precalculus

[blissful ridge]

Okay, okay

[viscid thistle]

I figured it out, thank you for helping!

[blissful ridge]

@viscid thistle post the solution

[narrow peak]

idk where to ask for partial fraction help anyway

[odd helm]

how would i solve number 8 and 9 i needed a number for an answer

[rare lava]

I would LHopital

[slender river]

@odd helm do you know the definition of the derivative

[viscid thistle]

Ngl looks like a test

@odd helm is it a test? Just out of curiosity you know

[odd helm]

oh lhopitals i guess i’ll do that

and yeah it was a test but i already completed it and i got these wrong

wait do i take the derivative of the top and bottom in terms of h

[patent beacon]

Recognize the definition of the derivative haha

[odd helm]

idk how i would do lhopital with this i thought it was a difference quotient problem

[patent beacon]

So 8 is asking "what is the derivative of cos(x) at x = π/3?"

[odd helm]

-sin(pi/3) = -sqrt 3/2

i don’t remember the answer choices but i guess that would be it

[soft verge]

How do I do ex2

I think the eq of motion are F_x=-kx F_y=F_y=-ky

But how do I show that they are circular orbits

[harsh cipher]

Hi

can I ask a java question here?

[willow bear]

maybe? you're most likely gonna be better off asking in a questions channel

[harsh cipher]

hi Ann

I'm very new to java and have a question on a lab I'm working on

Step description: The result of these computation shows the following results

Arithmetic operation: 10000 seconds equals: 2 hours, 46 minutes and 40 seconds

me: System.out.println("10000 seconds equals: 2 hours, 46 minutes and 40 seconds"); lol

[willow bear]

aight ok youve made your problem unclear as always

[harsh cipher]

ok

[willow bear]

are you trying to write a program that takes an input in seconds and converts it to hours/minutes/seconds

[harsh cipher]

Yes

I did that part

[willow bear]

so what's your issue?

you don't know how to output your results?

[harsh cipher]

the last part I typed

[willow bear]

it's not clear to me what the issue is

the issue you're having is that you do not know how to output your results. Y/N

[harsh cipher]

No.

[willow bear]

then what

[harsh cipher]

Do I need to put this sentence in println statement

[willow bear]

can i see the problem specification, exactly as stated?

[harsh cipher]

10000 seconds equals: 2 hours, 46 minutes and 40 seconds.

ok

[willow bear]

what part of "exactly as stated" do you not understand

[harsh cipher]

its coming

[willow bear]

also, you probably DO NOT want this EXACT line of code in your program

System.out.println("10000 seconds equals: 2 hours, 46 minutes and 40 seconds");

[harsh cipher]

wait

[willow bear]

this seems like only part of it

[harsh cipher]

I need to use chain assignment for the last part

[willow bear]

chain assignment?

[harsh cipher]

yea

[willow bear]

what's that meant to be

[harsh cipher]

example

a1 = a2 = a3 = 'B';

character data type

[willow bear]

i see no mention of the words chain assignment in the written problem specification you sent me

[harsh cipher]

yea...I think I'm wrong

[willow bear]

i would appreciate if you didn't pull this thing where at every step there are suddenly new previously unspoken rules that one somehow has to follow

[harsh cipher]

lol

ok

[willow bear]

anyway, i STILL don't understand what your issue is. am i misunderstanding you in your claim that you have already written the program?

[harsh cipher]

yes

I have

my question is

I don't get the very last step

[willow bear]

the very last step is the output

[harsh cipher]

ok

[willow bear]

presumably, you have your output format specifications on page 2

[harsh cipher]

Why is it under arithmetic operation?

It's just a description?

[willow bear]

i don't know, i didn't write the document, i'm not the one to ask

[harsh cipher]

The assignment instructions aren't very clear.

[willow bear]

indeed they are not

[harsh cipher]

I will submit and ask the instructor

As always, thanks for the help 🙂

[willow bear]

i provided none so don't thank me

[harsh cipher]

but you tried

😛

[heady jewel]

@soft verge well the solution to x"(t)=-C*x is the well known SHM solutions which basicaly corresponds to either the x or y axis of a circle

So with F_x=-kx and F_y=-ky you can get x=Asin(k) and y=Acos(k)

And (x,y) corresponds to points on the circle of radius A

[vague aurora]

i was watching khan academy and he drew the complex numbers specifcally as position vectors. I know what those are, but why did he do that instead of just points on the plane?

[blazing parrot]

i would guess that they were trying to relate position vector and complex numbers summation, subtraction and magnitudes

[vague aurora]

ah ok

also are magnitude and absolute value the same thing?

[lime bolt]

yea

[vague aurora]

neat

[acoustic harbor]

say what

[silver matrix]

I'm gonna share a screenshot better

is this correct so far?

My question is basically when deriving the e^... should I also derive the exponent of the argument when deriving this?

I hope i'm explaining myself correctly enough

[viscid thistle]

I mean not gonna lie

You know you can simplify it to make it easier

$e^{\ln{u}}=u$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Instead of having a staircase of exponents

@silver matrix

[silver matrix]

so I can basically say that e^ln u = u and then derive just u (in this case 7/2x^2) ?

and I would not have to deal anymore with that?

or I should anyways put it on the derivative?

[viscid thistle]

so I can basically say that e^ln u = u and then derive just u (in this case 7/2x^2) ?
Yes. Just differentiate u

[silver matrix]

🤔 Alright but i should also put the e^... and the 1/.., no?

[viscid thistle]

Is it a (7/2)^(x²) or a 7^(x²/2)?

[silver matrix]

neither of them (if I haven't done anything wrong)

[viscid thistle]

🤔 Alright but i should also put the e^... and the 1/.., no?
No. Literally

[silver matrix]

it would be 7/2x^2

[viscid thistle]

God please

[silver matrix]

oh no

sorry the first one

7^(1/2x^2 )

no, sorry sorry

yeah

the one you said

which is the same as saying 7^((x/2)^2)

[viscid thistle]

$(\frac72)^{x²}$ yeah?

[obsidian monolith]

Al𝟛dium:

[silver matrix]

no

[viscid thistle]

the one you said
...

[silver matrix]

it would be $7^{1/2x^2}$

[obsidian monolith]

Pa_u_los:

[viscid thistle]

Okay

[silver matrix]

that would be f(x)

it's just reordering the numerator with the root

in exponent form

and that would be u no?

alright...

the thing is

[viscid thistle]

Uh yes

[silver matrix]

that when x on the exponent that's the method I've learnt here

[viscid thistle]

it would be $7^{1/2x^2}=7^{x²/2}$

[obsidian monolith]

Al𝟛dium:

[silver matrix]

yes

[viscid thistle]

Okay

that when x on the exponent that's the method I've learnt here
@silver matrix wdym

[silver matrix]

but why don't use the e^ln

I mean

[viscid thistle]

Bc literally, its unnecessary

[silver matrix]

I was doing derivatives exercises for being introduced into vectorial cinematics when there was a problem where was x as an exponent. I didn't knew how to derive it and in this groupchat someone said me that's that how I had to do it (i did not found any video talking about it) and don't know, if I just have to say that is u and then do division differentiation, ok

oh I know why i can't do that

!

is because when applying the derivative power rule I can't subtract something to x

that's why

[viscid thistle]

$(2^{g(x)})'=2^{g(x)}\cdot g'(x)$

Okay, please don't be a "robot", there are gonna be times where you can simplify and save 15 minutes of your life and others not. So i wouldn't suggest setting a fixed methodology here

It's only gonna give you problems

[stuck lark]

why 2^f on LHS

[viscid thistle]

But like, literally $e^{\ln{u}}=u$ is a useful simplification to use here, that's it

[obsidian monolith]

Al𝟛dium:

[silver matrix]

alright

trust me that I never want to be a robot

thanks

[stuck lark]

mm still wrong

[viscid thistle]

Wdym

[stuck lark]

$(2^f)'=\ln(2)2^ff'$

[obsidian monolith]

RokettoJanpu:

[viscid thistle]

Oh i actually mixed up

Sorry yeah

[silver matrix]

where e is the base or with any base?

[stuck lark]

ln always refers to log base e

[viscid thistle]

Rokabe, jeez i still surprised i mixed both lol

[stuck lark]

wdym mix

[viscid thistle]

where e is the base or with any base?
Ambiguous, do you mean the $e^{\ln{u}}=u$ or the $2^{g(x)}=\ln(2) 2^{g(x)} g'(x)$

[obsidian monolith]

Al𝟛dium:

[stuck lark]

ahh (2^g)'

[viscid thistle]

wdym mix
I mixed the procedure with (x²)' and for example with the (2^x)'

[silver matrix]

I mean that if I have a 7 as a base (this case) should I say that e^... = ln(e) e^f · f' or 7^... = ln(7) 7^f · f' ?

[stuck lark]

i didn't read anything before (2^g)'

oh that's big uh oh

[silver matrix]

i'm starting to be so confused that I don't know if i'm being trolled

[viscid thistle]

Like

You may know what the .. means but i don't

Are we back at the beginning?

[silver matrix]

the function itself

i'm using the formula provided by rokabe

[stuck lark]

whatcha d/dx ing?

[silver matrix]

(e^f)' = ln(e) e^f · f'

[viscid thistle]

Uh fine i'll write it

[silver matrix]

should I use e or 7 in this case?

[stuck lark]

i mean the formula shouldn't be novel to you, it comes from chain rule

[silver matrix]

or in other words, the base of the problem or e?

i am not asking about the formula, i am asking about the base I have to use

because in that case is 2

and in this case I think it's 7

but

[stuck lark]

are you computing $(x\mapsto 7^{x^2/2})'$

[obsidian monolith]

RokettoJanpu:

[silver matrix]

what does even mapsto mean?

and no, im not

[stuck lark]

it just denotes a function or mapping

like x^2 itself isn't a function but x |-> x^2 is

[viscid thistle]

$$f(x)=e^{\ln{7^{x²/2}}}$$ $$f'(x)=(e^{\ln{7^{x²/2}}})'$$ $$f'(x)=(7^{x²/2})'$$ now the moment to use the $(2^f)'=\ln(2) 2^f\cdot f'$

[silver matrix]

that's the original function

it's not 7^(x^2/2) !!

it's 7^((x/2)^2)

[viscid thistle]

Bro

[silver matrix]

i told ya

[viscid thistle]

We already talked about this didn't we

[stuck lark]

so 7^(x^2/4)

[silver matrix]

yes we did

[viscid thistle]

it would be $7^{1/2x^2}=7^{x²/2}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

And you said yes

...

[silver matrix]

sorry yes

[viscid thistle]

Bruh

[silver matrix]

I misunderstood, i thought the /2 was dividing the exponent

my bad

(which wouldn't even make sense)

alright

so I've seen your last latex

and that's the question i had

then

[obsidian monolith]

Al𝟛dium:

[silver matrix]

so no e anywhere then?

so my last step from the screenshot would be wrong?

[viscid thistle]

You didn't even apply the $e^{\ln(u)}=u$? So we are talking different stuff

[obsidian monolith]

Al𝟛dium:

[silver matrix]

yes I did,

but not like that

i did f(x)' applying the chain rule

is similar to that formula

[viscid thistle]

Then you didn't ?!?!

No its not?

[silver matrix]

the first part of the division is e^ln(7 · (1/2)x^2)

so

is that even right?

that's still f(x)

[viscid thistle]

Huh

[silver matrix]

imaging that's right: then differentiate (u' · v - u · v') / v^2

so u' is

[viscid thistle]

Now you are using $e^{\ln((7x²)/2}$?!??

[silver matrix]

e^ln(7^(x/2)^2) · 1/7^((x/2)^2) · ((7/2)x^2)

no I'm not

If you think I have to go one step behind, please recommend me a specific resource and I will be happy to see it

[viscid thistle]

e^ln(7^(x/2)^2) · 1/7^((x/2)^2) · ((7/2)x^2)
The first one, e^(ln(7^(x/2)²)). You KEEP changing stuff

Like i don't know why you want to go that way

Also please

[silver matrix]

I don't want to go anyway. I just don't know how to do it and try to apply what I've seen

More parents than that? I don't know what I've missed..

have you seen my screenshot?

what are my errors?

[viscid thistle]

$(e^{f(x)})'=e^{f(x)}\cdot f'(x)$ which is different from $(a^{f(x)})'=a^{f(x)} \ln(a) \cdot f'(x)$

[silver matrix]

I think going back there can solve some issue s

[viscid thistle]

Setting that up

[silver matrix]

alright I'm gonna copy that

[obsidian monolith]

Al𝟛dium:

[silver matrix]

👍

so that e^ln..

is wrong

is 7^ln(argument)

no?

[stuck lark]

you forgot ' on LHS

[viscid thistle]

God im about to sleep, sorry for making so many typos

[silver matrix]

alright good night, nw, that last one was good

[stuck lark]

additionally you can extend a to be any nonnegative

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

no?
@silver matrix which argument? Im not yet sleeping lol

[stuck lark]

oh actually a in Z is pretty bad

[viscid thistle]

Actually

Hmm

[stuck lark]

very bad actually

[viscid thistle]

My brain is asleep lol

[stuck lark]

remember Z denotes integers

[silver matrix]

would it be then the rationals?

[stuck lark]

a being negative doesn't stop x |-> a^x being well defined over R, but you won't have luck defining a derivative anywhere

[silver matrix]

anyways, I'm going try tomorrow and post what i've done and see where's my possible error

thank you guys

[stuck lark]

the derivative of a^f is all good wherever f' is defined and if a is nonnegative, that's all you need to know

[silver matrix]

the derivative of a^f is all good wherever f' is defined and if a is nonnegative, that's all you need to know
alright, it would be cool to understand your language but sadly nowadays i'm a pleb

and I don't even understand what "well-defined" means

[stuck lark]

my last msg is to you, everything before to aledium

[silver matrix]

alright...

[viscid thistle]

Actually imma go sleep before i make another brainfart.

Cya 🚀

[stuck lark]

there shouldn't be an issue with my language

by a^f, i mean the function that you can view as, the function g defined by g(x)=a^f(x)

night al3

[jaunty spruce]

Given cos(A+B) = cos A cos B - sin A sin B, find cos(2x)

How do I approach this?

[upbeat bone]

Recall that 2x=x+x

From that, you can apply the formula and get what cos(2x) equal to

@jaunty spruce

[opaque olive]

https://gyazo.com/7b82178e0d910fac72263a8030e0304a

could anyone help me on part 3?

i dont see how i can do it, even using the previous parts

inflection points would be the same

[blissful ridge]

Determine all the critical points and their nature and work from there

[opaque olive]

i cant use a calculator

so the most i can do is calculate the points of inflection

and determine their nature

@blissful ridge

[regal kernel]

X = [0, 1, 2, 3, 4, 5] Input
y = [0, 0.3, 2.7, 5.77, 11.7, 17.4] output
What does X equal when y equals 0.6?
@opaque olive

[opaque olive]

something between 0 and 0.3

[blissful ridge]

That's enough to sketch a approximate graph

So if you have got your inflection points and maxima and minima
Sketch accordingly

[opaque olive]

i dont have maximas or minimas

@blissful ridge

[blissful ridge]

You know that at x=0 graph is at 0

Mark that point

[opaque olive]

ok

[blissful ridge]

Mark your inflection point

[opaque olive]

ok

[blissful ridge]

Determine the concavity at the neighborhood of concavity

[opaque olive]

what

[blissful ridge]

Determine if the graph is concave up or concave down

[opaque olive]

ok

okay

i guess ive got it

ty

[blissful ridge]

Np

[regal kernel]

@blissful ridge Did u explain how to solve the problem above?

because I have a graph and labeled that it's concave up, and it has a point of inflection at x=0, but i dont know how to predict datapoints

X = [0, 1, 2, 3, 4, 5] Input
y = [0, 0.3, 2.7, 5.77, 11.7, 17.4] output
What does X equal when y equals 0.6?

@stuck lark

[blissful ridge]

@regal kernel
Did you plot your points?

[regal kernel]

@blissful ridge No, but I sketched a line from the given points. It looks exponential

[nova trail]

uh

can someone help me

rn

yo
uh
can you review this answer?
The probability of Julia missing two shots in a row is 10 percent. We first need to convert 50% and 20% into decimal form because then we can multiply it, using the formula P(A intersection B) = P(A) * P(B). We then solve 0.10 = 0.5*0.2, 0.10 = 0.10 percent. 10 percent is the percent in which she will miss both shots.
question: Julia enjoys shooting paper balls into the wastebasket across her office. She misses the first shot 50% of the time. When she misses on the first shot, she misses the second shot 20% of the time. What is the probability of missing two shots in a row? (10 points)

like

does this answer make sense

[blissful ridge]

Post your question in a channel that is not occupied

[nova trail]

uh

ok

will you help me out

ok I pinged u

[blissful ridge]

@regal kernel
You know the nature of the graph x>0 plot the points using that nature

And what exactly do you wanna find?

[nova trail]

godfather

can you help me out

just for 5 mins

[regal kernel]

@blissful ridge Alright. I just plotted the positive side basically. I'm trying to find an the X value when y = 0.6

Would you know how I could do that? Should I just approximate

[jaunty spruce]

the center of the circle is at 0,0 so p and q are the lengths of the sides of the triangle. Find the hypotenuse using the picture (1) and then use the secant formula, I think it’s hypotenuse over adjacent, is that right?

[blissful ridge]

@regal kernel
The problem is we can't be sure if it's an exponential fucntion

[regal kernel]

@blissful ridge Could it be quadratic? And could I compare them>

?

[blissful ridge]

We can't keep comparing every non linear increasing fucntion

I don't see a other way than just approximate it

Do you have graph paper, the one with grid lines

[regal kernel]

Nope I just have regular notebook paper

But I have a ruler and printer paper

[ionic lichen]

@nova trail I can help with tah in the general chat

[blissful ridge]

@regal kernel try to plot your points as accurate as you can

This will increase the accuracy when you approximate

[regal kernel]

@blissful ridge Alright thanks a lot ill do that. I'll tell you what I get after if you're curious

[blissful ridge]

Sure

[regal kernel]

@blissful ridge I got X = 1.4 when y = 0.6

X = [0, 1, 2, 3, 4, 5]

y = [0, 0.3, 2.7, 5.77, 11.7, 17.4]

[blissful ridge]

Is it correct?

[regal kernel]

The options are between x= 1.3 and 1.5 now, but I ruled out 1.7 and 1.9 from that

Have an idea to check which value is more correct?

[blissful ridge]

Not really, if you don't mind can I see your graph

[regal kernel]

Sure

[blissful ridge]

Complete the graph, i.e join the points and then try to find x

[regal kernel]

Ahh alright. Thanks for taking time to help me

[blissful ridge]

Np

[regal kernel]

That was the issue there also so thanks. x = 1.3 seems like a way better choice then 1.5

Just to rundown, plot the datapoints neatly, connect the points, then approximate the answer

[jaunty spruce]

Can anyone help me with the answer to this? I tried solving with quadratic equation but somehow messed up along the way

[blissful ridge]

Show your work

[willow bear]

"with quadratic equation"

can yall PLEASE stop calling $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ the "quadratic equation"

[obsidian monolith]

Ann:

[blissful ridge]

I think he meant that he put cos(x) as y
And made a quadratic in y or something

[jaunty spruce]

Yes

[blissful ridge]

Show your work
@blissful ridge

[jaunty spruce]

$x = \frac{3 \pm\sqrt{-3^2-4(1)(4)}}{2a}$

[obsidian monolith]

fallen:

[jaunty spruce]

that’s how I set it up

[harsh smelt]

what is a

[jaunty spruce]

1

[blissful ridge]

What is that

[harsh smelt]

and why you have -4

[dry comet]

Should be (-3)²

With the

[harsh smelt]

show the quadratic equation you got?

[blissful ridge]

Just factorise it as you would any quadratic equation

[dry comet]

Also the 4(1)(4) should be 4(1)(-4) since c=-4

[harsh smelt]

i wanted him to see it by himself

[jaunty spruce]

I have a -4 because b^2-4ac is part of the discriminant part of the eq

[harsh smelt]

a = ?

c = ?

[jaunty spruce]

1

4

[blissful ridge]

Where did the minus go?

[jaunty spruce]

-4 , sorry why is that a negative?

[harsh smelt]

y^2-3y=4

thus y^2-3y-4=0

[jaunty spruce]

Oh okay so i change (4)(1)(4) to 4(1)(-4)

[harsh smelt]

so what you will get

[blissful ridge]

You can see the factors right away

[jaunty spruce]

-16

[harsh smelt]

[jaunty spruce]

isn’t it just 4 * -4?

[harsh smelt]

b^2-4ac

what is it

oh you got -16 as -4ac

ye

[jaunty spruce]

Yes

-9 - (-16)

[blissful ridge]

Why -9?

[jaunty spruce]

It’s -3^2 isn’t it?

Wait lol

[blissful ridge]

No, it not

[jaunty spruce]

Nvm

9 - (-16)

[blissful ridge]

Yeah

But why aren't you factoring it?

[jaunty spruce]

Yes so I have to solve for the 2 y’s it’ll give me and then plug it into cos(x)?

[blissful ridge]

Yeah

[jaunty spruce]

So plug in -1.5 and 12.5 into cos(x)?

[harsh smelt]

are u sure with answer

neither of this is in range of cos

[jaunty spruce]

-3/2 and 25/2

[harsh smelt]

nah

these are not roots

[jaunty spruce]

don’t you just do 9-(-16) = 25
and then you’re left with -3+-sqrt25/2

[blissful ridge]

2 divides whole numerator not just sqrt 25

[harsh smelt]

and not -3

[jaunty spruce]

-3+-5/2?

[harsh smelt]

[blissful ridge]

Commander just told you

[jaunty spruce]

Sorry, in a khan academy I watched they divided both the “-3” and what was in the square root

So I’m only dividing -5 by 2?

[blissful ridge]

No, you are dividing the whole numerator

And fix that -3

[jaunty spruce]

sqrt9-16/2

[harsh smelt]

what

[jaunty spruce]

Divide the entire numerator besides -3 is what he said Isn’t it

[harsh smelt]

idk, i never watched khan academy

[jaunty spruce]

No I’m referring to Godfather lol

[harsh smelt]

but you divide the whole expression

And fix that -3
"fix" in the sense correct mistake

repair

[jaunty spruce]

I did and got 3^2 (9)

[harsh smelt]

what that means

[blissful ridge]

For the love of god, show your working

[jaunty spruce]

Y= -3 +-sqrt3^2-4(1)(-4)/2(1)

-3=sqrt9-(-16)
/2

-3=25/2

[harsh smelt]

[jaunty spruce]

What exactly am I doing wrong

[blissful ridge]

Only ann can help you now

[jaunty spruce]

Lol sorry, math has never been my strong suit which is why I’m trying to practice and improve

[willow bear]

please tell me youre not just writing your work out in plaintext only

[jaunty spruce]

$x = \frac{-3 \pm\sqrt{3^2-4(-1)(4)}}{2(-1)}$

[obsidian monolith]

fallen:

[jaunty spruce]

Oops hold on

[willow bear]

also can i see the original equation

because i think x is decidedly not equal to that

[jaunty spruce]

Yes

$y = \frac{-3 \pm\sqrt{3^2-4(1)(-4)}}{2(1)}$

[obsidian monolith]

fallen:

[willow bear]

ok so you put y = cos(x) alright

can you write out what EQUATION you got after putting y = cos(x)
i want you to be explicit so that all mistakes can be caught and weeded out, including sign errors

[jaunty spruce]

(2.5)^2 - 3(2.5) =

[willow bear]

no

that is not what i am asking you

the equation you start with is cos^2(x) - 3 cos(x) = 4

you substituted y = cos(x)

[jaunty spruce]

Yes

[willow bear]

what is the new equation

im trying to get you to do this step by step and not jump ahead at every opportunity

[jaunty spruce]

y2-3y=4

[willow bear]

y**^**2 - 3y = 4.

[jaunty spruce]

y^2

yes

[willow bear]

yes, y^2 - 3y = 4.

or in other words, y^2 - 3y - 4 = 0.

[jaunty spruce]

Okay

[willow bear]

it doesn't sound like you're explicitly required to use the QUADRATIC FORMULA here

and this one in particular factors easily into (y-4)(y+1) = 0

[jaunty spruce]

Okay, so what would be the next step?

[willow bear]

do you know how to solve a quadratic equation by factoring

specifically once the factorization itself has already been done

[jaunty spruce]

No I do not

[willow bear]

...

[jaunty spruce]

sigh

[willow bear]

that's a shame bc it's something you rly should know by now but whatever

i guess i can take you through the QF solution since that seems to be sth you DO know

[jaunty spruce]

I’ll add solving QF by factoring to my watch list

[willow bear]

solving quadratic equations*

give me a minute

i'll get back to you

[jaunty spruce]

Oh wait lol

I think I do know how

[willow bear]

oh you do?

so can you tell me the solutions for y of the equation (y-4)(y+1) = 0?

[jaunty spruce]

Nevermind lol I only remember doing it when “y” was ^2

[willow bear]

there are two options here

either i explain to you, in my own words, how to solve equations by factoring - this will take some time and will also require you to set aside your problem for a while until we are ready to come back to it

[jaunty spruce]

Okay

[willow bear]

or i take you through the solution that involves the quadratic formula without all the sign errors

[jaunty spruce]

Can we do the QF

And then factoring

[willow bear]

okay

[jaunty spruce]

I get up in 2 hours for work but I really want to get this sorted before then

[willow bear]

so your equation is 1y^2 + (-3)y + (-4) = 0

[jaunty spruce]

Yes

[willow bear]

i am writing it this way to make it 100% explicit exactly what the coefficients are

so using the quadratic formula, you get:

$y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times (-4)}}{2 \times 1}$

[jaunty spruce]

failed to render

[obsidian monolith]

Ann:

[willow bear]

there

missed a }

[jaunty spruce]

They told me rhe (-3)^2 was wrong

But carry on

[willow bear]

who

where

i see no instance of "the (-3)^2 is wrong"

[jaunty spruce]

I said “it’s -3^2 isn’t it”

And they said “no”

[willow bear]

there is a difference between -3^2 and (-3)^2

[elfin dirge]

-3^2 ≠ (-3)^2

[jaunty spruce]

Okay

[viscid thistle]

-3^2 = -9, (-3)^2 = 9

[willow bear]

okay so like

[jaunty spruce]

Okay got it

[willow bear]

now it's time to simplify this

carefully, step by step

$y = \frac{3 \pm \sqrt{9 - (-16)}}{2}$

[obsidian monolith]

Ann:

[jaunty spruce]

Yes

[willow bear]

$y = \frac{3 \pm \sqrt{9 + 16}}{2}$

[obsidian monolith]

Ann:

[willow bear]

$y = \frac{3 \pm \sqrt{25}}{2}$

[obsidian monolith]

Ann:

[willow bear]

$y = \frac{3 \pm 5}{2}$

[obsidian monolith]

Ann:

[willow bear]

which gives you your solutions: y=4, y=-1

can you now go back to x and solve for x

[jaunty spruce]

Wait you divided 3 and 5 by 2 right?

[viscid thistle]

No you add and subtract the 3 +− 5

And then divide

[willow bear]

i evaluated $\frac{3\pm 5}{2}$ first taking the $\pm$ as plus (giving 4) and then as minus (giving $-1$)

[obsidian monolith]

Ann:

[jaunty spruce]

Oh okay

[viscid thistle]

Im new to the server how do you use the bot thing?

[jaunty spruce]

That makes sense

[willow bear]

there is a cheat sheet in #resources @viscid thistle

[viscid thistle]

Thank you Ann!

[willow bear]

anyway, fallen,
now that we have solved the equation for y

it's time to go back to x

we have two equations to solve for x

cos(x)=4
and cos(x)=-1

one of these happens to have no solutions at all

[jaunty spruce]

Okay

(4)^2 + (-3)(4)+ (-4) = 0

(-1)^2 + (-3)(-1)+ (-4) = 0

[harsh smelt]

||hint: what is the range of cosine||

[jaunty spruce]

-1 < cosx < 1

[willow bear]

-1 ≤ cos(x) ≤ 1

[jaunty spruce]

How do you do the less than or equal to in chat

[timid lagoon]

<=

[jaunty spruce]

Thank you

[timid lagoon]

says the same thing

[viscid thistle]

Or you could use a scikeyboard on your phone

[jaunty spruce]

So how do I go about finding all the solutions?

Do I do that but substitute my “y’s” for cos(x)

[willow bear]

Do I do that but substitute my “y’s” for cos(x)
well yes

you've done the substitution and are now ready to undo it

and in fact that's what i did

[jaunty spruce]

-1 <= 4 <= 1

-1 <= -1 <= 1

so 4 is out because it’s not less than or equal to 1

[harsh smelt]

4 is out because |4| > 1 if to be precise

[willow bear]

|4| > 1

[jaunty spruce]

Yes

[willow bear]

ok yeah i mean. yeah congrats you've identified the one that has no solutions

now the only hurdle is to solve cos(x) = -1

[jaunty spruce]

And I do that by plugging -1 into the original equation?

[harsh smelt]

no

-1 solves original equation

[willow bear]

forget about your original equation

[harsh smelt]

plug -1 to y = cos(x)

[jaunty spruce]

Okay

.5403023059

[willow bear]

what the fuck

[echo wagon]

Did you take cos(-1)?

[willow bear]

they might've, tbh

[jaunty spruce]

Yes

[echo wagon]

If so, insert Ann's response

[willow bear]

@harsh smelt please refrain from giving unclear/easily misinterpretable instructions

[harsh smelt]

sure, no problem

[willow bear]

fallen, you are not asked for the value of cos(-1).
you are asked to solve cos(x) = -1

i.e.

cos(what) = -1

[jaunty spruce]

okay I’ve got it

Thank you

seriously, thank u

Just to be 100% sure the only solution was pi, yeah?

[willow bear]

yes

[jaunty spruce]

I’m going to imprint this process literally onto the side of my brain I’m sorry I wasted so much time. Thank you

[narrow peak]

hi

me need help

Given that $\frac{4}{n}(3x²)({\frac{2}{9x²}})^{n-2})=\frac{m}{x²}, where x≠0,$ find the values of the constants $m$ and $n$.

fuck latex

[obsidian monolith]

Hmm:

[narrow peak]

there

ping me if u wanna help

[fleet yew]

@narrow peak (a/b)^n = a^n/b^n

[narrow peak]

i split everything up and it became very fucked up

[fleet yew]

Ok

what you should do is mutliply everything together then cross multiply

[narrow peak]

wot

[fleet yew]

After you use that rule of raising a fraction to a power

[narrow peak]

multiply what

[fleet yew]

Multiply together the terms on the LHS after you expand that fraction raised to the power n - 2

[narrow peak]

so i only expand the fraction raised to (n-2)?

then multiply?

[fleet yew]

(a/b)^n = a^n/b^n

Do that first

[narrow peak]

aight

fuck

[fleet yew]

I would suggest leaving the negative powers as fractions

[narrow peak]

Multiply together the terms on the LHS after you expand that fraction raised to the power n - 2
@fleet yew

im confused

[fleet yew]

Basically have an equation in the form a/b=c/d

[narrow peak]

ohh

@fleet yew

[fleet yew]

So uh

Mind posting the original problem

There has to be more context

This is not solvable as is

[willow bear]

hold on what

can i see the original prob again

[spark sorrel]

Suppose we know $A=a^2+b^2+c^2+d^2$, $B=a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2$, $C=a^2b^2c^2+a^2b^2d^2+b^2c^2d^2+a^2c^2d^2$, and $D=a^2b^2c^2d^2$. How can we find $a+b+c+d$?

[obsidian monolith]

MathPhysics:

[willow bear]

you can't

[viscid thistle]

Occupied channel too

[spark sorrel]

Why?

[willow bear]

replacing all instances of a with -a will not affect A, B, C and D but will affect a+b+c+d

ditto for replacing any other lowercase letter with -1 times itself

[spark sorrel]

Ok. Consider only positive values

[willow bear]

a, b, c, d > 0?

[spark sorrel]

Yes

[willow bear]

ok

solve the equation $x^4 - Ax^3 + Bx^2 - Cx + D = 0$ for $x$, then sum the square roots of the solutions

[obsidian monolith]

Ann:

[spark sorrel]

There is not an easier way?

[narrow peak]

@fleet yew find values of m and n

[willow bear]

this is the first thing that my mind jumped to

[narrow peak]

i legit didnt know what to do

my idea was to split everything up and see if i could form 2 equations

[uncut mulch]

simplify your terms and first equate the exponent of x to determine n

[willow bear]

can i see the original question exactly as it was stated

[uncut mulch]

find $m,n$ such that :
$$\frac4n (3x)^2{\br{\frac{2}{9x^2}}^{n-2}}\equiv \frac{m}{x^2}$$

[obsidian monolith]

ramonov:

[spark sorrel]

You are asking me?

[willow bear]

no, i was asking AMD

mathphysics, if you want to continue your question, we should move to #help-9

[spark sorrel]

Yes, for sure.

[willow bear]

$\frac{4}{n} \cdot 9x^2 \cdot \frac{(2/9)^{n-2}}{x^{2n-4}} = \frac{4 \cdot 2^{n-2}}{n \cdot 9^{n-1}} \cdot \frac{1}{x^{2n-2}}$

[obsidian monolith]

Ann:

[willow bear]

idk why yall are overblowing this it's lit that simple lol

[fleet yew]

Not (3x)^2

3x^2

[willow bear]

ok sure but again

it's not as hard as y'all are making it out to be

n is obviously 2, and m can be expressed entirely in terms of n

[uncut mulch]

(I went off the page instead of the original tex)

[narrow peak]

@willow bear how n=2 wot

[willow bear]

your expression is (some constant)/x^(2n-2)

and you want it equal to m/x^2

so 2n-2 = 2

[narrow peak]

ur latex went too fast for my dog brain to follow

can u uh show more steps

pls

xd

ok wait nvm

uh

ok rhs is weird

[willow bear]

it's a bit messy

[narrow peak]

yes it is

how did u get ur right side

of the eqn

just ping me if u come back

[willow bear]

ok so just to be 100% clear on this matter

your original LHS is $\frac{4}{n} \cdot 3x^2 \cdot \paren{\frac{2}{9x^2}}^{n-2}$

y/n

[obsidian monolith]

Ann:

[narrow peak]

no

$(3x)^2$

[obsidian monolith]

Hmm:

[narrow peak]

@willow bear

[willow bear]

ok so amd lied

aight so do you want me to be as verbose as i can possibly go, writing out every single step with meticulously detailed annotations

[narrow peak]

errrrrrrrrrrrrr

no

ill tell u if i dont understand

@willow bear

[willow bear]

yeah ok i found a fuckup

lmfao

[narrow peak]

xd

fuckup where

[willow bear]

the exponent on the x should be 2n-6, not 2n-2

[narrow peak]

er

ok

start from beginning?

my original plan was to split everything and (somehow) form 2 equations

didnt work out

@willow bear

[willow bear]

bruh

what you have there is basically a big messy product

sort the x's from the everything else, clean it all up, find the value of n, plug it in, find the value of m

[narrow peak]

what how

[willow bear]

$\frac{4}{n} \cdot 9x^2 \cdot \paren{\frac{2}{9x^2}}^{n-2} \ \ = \frac{36x^2}{n} \cdot \frac{2^{n-2}}{9^{n-2} x^{2n-4}} \ \ = \frac{36 \cdot 2^{n-2}}{n \cdot 9^{n-2}} \cdot \frac{x^2}{x^{2n-4}}$

[obsidian monolith]

Ann:

[narrow peak]

rip

[willow bear]

literally just this...

like

[narrow peak]

let me look

D:

[willow bear]

dont overthink it

you are massively overthinking it

seriously

this is a routine algebraic simplification

[narrow peak]

i do that to every problem

[willow bear]

you might even say

alge-bruh

[narrow peak]

wow

xd

ohhhhhhhhhhhhhhhhh

shit

the x dont cancel

oh no

@willow bear

[willow bear]

wym they don't cancel

they arent meant to cancel

you should be left with $\frac{1}{x^{2n-6}}$

[obsidian monolith]

Ann:

[narrow peak]

wtf

i got

$\frac{72}{n³9^{n-2}}\cdot{x^{6-2n}}$

[obsidian monolith]

Hmm:

[willow bear]

n^3? excuse me what

where are you getting an extra n^2 from

[narrow peak]

ok its just n

i

braincell glitched

$\frac{72}{n\cdot9^{n-2}}\cdot{x^{6-2n}}$

[obsidian monolith]

Hmm:

[narrow peak]

@willow bear

[willow bear]

where did the 2^(n-2) go tho

and how did you end up with 72

[narrow peak]

https://cdn.discordapp.com/attachments/363224154469826562/742322439677280316/183668144404037632.png

[willow bear]

$2^{n-2} \neq 2n^{-2}$!!!

[obsidian monolith]

Ann:

[willow bear]

what the fuck!!!

[narrow peak]

wait where

im braindead as fuck today

[willow bear]

second line in your work

$2^{n-2} \neq 2n^{-2}$!!! \$2^{n-2} \neq 2n^{-2}$!!! \$2^{n-2} \neq 2n^{-2}$!!! \$2^{n-2} \neq 2n^{-2}$!!! \$2^{n-2} \neq 2n^{-2}$!!! \

[obsidian monolith]

Ann:

[narrow peak]

oh fuck handwriting

LOL

dis better?

@willow bear

[willow bear]

sure now it's ok

[narrow peak]

ok what now

[willow bear]

you want this to equal some constant times x^-2

match the exponents on the x

6 - 2n = -2

solve for n

[narrow peak]

ohhhhhh

ok done

this took 10 years because i am braindead as fuck

ok next

The equation of a circle, $C$, is $x²+y²-4ux+2uy+5(u²-20)=0$ where $u$ is a positive constant. Determine the value of $u$ for which the line is a tangent to the circle, $C$.

[obsidian monolith]

Hmm:

[narrow peak]

first part of the qn asked me to find center which i did

aka (12,-6)

i vaguely remember a circle property

something about tangent perpendicular to radius??

am i right

[willow bear]

what line.

[narrow peak]

huh wym

[willow bear]

7uhgermhng

ugh

nvm

[narrow peak]

ill just take pic

,rccw

[narrow peak]

5b(ii)

[willow bear]

the line x=2

why......... did you not say this

[narrow peak]

because i braincell glitched again

fuck me

ok idk what to do

help

[blissful ridge]

How do you determine the centre of the circle from the given equation

[narrow peak]

complete the square

also sub in u=6

[blissful ridge]

Yeah, but there is a other method as well

[narrow peak]

the general form

i found centre already

its 5b(ii)

[blissful ridge]

In part B you don't have the value of u

You have to determine u

[narrow peak]

yes

[blissful ridge]

So again I'm asking,
How do you determine the centre from the general equation of circle

[narrow peak]

i mean theres only 2 ways

complete the square/general form

[blissful ridge]

Yeah, general form

So what's the centre

[narrow peak]

(oh)

its (-12,-6)

[blissful ridge]

What part are we doing?

[narrow peak]

b(ii)

they asked for centre in a

[blissful ridge]

Yeah, then write the centre in the form of u

What's the coordinates of centre

[narrow peak]

er

wait

[blissful ridge]

It's variable in u

[narrow peak]

(-2u,u)

??

[blissful ridge]

The centre is (-g,-f) in the general equation

[narrow peak]

oh shit

(2u,-u)??

[blissful ridge]

Yeah

Good

Now you have the equation of tangent, what does that tell you

[narrow peak]

do i use circle property

the tangent perpendicular to radius

[blissful ridge]

What is the condition for tangency for circle

[narrow peak]

radius is perpendicular to the tangent

i think

[blissful ridge]

Yeah, now perpendicular distance from centre to tangent is equal to the radius of the circle

Use that condition to solve for u

[narrow peak]

ohh

[silver matrix]

Source: https://discordapp.com/channels/268882317391429632/363224154469826562/742131708295643228

This does not make sense for the example I have.

$f(x) = \frac{7^{\frac{1}{2}x^2}}{x^\frac 32}$

[obsidian monolith]

Pa_u_los:

[silver matrix]

Then f'(x) applyin that logic would be..

$f'(x) = 7^{\frac{1}{2}x^2} \cdot \ln(7) \cdot x$

[obsidian monolith]

Pa_u_los:

[silver matrix]

well

that's not that bad, i misconfund myself and thought that ln(7) was 0. My bad, let's continue.

(that's the derivative of the numerator, not the whole function)

[harsh smelt]

m?

[viscid thistle]

Hello?

You didn't even apply correctly the quotient rule

[silver matrix]

yeah sorry, I meant u'

now I'm gonna share the screenshot

of what I've done

,rotate

[silver matrix]

sorry, but I might have send the same thing multiple times bc of my monkey wifi

is this correct?

[harsh smelt]

seems to be alright

[silver matrix]

great thanks

[harsh smelt]

simplify tho

[viscid thistle]

There is a mistake unless your 6 is a 5 irl

I can't even tell if its a 6 or a 5

[harsh smelt]

i would say it is 5

since there is small gap

[viscid thistle]

@silver matrix is it a 5 or a 6?

[harsh smelt]

it is 8

[viscid thistle]

Lmao

[harsh smelt]

it is integer between 5 and 6

[silver matrix]

yeah it's a 5, thanks

[jaunty spruce]

Is this correct?

[opaque olive]

can someone remind me what the formula was to evaluate this $\sum_{r=0}^{a-1}2^r$

[obsidian monolith]

Yes:

[opaque olive]

<@&286206848099549185>

[opaque olive]

ok thanks

wait

i thought a geometric sequence is one that has a common ration between each term'

@viscid thistle

[viscid thistle]

http://www.phengkimving.com/calc_of_one_real_var/09_the_intgrl/09_01_summ_notatn_and_formulas.htm

I thought this was what you wanted, if not sorry about that

At the bottom of the page

[opaque olive]

ah this is what i wanted lol ty

[viscid thistle]

np

[wide rampart]

quick question how do i in (√x)'

derivative of root

nvm im dumb

i can go x^1/2

[ember remnant]

What is the difference between Sin and Cos

why is t-15 for cos?

and why is it -10 not 10

all questions are for #3

@viscid void

someone help please

[vivid glacier]

hello Mr. Lemons I'll try to help you

I'll answer the t and t-15 first

inside the sine, (pi/30)t is just (pi/30)t

inside the cosine, (pi/30)(t-15) = (pi/30)t-(pi/30)15 = (pi/30)t-pi/2

the -pi/2 inside the cosine is the amount that the cosine function have to be shifted to be the same as sine

example

if t = 0 we have sin(0) and cos(-pi/2) = 0

if t = 15 we have sin(pi/2) and cos(0) that are both equal to 1

that will work for all values of t

and the -10 in the front say like the upper and lower bound of the range of the function

and it's negative so it begins decreasing and not increasing

@ember remnant I hope I've helped you

I really wish you don't get annoyed with pings

[narrow peak]

hellO

,rccw

[narrow peak]

found a

cant find the rest

idk how u can get 3 equations from this when they only gave 2 points

[elfin dirge]

Think about if there was no c

The graph would be on the x axis

[narrow peak]

how 2 get b

[elfin dirge]

But clearly it isn't, so c must control the height

So c must be the height of the vertex

[narrow peak]

let me process what u just said gimme a sec

[elfin dirge]

Think about what each variable does

a changes the slope, b changes the horizontal offset, c changes the height

[narrow peak]

i think a controls gradient

whats horizontal offset

[elfin dirge]

Left and right translation

[narrow peak]

is it shifting graph right and left

[elfin dirge]

Yes

[narrow peak]

oh

then c controls (12,2) aka vertex

[shadow plaza]

a changes the slope, b changes the horizontal offset, c changes the height
@elfin dirge b contributes in variance of height along with c

[narrow peak]

what is happening

[elfin dirge]

? No it doesn't

[narrow peak]

!!!!!!!!!

[elfin dirge]

Sidharth think about it. It doesnt

[narrow peak]

so what is what

[elfin dirge]

so yes b is the left and right shift

[shadow plaza]

b doesnot change horizontal offset. I mean you cannot be certain that it does until you have the equation of line in standard form

[narrow peak]

whats a standard form

y=mx+c?

[elfin dirge]

Sidharth it's not a line it's an absolute value function

Sidharth is confused, ignore him

[narrow peak]

er

sure

[elfin dirge]

So

Well we know how much the graph is shifted by

12 units

so we have to solve 12 = b/a

And you said you found a

[narrow peak]

why b/a huuhhhhh

[elfin dirge]

Because usually when translating functions, it's af(x-b)+c

[narrow peak]

never heard of that one before

[elfin dirge]

that's the general equation for that sorta thing

[narrow peak]

thats not in my syllabus i think

[elfin dirge]

Oh ok

So since a is only applied to x and not the whole function, we have to divide by a when solving for b

[narrow peak]

what i am confused why

[elfin dirge]

Because for a to be the slope, the equation would have to be a|x+b|

But our equation is |ax+b|

Which is equivalent to a|x+b/a|

so b/a is our shift term

[narrow peak]

u factor a out oh

seems legit

[elfin dirge]

Alright

So you said you found a, correct?

[narrow peak]

yes

[elfin dirge]

Did you get -1/2?

For a

[narrow peak]

yep

WAIT

[elfin dirge]

What

[narrow peak]

i got a=1/2

[elfin dirge]

Oh so slope is rise/run as you know

[shadow plaza]

Jack papel, i agree it's an absolute value function but it is made of two lines right? So suppose i try to break the modulus and write the equation for say left part (having the gradient positive), that would go as:
y = c-(-(ax+b))= c+ax+b= ax+(b+c)
Comparing it with y=mx+C we get,
C= b+c which is altogether responsible for the intercept on y axis.

[elfin dirge]

The rise is negative

Sidharth I swear to God graph the function and you'll see I'm right

[narrow peak]

i sent pic of graph anyway

[elfin dirge]

Anyway so if a = -1/2 let's solve this then

[shadow plaza]

Jack, explain to me where do you think i am going wrong.

[narrow peak]

i got a=1/2 tho

[elfin dirge]

Oh so slope is rise/run as you know
The rise is negative

[narrow peak]

oh