

so the number next to it is the graident?

[bitter epoch]

That's correct

[severe flame]

i just cant figure out how to split them up ivetried

[bitter epoch]

Well, it's the same as the 1/3 + 2/3 = 3/3

We have 3x + 6 all divided by 4

[severe flame]

yes

[bitter epoch]

So we can split it where the + sign is

[severe flame]

okay but how do we do that

[bitter epoch]

How? Well it turns into two separate fractions, being 3x/4 + 6/4

[severe flame]

oh i dont know if i understand it fully could u help me through a question please

[bitter epoch]

Sure

[severe flame]

perpendicular to y=-2x+9 and has a y intercept of 4

first we find the graident right

y=mx+b

[bitter epoch]

Yep

Have you learnt the method to do so?

[severe flame]

yes a little i dont understand it

[bitter epoch]

There's an easy way to find the gradient of a perpendicular line

[severe flame]

what is it

[bitter epoch]

m_1 * m_2 = -1

[severe flame]

how do we do that

[bitter epoch]

If this isn't method your teacher wants you to use, then there might be another method

How were you taught to solve this problem?

[severe flame]

i wasnt there rhe day it was taught however on the formula sheet given

it says m^1 times m^2= -1

i dont know what that means though

[bitter epoch]

m1 and m2 are the gradients of the lines

[severe flame]

yes

[bitter epoch]

For example, m1 = -2 in the problem you gave, and m2 is what we want to find

How would you solve for m2?

[severe flame]

i dont know

would we try to isolate the x with 9..?

[bitter epoch]

Not quite, we only need m1 * m2 = -1

[severe flame]

then how do we find m^2

[bitter epoch]

Do we know what m1 is?

If we have 1 equation, we can solve it if there is only 1 unknown

[severe flame]

u said it was -2 right

so -2 times m =-1

[bitter epoch]

Correct

[severe flame]

so -1 divided by -2 is m?

[bitter epoch]

Yes, that's right

What does that number equal?

[severe flame]

1/2

0.5

[bitter epoch]

Yes

That's the first part of the question done

[severe flame]

but how do we find the graident with that

so whats the graident

[bitter epoch]

That is the gradient

1/2

m1 and m2 are the gradients of two different but perpendicular lines

[severe flame]

so we need to find m^2 to find the graident?

[bitter epoch]

The gradient is m2

[severe flame]

ohhhhh

so then the equation would be

the point graident formula

[severe flame]

how do we find the equation

[bitter epoch]

Well, in the equation y=mx+b, what does "b" represent