precalculus

[valid violet]

Before you agreed

[harsh cipher]

type it w $

[valid violet]

$\log_{a^b}(x)=\frac{\log_2(x)}{\log_2(a^b)}$

[obsidian monolith]

gfauxpas:

[harsh cipher]

yes I agree

[valid violet]

$y = \log_{a^b}(x)$

[obsidian monolith]

gfauxpas:

[valid violet]

$r= \log_2(x)$

[obsidian monolith]

gfauxpas:

[valid violet]

$s=\log_2(a^b)$

[obsidian monolith]

gfauxpas:

[valid violet]

$y = \frac rs$

[obsidian monolith]

gfauxpas:

[valid violet]

Yes or no

[harsh cipher]

Yes

[valid violet]

$t=b \log_2(a)$

[obsidian monolith]

gfauxpas:

[valid violet]

$s=t$

[obsidian monolith]

gfauxpas:

[valid violet]

Agree?

[harsh cipher]

yep

[valid violet]

$y = \frac rt$

[obsidian monolith]

gfauxpas:

[valid violet]

Agree?

[harsh cipher]

k

[valid violet]

That's what you were having trouble with, idk why

Now substitute back in y, r, and t

[harsh cipher]

https://cdn.discordapp.com/attachments/363224154469826562/656569590633201664/205675981845954560.png

that's what you wrote here

[valid violet]

Yes, and you were skeptical

[harsh cipher]

and this

https://cdn.discordapp.com/attachments/363224154469826562/656569619166920704/205675981845954560.png

[valid violet]

Yes

That's what we just did

[harsh cipher]

you just made it with s = t

to make it simpler

or cuz i didn't get it....

😦

[valid violet]

Now do you agree you can substitute back in y, r, and t for the log expressions?

Khan academy uses this technique of introducing variables for everything

[harsh cipher]

I watched the vid yesterday...

[valid violet]

I meant in general in all subjects

He introduces lots of temporary variables

[harsh cipher]

okay can you show me that

[valid violet]

I just did

[harsh cipher]

how to substitute back in y,r and t

[valid violet]

Oh

What was y equal to

$y = \log_{a^b}(x)$

[obsidian monolith]

gfauxpas:

[valid violet]

$r= \log_2(x)$

[obsidian monolith]

gfauxpas:

[valid violet]

$t=b \log_2(a)$

[obsidian monolith]

gfauxpas:

[harsh cipher]

and you stated that t = s

[valid violet]

We don't need s anymore

Disposable plasticware

Use and discard

[harsh cipher]

okay

let's suppose I agree that y= r/t

and I substituted the log expressions

[valid violet]

We proved it D:

Why wouldn't you agree, we proved it together

[harsh cipher]

Okay

$t=blog_2(2)$

[obsidian monolith]

אewb64:

[valid violet]

[harsh cipher]

i mean

$t=blog_2(a)

$t+blog_2(2)$

[obsidian monolith]

אewb64:

[harsh cipher]

ahhh

[valid violet]

Yes. Usse \log

b\log_2(a)

[harsh cipher]

$t=b\log_2(a)$

[obsidian monolith]

אewb64:

[harsh cipher]

I got another question...

[viscid thistle]

Alec

[runic cradle]

Can someone confirm that you get 28/15 for the answer?

[uncut mulch]

since 28/15 > 1, its already wrong

@runic cradle

[runic cradle]

:(

[uncut mulch]

show your work so i can see what you messed up

[runic cradle]

i got -56/65 on the actual test that im redoing to study and i cant figure out how i got that

ye hold on one second

Sorry I sent them backwards but hopefully it’s still intelligible @uncut mulch

Ignore the red text above the second image. They’re some unrelated notes

[uncut mulch]

how are you getting 30 in the denominator?

[runic cradle]

im an idiot

it should be 65

[uncut mulch]

yeh

[runic cradle]

but my friend said -56/65 was the right answer and i would get positive that

am i correct or is he?

[uncut mulch]

56/65 would be correct

[runic cradle]

alright thank you!

i always make mistakes like that sorry for bothering

[uncut mulch]

np

[jaunty mason]

a + b = -3, a^2 + b^2 = 7

find ab

How is this not -1?

[fading token]

Hint: calculate (a + b)²

[jaunty mason]

i know

[fading token]

But then you should know that $9=(a+b)^2=a^2+b^2+2ab=7+2ab$ which means that $2=2ab$, i.e. $ab=1$

[obsidian monolith]

Rijinaru:

[jaunty mason]

NVM i had expanded (a+ b)^2to a^2 + b^2 - 2ab for some reason lmao

my bad

[fading token]

That's indeed the incorrect expansion

[jaunty mason]

edited

[fading token]

edited

[jaunty mason]

The equation $x^2 + bx + c = 0$ has roots $\alpha$ and $\beta$

[obsidian monolith]

Tāhā:

[jaunty mason]

Prove that if $\alpha = \beta - 2$, then $b^2 = 4(c + 1)$

[obsidian monolith]

Tāhā:

[jaunty mason]

i don't think you have to touch the discriminant

$\Sigma\alpha = -b$ and $\alpha\beta = c$

[obsidian monolith]

Tāhā:

[jaunty mason]

You get a simultaneous system of equations but I'm not sure how you solve from there

[fluid shore]

The discriminant is unnecessary. Just use the equations above and you should get it

$\alpha + \beta = 2\beta - 2 = -b$

$\alpha \beta = (\beta -2 ) \beta = \beta^2 - 2\beta = (\beta - 1)^2 -1 = c$

[obsidian monolith]

Abhijeet Vats:

[jaunty mason]

and then substitute the value for beta from the first equation?

[fluid shore]

Well, from the first equation, you have:

$\beta - 1 = -\frac{b}{2}$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So you don't have to solve for $\beta$. All you need to do is recognize the above relation.

[obsidian monolith]

Abhijeet Vats:

[jaunty mason]

gotcha

[naive sage]

yo
anyone free i need help pls
❤️

[fluid shore]

What does it mean for y to vary inversely with x?

[naive sage]

idk lol, fkin hate there questinos

i mistakely attempted my quiz online

now i gott afinish it 😭

[fluid shore]

Well, okay, idk if this actually breaks the rules since we're not supposed to provide help to you during a test or examination :/

[naive sage]

this just a quit hehe

[viscid thistle]

should've cropped it out lmao

[naive sage]

hehe

[jaunty mason]

Ok, how about this

A quadratic equation has roots $\alpha$ and $\beta$. Given that $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{2}$ and $\alpha^2 + \beta^2 = 12$, find two possible quadratic equations that satisfy these values.

[obsidian monolith]

Tāhā:

[fluid shore]

Well

[jaunty mason]

I cannot seem to solve the simultaneous equations I get 👀

[fluid shore]

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1}{2}$

[obsidian monolith]

Abhijeet Vats:

[jaunty mason]

should i tell you where i got?

[fluid shore]

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Sure

[obsidian monolith]

Abhijeet Vats:

[jaunty mason]

$2\alpha\beta(2\alpha\beta - 1) = 12$

[obsidian monolith]

Tāhā:

[jaunty mason]

and $\alpha\beta = 2\sqrt{12+2\alpha\beta}$

[obsidian monolith]

Tāhā:

[fluid shore]

Well, okay, here's the thing right

Your quadratic equation has roots $\alpha$ & $\beta$, correct?

[jaunty mason]

mhm

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So, you need to form something like:

$(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Can you form something like the above from what you have?

[jaunty mason]

i dont think so, no

I mean, I cannot do it 😛

nvm, just realised https://discordapp.com/channels/268882317391429632/363224154469826562/656859848847261707 is a quadratic equation 🤦‍♀️

idk why i'm having such mental slowdown today

[fluid shore]

Lel

[jaunty mason]

The roots of the equation $x^3 + 4x - 1 = 0$ are $\alpha$, $\beta$ and $\gamma$. Use the substitution $y = \frac{1}{1 + x}$ to show that the equation $6y^3 - 7y^2 + 3y - 1 = 0$ has roots $\frac{1}{1 + \alpha}$, $\frac{1}{1 + \beta}$ and $\frac{1}{1 + \gamma}$.

[obsidian monolith]

Tāhā:

[jaunty mason]

How do I make the substitution?

$(x -\alpha)(x -\beta)(x -\gamma) = 0$ ,basically this?

[obsidian monolith]

Tāhā:

[willow bear]

$6\paren{\frac{1}{1+x}}^3 - 7\paren{\frac{1}{1+x}}^2 + 3\paren{\frac{1}{1+x}} - 1 = 0$

[obsidian monolith]

Ann:

[jaunty mason]

Context? I don't know what's that supposed to tell me ^^

[willow bear]

you asked how to make that substitution

i made it for you

[jaunty mason]

nah, just figured it out you're supposed to make the substitution in the first equation and that'll give you the second equation

you're supposed to make $x$ the subject in $y = \frac{1}{1 + x}$ first

[obsidian monolith]

Tāhā:

[jaunty mason]

But how do I assert that the equation $6y^3 - 7y^2 + 3y - 1 = 0$ has roots $\frac{1}{1 + \alpha}$, $\frac{1}{1 + \beta}$ and $\frac{1}{1 + \gamma}$.

[obsidian monolith]

Tāhā:

[jaunty mason]

Without actually calculating the values of $\Sigma\frac{1}{1 + \alpha}$, $\Sigma\paren{\frac{1}{1 + \alpha}\paren{\paren{\frac{1}{1 + \beta}}}$ and $$?

[fluid shore]

Okay wait

Calm down, yea

Have you used the substitution to check if the two equations are equivalent to each other?

[jaunty mason]

mhm

[fluid shore]

Okay, so solving one will give you the roots to the other, yes?

Let $\alpha$, $\beta$ & $\gamma$ be the roots of the original cubic. Then:

$x^3 + 4x - 1 = (x-\alpha)(x-\beta)(x-\gamma)$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Now, use your substitution for that right-hand side and solve for y.

You'll have 3 possible answers for y.

[jaunty mason]

Not sure I follow

You want me to actually solve the cubic?

[fluid shore]

Nope. Since you've shown that the two equations are equivalent through that substitution, you've basically shown that:

$\frac{1}{y} - 1 = \alpha \lor \frac{1}{y} - 1 = \beta \lor \frac{1}{y} - 1 = \gamma$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So now, solve for y in each case.

[jaunty mason]

what does \lor mean 👀

is that boolean algebra?

[fluid shore]

Yea don't worry about that for now

It just means 'or'

[jaunty mason]

a-ha

[fluid shore]

Basically, i'm listing the possible values of x

[jaunty mason]

can ab + ac + bc be written in another way?

[willow bear]

with this little context, no it can't

[jaunty mason]

general context ig

[willow bear]

what do you need this for

[jaunty mason]

to simplify $\paren{ab + a + b + 1}^2 + \paren{ac + a + c + 1}^2 + \paren{bc + b + c + 1}^2$

[obsidian monolith]

Tāhā:
Compile Error! Click the reaction for details. (You may edit your message)

[stuck lark]

$\br{ab + a + b + 1}^2 + \br{ac + a + c + 1}^2 + \br{bc + b + c + 1}^2$

[obsidian monolith]

RokettoJanpu:

[jaunty mason]

yeah

[willow bear]

you're gonna have to type the parentheses yourself without a custom command

anyway w/e

hhh

[jaunty mason]

custom commands? 👀

[willow bear]

how familiar are you with LaTeX

i'm guessing not very

[jaunty mason]

i don't mean to get political

[willow bear]

wha

[jaunty mason]

but what in f r i c k is a latex

[willow bear]

https://en.wikipedia.org/wiki/LaTeX

it's the language this bot uses

[jaunty mason]

a-ha

anyways

[willow bear]

uhhhhhhh

well you COULD write your thing as (a+1)^2(b+1)^2 + (b+1)^2(c+1)^2 + (c+1)^2(a+1)^2?

i guess?

idk.

i don't really want to spend time fucking around with symmetric polynomials rn

[jaunty mason]

yes lmao, that's what i expanded to get this 👀

[willow bear]

,,,

[jaunty mason]

I'm not sure I have to do this in the first place, the question asks me to find the value of $\frac{1}{(\alpha + 1)^2} + \frac{1}{(\beta + 1)^2} + \frac{1}{(\gamma + 1)^2}$

[obsidian monolith]

Tāhā:

[willow bear]

h

i'm gonna tab out

i have other shit to do

[jaunty mason]

ok

part of this question https://discordapp.com/channels/268882317391429632/363224154469826562/656870118277775382

[jaunty mason]

Can someone look into this ^^

[tardy crescent]

Hey @jaunty mason , to solve your problem you can substitute (1/1+x) for y in the second equation, and then find common denominators. What you need to do is rewrite the second equation in terms of x, that will make things clear

$6\paren{\frac{1}{1+x}}^3 - 7\paren{\frac{1}{1+x}}^2 + 3\paren{\frac{1}{1+x}} - 1 = 0$

[obsidian monolith]

VitalSine:
Compile Error! Click the reaction for details. (You may edit your message)

[tardy crescent]

So you do that substitution, and what you get is: 6/(1+x)^3 - 7/(1 + x)^2 + 3/(1 + x) - 1 = 0

Now make a common denominator, (1+x)^3 in all your terms. So you multiply the second term by (1+x) / (1+x), the third term by (1+x)^2 / (1+x)^2, and the last term (-1) by (1+x)^3 / (1 + x)^3

Once you have the common denominator, just multiply out the parentheses in the numerator, combine like terms, and you'll have an equation in x

It will be the negative of your first equation, so -x^3 - 4x + 1, divided by (1 + x)^3. It works out real nice like that, as The zeroes of a rational function are the zeroes of its numerator (with denominator not equal to zero), so the zeroes of this equation are the same as the zeroes of your 1st equation, a and b and c

However, your second equation was originally in y,

and y = 1/1+ x, so whatever x values resulted in a zero when the 2nd equation was in x, you're going to want to plug them into 1/(1 + x) to find the y values that will also result in a zero

And then that gives you 1/(1 + a), 1/(1 + b), and 1/(1 + c)

[viscid thistle]

Am I on the right track

[tardy crescent]

Hey @viscid thistle , number 3 looks good to me, and I couldn't see number 6 on your page. Number 1 also looks good to me. But for number 2, the two cylinders are not similar. If they really were similar by a factor of k = 2 cm, then the bigger cylinder would have dimensions 10 and 4, not 10 and 5, as 2 * 5 = 10 and 2 *2 = 4

[viscid thistle]

hey guys

so i’m graduating college and im ganna have to take calculus.. i haven’t done pre calculus.. i wanna learn.. recommendations?

[jaunty mason]

How to use the Newton identities for n < 0 ?

i.e. how do I use the Newton identities to find $\frac {1}{\alpha^n} + \frac {1}{\beta^n} + \frac {1}{\gamma^n}$

[obsidian monolith]

Tāhā:

[surreal prism]

hey can anyone help with this logarithm question: 6.44^{3x-2}+4.34^{x+3}=45

i'm in high school taking honors precalculus and we had a sub today so i did not understand how to do this

[timid dock]

all i want is the work and answer

i wanna find easier ways to do this problem

[willow bear]

what's int()

also we don't exactly give out answers here

[timid dock]

greatest integer function

or step function

or floor function

heres my problem

the way my teacher taught it, I don't really get why anything happens

[willow bear]

can you show what your teacher did

[timid dock]

why does x ∈ [-3,1) determine the domain of the composite function

[willow bear]

wow there sure are a lot of things going on here

ok so this is for h(g(x))

[timid dock]

yes

[willow bear]

alright

well

in order for h(g(x)) to be defined, you need two things

1. x needs to be in the domain of g
2. g(x) needs to be in the domain of h

does it make sense to you why this is what you need to have h(g(x)) be defined

[timid dock]

not really

[willow bear]

which part doesn't make sense

[timid dock]

what happens if x is not in the domain g

[willow bear]

if x isn't in the domain of g, then g(x) is undefined

[timid dock]

and same way with h(x)

[willow bear]

if g(x) isn't in the domain of h, then h(g(x)) is undefined

does it make sense now

[timid dock]

yes

so if the domain of g is [-3,3]

and g(x) ∈ domain h is [-3,1)

wait shit

am i retarded

[willow bear]

no

g(x) is in the domain of h iff x in [-3, 1)

so now take intersection, since you want both of these to be true at once

[timid dock]

ok but why does [-3,1) restrict [-3,3]

[willow bear]

??

$[-3,3] \cap [-3,1) = [-3,1)$

[obsidian monolith]

Ann:

[timid dock]

ok i never thought about it that way

[willow bear]

so now take intersection

[timid dock]

could you explain it on a number line since i might have to do that

[willow bear]

i mean

can you draw those two intervals on a number line

[timid dock]

yes

[willow bear]

do that and show me what you get

[timid dock]

wait im actually so stupid

please tell me this happens to everyone

ive been thinking about this wrong for so long

[willow bear]

this happens a lot

can't say if it's universal but it is common

[timid dock]

phew

btw is learning latex easy

in your opinion

[noble halo]

hello everyone

what's throwing me off is the negative powers

[timid dock]

that's just inverse

[noble halo]

and i was never taught the t=x^-1

[viscid thistle]

make them into positive powers then

[willow bear]

what do you mean "never taught the t=x^-1"

have you not been taught the concept of substitutions

or are you explicitly forbidden from using that substitution here

[noble halo]

i haven't been taught the concept of substitutions

[willow bear]

,,,,

wow.

you're missing out.

[noble halo]

😭

why'd that turn into an emoji

anyway

[willow bear]

look up "algebra substitutions"

[noble halo]

so how would i change it to positive power

oh ok

[willow bear]

but if you want to solve it without... you can multiply both sides by x^2

which you can do since x isn't allowed to be zero anyway

[viscid thistle]

isnt that simpler than using substitutions anyway

[noble halo]

is there another way to do it besides substitution

[viscid thistle]

yes the method ann just said

[noble halo]

oh what you just said

-8 & -6?

[viscid thistle]

sounds correct

[noble halo]

also another thing

why would i specifically want to use x^2 to multiply by both sides

[viscid thistle]

to get an easy to factor quadratic

[willow bear]

to get a quadratic in the first place

x^2 * x^-2 = 1

x^2 * x^-1 = x

[jaunty mason]

https://discordapp.com/channels/268882317391429632/363224154469826562/657068811362500628

^^ Can someone help me with this?

[jaunty mason]

<@&286206848099549185> ^

[jaunty mason]

@ Helperzz :>

[obsidian monolith]

FNDN:

[willow bear]

h

did you mean

$\tan(a) + \tan(b) = \tan(\frac{a+b}{2})\tan(\frac{a-b}{2})$

[obsidian monolith]

Ann:

[willow bear]

bc this does not seem like it would be true in general

...

[serene heath]

[willow bear]

that's even worse??

that doesn't hold in general either??

wtf do you want lmao

what do you mean by "right"

neither of your two identities are true in general

[serene heath]

[echo plaza]

lol

[viscid thistle]

That’s easy

[viscid thistle]

ok..?

[odd helm]

Ok

[pseudo sonnet]

just came in here to say thanks to all the <@&286206848099549185>

got an A in my college course for precalc

thanks yall 🙂

[clever inlet]

😍

[heady jewel]

😍 nh

[zenith bough]

why the asymtotes of this is 3?

if I make the denominator = 0 then isnt it 2?

[fluid shore]

Well, think about what happens as $x \to \pm \infty$

[obsidian monolith]

Abhijeet Vats:

[zenith bough]

do i have to continuously add numbers in for x?

[fluid shore]

? I mean, you are making x arbitrarily large in both directions. You have to manipulate your expression in order to get the limit

[zenith bough]

is there formula to find the asymptotes for that equation?

[fluid shore]

Well:

$\lim_{x \to \pm \infty} \frac{3x}{\sqrt{x^2+4}}$

Fk

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Try calculating that. There's no formula that'll tell you how to do it. Algebraic simplications to the given expression can get you close to the answer

[zenith bough]

Thank you

[heady jewel]

Np

[fluid shore]

Wow lel

[odd helm]

I’m lost on c

[odd helm]

This is what I drew

<@&286206848099549185>

[grizzled turret]

How do I prove the following?

[fluid shore]

@odd helm Here's a hint: work with variables instead

Also, draw a big diagram

Your diagram is way too small

@grizzled turret What have you tried?

[grizzled turret]

If I understand correctly,

I would have to use the following identity:

Solving 6 / (3 * sqrt(5)) = x / sqrt(1+x^2) gives me x = 2

[harsh hamlet]

How can I solve for x here? I tried sum difference formula, multiplying by conjugate, similar denominators, but I never get a factorable solution

[fluid shore]

I would, personally, not recommend using identities you've memorized. Solve it from first principles.

No, you have arctan(4/3). So, x = 4/3

[fluid shore]

@harsh hamlet Please go to #help-2

Ah okay you're simplifying the arcsine

Idk why you'd do that though

If you apply a tangent on both sides, you'll get a pretty nasty infinity on the other side

[grizzled turret]

Then how would you proceed

[fluid shore]

I'd simplify the arctan instead

I'd turn that into an arcsine

[fluid shore]

Mmh go on

Also, $\frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}}$

[obsidian monolith]

Abhijeet Vats:

[grizzled turret]

I've no idea what to do next ;U; I'm trying to find a way to get rid of one of the sin by cancellation or by combining or something

[fluid shore]

Well, let us have the following notation:

$\theta = \arcsin(\frac{2}{\sqrt{5}})$

$\phi = \arcsin(\frac{4}{5})$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Where I've simplified the stuff in the second arcsine

[grizzled turret]

Yes

[fluid shore]

So, you can see that:

$\frac{\pi}{2} = \theta + \phi$

That's what you have to prove. So, the right thing to do would be to apply a sine to the right side and see what it gives you

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

At least, that's the approach that i can think of from the top of my head

[grizzled turret]

hold on, you removed one of the division by two

\theta + 1/2 \phi$

[fluid shore]

Ah yes, forgive me

I just woke up lol

But anyways, try to approach it from that angle

See what you get

[grizzled turret]

Welp, stuck..
If this can help solve the problem :

[fluid shore]

:/

[uncut mulch]

well that form does make things easier to calculate

[native sequoia]

Abhijeet suggested to take sine of both sides

[uncut mulch]

have you ever used
sin( a + b)?

[grizzled turret]

Taking sine of both sides I'd get:

2 * Θ + Φ = 0

wait no

[uncut mulch]

nope, found another method of killing puppies

[fluid shore]

???????

[patent beacon]

That was indeed a puppy death right there

[fluid shore]

Ramonov......you okay?

[grizzled turret]

How would you simplify sin(2 * Θ + Φ)

[fluid shore]

Use your addition formula

[uncut mulch]

do you know the identity involving:
sin(a+b)

[patent beacon]

You are assuming that sin(a + b) = sin(a) + sin(b)

[grizzled turret]

Sin(a+b)=sin(a)cos(b)+sin(b)cos(a) ?

[uncut mulch]

lowercase s but yes

[native sequoia]

that gives me an idea, "all functions are linear" would be a good nickname

[grizzled turret]

And how do you go from sin(a)cos(b)+sin(b)cos(a) = 0 ;s

[uncut mulch]

apply the identity to
sin(2Θ + Φ)

[grizzled turret]

sin(2Θ)cos(Φ) = -sin(Φ)cos(2Θ)

[uncut mulch]

that isn't what you want

you have the values of theta and phi

and you want to show that
sin(2Θ)cos(Φ) - sin(Φ)cos(2Θ) simplifies to 0

[grizzled turret]

yes

But how

[uncut mulch]

first apply the double angle identities for sin(2Θ) and cos(2Θ)

[grizzled turret]

sin 2Θ = 2 sin(Θ)cos(Θ)

[uncut mulch]

given:
$$\theta = \arcsin(\frac{2}{\sqrt{5}})$$
$$\phi = \arcsin(\frac{4}{5})$$
can you determine the values of: $\sin(\theta),\cos(\theta),\sin(\phi),\cos(\phi)$?

[obsidian monolith]

ramonov:

[uncut mulch]

um where are you getting that?

[grizzled turret]

half angle identity

[uncut mulch]

oh... here, we are using phi to represent arcsin(4/5)

[uncut mulch]

that's different from the original question you posted

[uncut mulch]

and earlier we used, theta and phi to represent:

[grizzled turret]

Yes, because:

A mistake was done there

[uncut mulch]

makes writing less painful, just a typo

currently:

https://cdn.discordapp.com/attachments/363224154469826562/658140876815597586/unknown.png

is fine

values of theta and phi are what they should be

i.e.
2(theta) + phi = pi

[grizzled turret]

alright

So we get

[uncut mulch]

so currently we are working on showing
sin(2Θ)cos(Φ) - sin(Φ)cos(2Θ)
simplifies to 0

yes, and your values for cos?

[grizzled turret]

Well, I know I could write

[uncut mulch]

think of something less complicated,
what's a well known relation between sin and cos, OR draw a triangle

[grizzled turret]

errr, except for the opposite of the above where we add pi/2 in sin, I'm not very sure

[uncut mulch]

draw a right triangle where sin(theta) = 2/sqrt(5)

||sin^2(x) + cos^2(x) = 1||

what can you use to find the third side, hence what is cos(theta)

[uncut mulch]

horrid/wrong notation

cos**(x)** =

also simplify the numerator, and use \theta instead of x

[uncut mulch]

similarly, what is cos(phi)?

[uncut mulch]

you only mentioned the double identity for sin(2theta) earlier,
do you know the identity for cos(2theta)?

[grizzled turret]

I should get this far without the double cos identity right?

[uncut mulch]

did you switch your theta and phi?

[grizzled turret]

I'm not sure

[uncut mulch]

sin(2Θ)cos(Φ) - sin(Φ)cos(2Θ)

there isn't cos(2phi) in that

earlier you had the identity for sin(2Θ), which leads to
2sin(Θ)cos(Θ)cos(Φ) - sin(Φ)cos(2Θ)

also there are 3 forms for the double angle identity for cosine, all of which will work here

[grizzled turret]

Nasty problem but it worked out in the end

Never been asked this kind of problem before and most probably never will

I was comparing two copies and noticed that both had the exact same answer, but both with different formulas so that got me curious

[uncut mulch]

wdym by exact same answer...?

[grizzled turret]

One copy had the answer circled in green

While the other used the answer in red

[uncut mulch]

still not sure what you mean

[grizzled turret]

In a geometry test, two students arrived at the exact(and correct) solution, but by using two different formulas

[fluid shore]

You mean the other guy converted the arcsine into an arctangent?

[uncut mulch]

but wdym by the green and red answer?

^

[grizzled turret]

No, they didn't do any conversion to their answer, this is the formulas they derived

[uncut mulch]

i mean the "answer" for this question is confirming that the identity is true...

so what formula did they use instead?

[grizzled turret]

Yes, this was my question that I asked after seeing the two answers

One student answered that the area was

[fluid shore]

?????

[grizzled turret]

While the other answered

Both answers yielding the same result

Hence where my question comes from

I thought that was pretty interesting

[uncut mulch]

,w 5arcsin(6/(3sqrt(5))) = 5/2 arctan(4/3)

[undone pawn]

F

3sqrt(5)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[undone pawn]

oof

@grizzled turret

[fluid shore]

Yer mum is false

OHHHHHHHHHHH

REKT

[undone pawn]

🔥 🔥

[uncut mulch]

it may help if there was context, the original question, and/or the 2 different methods being used

[grizzled turret]

This wasn't a question

Are the actual two answers from the students(one on the left of the equal sign and one on the right)

When computed the answer is the same : 44.4642...

And that got me curious on how I could prove that they are indeed equal

[uncut mulch]

ic

[hoary valley]

nvm

[rocky bison]

,$ \sqrt{x^x}=\left(x^x\right)^\frac12=x^\frac{x}2

[obsidian monolith]

pseudo:

[rocky bison]

You can then apply the same principles to your other example.

And see they work out to the same

[native sequoia]

for positive reals they are equal

[hoary valley]

apparently the x is an exponent to the whole square root

[rocky bison]

@hoary valley irrelevant

You can follow the same working

Just your intermediary step will be

,$ \left(x^\frac12\right)^x

[obsidian monolith]

pseudo:

[hoary valley]

I can't just follow the steps, If I didn't know where the x is..

[rocky bison]

Yes you can

You also know where the X is 🤔

[hoary valley]

It's weirdly written above the X and outside the square root

[rocky bison]

And even if you didn't there's only two possibilities that are reasonable

And we just demonstrated how both possibilities are identical anyway

[hoary valley]

Should have been written like that. but thanks anyways ❤️

[rocky bison]

Tbh it was perfectly clear because it was outside the line

[uncut mulch]

depends on the value of x

[rocky bison]

Or wasn't under the line

@uncut mulch given its precalc it's a reasonable assumption that they'll only be considering positive reals.

[paper temple]

So I have my college precalculus final tomorrow, and we get to bring a half sheet of paper for notes and formulas

anything that people here think I should definitely include?

[valid violet]

Trig identities

[paper temple]

for sure

[valid violet]

Do you have a list of topics that might be on the exam?

[paper temple]

I have a practice final

[valid violet]

Laws of exponents and logarithms if you haven't memorized those perfectly

But you really should hav e :P

[paper temple]

I have memorized nothing

lol

I need a 40% on this exam to pass the class. That's what we're shooting for minimum

[valid violet]

Damn

One sec

https://proofwiki.org/wiki/Laws_of_Logarithms

https://proofwiki.org/wiki/Exponent_Combination_Laws

These are very important

[paper temple]

okay

thank you

[valid violet]

Np

and you want the formula for things like question 9

Oh! The interest formulas

[paper temple]

I have the interest formulas

thankfully

[valid violet]

Good

[valid violet]

@paper temple

It's difficult to write down the formula for polynomial long division

So make sure you can do it

[paper temple]

I'll write down an example

[valid violet]

[paper temple]

thank you so much for the help

[valid violet]

Np

Funny story

I had a professor that allowed one piece of paper

And he specified it had to be flat because

One time a student did an origami thing

And fit multiple pages into one page that expanded like a pop up book

[valid violet]

Lol

[paper temple]

I've seen one where they got a massive sheet of paper

like, bed sized

and unfolded it in class

[valid violet]

Nice

[paper temple]

unfortunately my professor specified dimensions

[valid violet]

If you're doing that you probably either know the material completely or definitely fail

[paper temple]

he said standard 8 and a half by 11, I guess you could do that in feet, but it wouldn't work so well

xP

[valid violet]

But did he say flat

[paper temple]

no

[valid violet]

Ahah

Origami time

[paper temple]

the exam is tomorrow, and I'm not the star pupil to get away with such a thing

I got a 36 on the last exam lol

[odd helm]

I legit cannot understand bearings

I have no idea why I don’t think they are meant to be all that hard

But the bearing word problems just make me feel confused

[fluid shore]

It's okay

[odd helm]

Any tips on how to solve them?

[fluid shore]

You just need to.......get your bearings straight 😄

[odd helm]

Ha.

[fluid shore]

I used to wing them

And got them correct lmao

It's basically just trigonometry

[odd helm]

Smh

[fluid shore]

Don't treat it as a separate topic entirely

[odd helm]

I mean ig it does include trig

[fluid shore]

It doesn't include trig

It is trig

Draw big diagrams of the situations

Try to make them decently accurate

Then, use trig to simplify your work

[odd helm]

Alright and also like I can understand how the problems work kind of if the problem says go NE at the beginning I can easily do that

[rigid sun]

lol

ok then

[odd helm]

But then it’s like “ok now go 30 Sw”

[fluid shore]

Actually here's a secret

Do you want a little secret?

[odd helm]

Yeah

[fluid shore]

Okay here it is: nearly 100% of school problems can be solved if you just stare at them hard enough

Like, you won't understand what the question may want immediately

But if you stare at it long enough, you will get it

[odd helm]

Uh .... I mean that has happened to me before

Yeah I see what you mean lol

[rigid sun]

using a diagram

[fluid shore]

So, for you, you need to introduce enough details into the problem to make it easier for your brain to stare at the problem for you

For example

Draw diagrams

Write down certain obvious relations

Use variables

Shit like that

[odd helm]

Yeah I’ll try doing that my diagrams kind of suck but I’ll take the time to make them neater

[fluid shore]

Put in effort now and it won't take long to make them neat

All of it is contingent on the effort you put in

[odd helm]

Alright thank you

And when you say add variables

What do you mean exactly?

[fluid shore]

Well, if the problem throws a whole bunch of numbers at you

Then use variables instead

[odd helm]

Yeah that might help

Thank you

[odd helm]

For this problem I did law of sine but it didn’t work could someone explain?

Why can’t I do 55 degrees and 50 degrees

Since 180-125= 55

[uncut mulch]

are you referring to angle BAC?

[odd helm]

Yes

[odd helm]

I’m starting to understand bearings

And it only took 3 hours! 😃😔

I just need help with the question I posted above

And I think I’ll be in pretty good shape

[uncut mulch]

because you aren't told that AC goes from N to S

i.e. the bearing of C from the original position isn't known to be 180°

[odd helm]

So only when you are going straight between one point and another you can assume that if you go theta degrees in one direction, that means you do 90- theta for other part of angle

Because I saw another example when the two points were on an east-west line so directly straight from each other

And it worked that time

[uncut mulch]

that means you do 90- theta for other part of angle
?

it may help if you draw the full compass

[odd helm]

What I’m trying to say is that if you need to turn a direction to go from one point to another

You can’t say that the two angles are complementary

But if you don’t need to turn then you can say that

Is that correct? Or am I messing up

[uncut mulch]

what angles are complementary?
can you draw a diagram?

[odd helm]

Like this

They are directly straight from each other so 180 degrees right?

So you can say that the two angles are complementary?

[uncut mulch]

diagrams too blurry

[odd helm]

Lol

[uncut mulch]

are the things at the top supposed to be the bearings of the point at the point (let it be C)
from A and B?

because that doesn't tell you whether B is directly east of A

however from those bearings, angle C IS 90° and
angCAB and CBA would be complementary.

(but not necessarily 45° each)

[odd helm]

Yeah it doesn’t tell if it is direct but like if the problem said that you could assume they are complementary

[uncut mulch]

don't assume, draw a more general diagram and show that they ARE regardless of whether B is directly east of A. (for these bearings)

[fluid shore]

Tenemos, your diagram is way too small

Legit, it's way way way too small

Draw something that's half a page big

Then you'll be able to see everything that's going on

[still yew]

I thought the Precalc exam was going to be like 40-50 questions long, it was 17...

Ending on Identies...

[warm crescent]

Can anyone explain me definattion of inductive set with soem basic examples

some*

[fluid shore]

?

Let S be a set. Then, S is an inductive set if it satisfies the following properties:

1. $1 \in S$, where $1$ is the multiplicative identity.
2. $\forall x \in S: x+1 \in S$

[obsidian monolith]

Abhijeet Vats:

[warm crescent]

Multiplicative identity?

[willow bear]

it's the number 1

[fluid shore]

So, for example, the positive integers form an inductive set. In fact, they form the smallest inductive set.

[willow bear]

it's called the multiplicative identity because multiplying something by 1 is the same as doing nothing

[fluid shore]

^

[willow bear]

likewise 0 is the additive identity

[warm crescent]

Oh Ok

[fluid shore]

The term comes from the field axioms. It's one of the axioms that's listed. These axioms basically tell us how to work with numbers.

[warm crescent]

$\b{Z^{+}}=\b{N}$

[obsidian monolith]

Raftaar:
Compile Error! Click the reaction for details. (You may edit your message)

[willow bear]

i wouldn't describe it as that abhi

[fluid shore]

How would you describe it?

[willow bear]

the field axioms tell us nothing of how to work with numbers, not directly at least.

they describe the behavior of numbers

i.e. how they interact, in a sense

[fluid shore]

Oh okay yea fair enough

[willow bear]

it's on the mathematician to be mindful of the axioms and their consequences

[fluid shore]

I guess i phrased it a little badly there. Oof, still have a lot to learn, I suppose

[languid crane]

Don’t worry, everyone of us has a lot to learn

Some more than others (me, for example)

[warm crescent]

[fluid shore]

I shall strive to become more intelligent, regardless of how much i suck

[warm crescent]

Kabhi Kabhi lagta hai ki Apun hi Nobita hai.

Thanks btw

[willow bear]

please translate

i don't even know how to say "i don't speak hindi" in hindi

[warm crescent]

Sometimes I think I am Nobita.

[fluid shore]

Main hindi nahi bol sakti hoon

@willow bear

[warm crescent]

Abhi kya aap doreamon dekhte ho?

[fluid shore]

Nahi

[warm crescent]

kya aapne kabhi bhi doreamon nahi dekha

[willow bear]

i need to learn basic hindi

so far the only word i know is bhenchod

[warm crescent]

lol

[willow bear]

[warm crescent]

Lmao

[fluid shore]

Nahi, main kabhi nahi dekha hoon.

Main Lund University jane wala hoon 🙂

[warm crescent]

Aap12th me ho?

[fluid shore]

Fuck yes, that seems to be the only word from hindi that anyone knows

Nahi, main army mein hoon. Main agle saal university jaunga

[torn swift]

I only see Doraemon, that’s all I can understand

[warm crescent]

12th pass?

I am Nobita without Doraemon

[fluid shore]

Yea I did IB and did pretty well

[warm crescent]

Oh sahi hai

Ab aap college jao

[undone pawn]

cool

[torn swift]

Pre Doraemon Nobita, tough times

[warm crescent]

Yes

evryone bbully me in the school

[fluid shore]

Han, main lund main jaane wala hoon 🙂

[undone pawn]

#chill

[warm crescent]

Oh sahi hai

bye abhi ke liye

aap loda-lasan school me jaat hun

[willow bear]

does nahi mean no

[fluid shore]

Yes

[willow bear]

what's the word for yes

[warm crescent]

Haan

[fluid shore]

Haan or haanji

The latter is used when you wanna be respectful in your speech

[warm crescent]

Ree

[willow bear]

[fluid shore]

If you're curious, lund means 'penis' in hindi when pronounced in a certain way. That's why i like making jokes cos it's actually a university i'm considering

[willow bear]

haha dick university

[fluid shore]

Yea lmao

[warm crescent]

I goto loda lasun high school

#chill

[tall cedar]

Hey guys, I'm confused about something

lets say there is something like

log x * log y

how do I simplify that?

[fluid shore]

You don't

I assume they have the same base?

[tall cedar]

hmm..

then is there a way to simplify like

(log x)^2?

[fluid shore]

You can't simplify further. However, it also depends on the context of your question

Well, no.

[tall cedar]

i c

[fluid shore]

It might be better for you to post an example of a question you've attempted, where you've had to deal with those sorts of things.

[tall cedar]

nahh

I was thinking

and was like hmm

so log x + log x = log x^2

so what if there's log x * log x

:P

thanks though <3. have a nice day!

[fluid shore]

Unfortunately, i don't think there are any rules for simplifying that sort of thing.

🙂

[fleet yew]

@tall cedar

[fluid shore]

😦

[tall cedar]

@fleet yew can u ummmmmm explain it further?

What does log^2(x) mean? Isn't the same as (log x)^2?

Anyhow, thanks for helping me. I really appreciate it. :)

[stuck lark]

What does log^2(x) mean? Isn't the same as (log x)^2?
ye

[keen timber]

log^2(x) = logx • logx
logx^2 = 2logx = logx + logx

[valid violet]

I've seen a context where log^2(x) = log(log(x))

a better notation for that would be

$\log^{\circ 2}(x)$

[obsidian monolith]

gfauxpas:

[valid violet]

like the composition symbol

[odd helm]

I used sin to solve this but got +71 degrees

Why wouldn’t it work with sin?

(I may have posted this earlier in a different server tbh I don’t even remember if I did and I don’t remember which server it was in)

[rugged nimbus]

Anyone know how to intersect a polar equation with straight line y=x ??

[harsh cipher]

I got a question!

[undone pawn]

@rugged nimbus

[harsh cipher]

when finding one coterminal between 0 and 2pi

19pi/6

why do we subtract -2pi

[undone pawn]

y=x in polar is just theta = pi/4, 5pi/4

[rugged nimbus]

@undone pawn what

[harsh cipher]

and not add 2pi...

[rugged nimbus]

@undone pawn how

[undone pawn]

literally just plug in polar coords

x=y

rsin(theta) = rcos(theta)

tan(theta) = 1

[rugged nimbus]

Why sudden 1

[undone pawn]

divide both sides by rcos(theta)

[rugged nimbus]

Ohh

I get that tan tetha = 1

What about that part you said earlier pi/4?? How u get that nuber? Is it in formula??

[undone pawn]

no I just saw what angle tan theta =1

[rugged nimbus]

I see @undone pawn

Thanks2

Like that?? @undone pawn

[warm crescent]

$f(x)=|x-2|+|x+2| , x \in [-3,3] $
Write this function as a piecewise function

[obsidian monolith]

Raftaar:

[fluid shore]

What have you tried?

[undone pawn]

@fluffyy#2906 seems right

@Raftaar#1096

[rugged nimbus]

Thanks

[undone pawn]

ok why can't I tag anyone

[rugged nimbus]

@undone pawn

[undone pawn]

@rugged nimbus

aight works now

[rugged nimbus]

Can

[undone pawn]

👍

[rugged nimbus]

Btw i want to ask again

Still abt intersecrion

[undone pawn]

ye?

[fluid shore]

,rotate

[fluid shore]

Dafuq

[undone pawn]

lmao

[fluid shore]

Okay, what have you written

[rugged nimbus]

Im stuck

[fluid shore]

You can't do cancellations like that in the middle of the computation

[undone pawn]

miscalculations

[fluid shore]

It makes your work very messy

[undone pawn]

^

[rugged nimbus]

Im tired using pencils, hand hurt haha

[fluid shore]

You have:

$3\sin(\theta) = 1 + \sin(\theta)$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So, you just have to solve for $\sin(\theta)$

[obsidian monolith]

Abhijeet Vats:

[rugged nimbus]

2 = sin theta?

[fluid shore]

What

How

Look very closely

[rugged nimbus]

How to solve sin tetha

So i put the 3 and one at the left side right

[undone pawn]

ok

what's the next step after what abhijit wrote

[rugged nimbus]

In left side it become sin tetha /sin tha which is 1 ??

[fluid shore]

How.....?

[undone pawn]

write out the next step

don't skip

[rugged nimbus]

Its wrong??

[undone pawn]

yes

[rugged nimbus]

Sin tetha = 2???

[undone pawn]

no

how tf do you get that

[rugged nimbus]

Sin ttha - sin etha = 2??

[undone pawn]

wtf dude

[uncut mulch]

still, how tf do you get that...

[undone pawn]

stop guessing

You have:

$3\sin(\theta) = 1 + \sin(\theta)$

[obsidian monolith]

FlynnXD:

[rugged nimbus]

I im stuck

[undone pawn]

You have:

$3\sin(\theta) - \sin(\theta)= 1$

[obsidian monolith]

FlynnXD:

[fluid shore]

Let $x = \sin(\theta)$

[obsidian monolith]

Abhijeet Vats:

[undone pawn]

this is basic algebra..

[fluid shore]

So, now you have:

$3x = 1 + x$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

How do you solve for x?

[rugged nimbus]

Oh

I see

Hahahahaha

So no need to put on left sideeee

[undone pawn]

[fluid shore]

What

Okay wait

How do you solve for x?

Everyone else shhh

[undone pawn]

@Abhijeet Vats you need to put it on the left sideee

[rugged nimbus]

3x-x

[fluid shore]

Okay

[rugged nimbus]

-_-

[fluid shore]

So what is that equal to?

[rugged nimbus]

1/2

[willow bear]

0

[fluid shore]

Wtf

3x-x =1/2?

STOP SKIPPING STEPS

[uncut mulch]

pls write out the whole equation

[fluid shore]

STOP IT

What is 3x-x?

[rugged nimbus]

Why

2x

[fluid shore]

Okay

So 2x = 1

What's x?

[rugged nimbus]

1/2

[undone pawn]

25/50

[rugged nimbus]

Do i get reward

[undone pawn]

yes

[fluid shore]

Okay, so we have:

$\sin(\theta) = \frac{1}{2}$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Can you solve that?

[rugged nimbus]

Yess dun worry i can take care of it

Thanks guys

[fluid shore]

I am very worried when you are struggling with basic algebra.

[rugged nimbus]

I hate algebra haha

[undone pawn]

._.

[willow bear]

.-.

[rugged nimbus]

Hey

[undone pawn]

you're goin to have a bad time..

[uncut mulch]

😰

[fluid shore]

YOU CANNOT

BE SOLVING SUCH PROBLEMS

[rugged nimbus]

Im actually sqrt(cos(x))cos(300x)+sqrt(abs(x))-0.7)(4-x*x)^0.01 sqrt(6-x^2) you guys

[fluid shore]

WITHOUT IMPROVING YOUR ALGEBRA

[undone pawn]

@Abhijeet Vats is Ann vers. 2.0

[rugged nimbus]

I will improve when the time comes

Dun worry xd

[willow bear]

sadly the title of Ann à prix réduit is already taken

wait

[fluid shore]

What do you mean 'when the time comes'

[uncut mulch]

algebra is pretty much present everywhere,
without it you're going to struggle in all aspects

[willow bear]

where is onion

[fluid shore]

You can't 'improve when the time comes'

[willow bear]

the time is NOW

[fluid shore]

This isn't some harry potter dumbledore shit where 'You will understand when the time comes'

This is important stuff

[undone pawn]

fluffyy has left the chat.

[fluid shore]

I gave entirely valid advice 😦

[willow bear]

[rugged nimbus]

@fluid shore thanks

I know la

[fluid shore]

You're Singaporean?

[rugged nimbus]

No

Why

[fluid shore]

Well 'la' is typically used in Singapore

[rugged nimbus]

Really? Im live nearby that country

La also use often hahaha

U know alot @fluid shore

[fast tinsel]

Does anyone have an idea how they got to the right hand side of this equation?

In the book they say: evidently the roots are then: ^

I'm afraid I don't see how this is evident