precalculus

[echo wagon]

Or codomain

That just says the codomain is R, not the image necessarily

[shut shuttle]

Yes

[echo wagon]

So f(x) = e^(-x) would be an example of a continuous strictly decreasing function that is not bijective from R to R.

[muted steeple]

okay, figured it out x = 1/ln(6)-1

right?

[echo wagon]

Bracketssss

1/(ln(6)-1) is right

[shut shuttle]

Okay if we change the codomain to R^+ it'd be a bijective function right?

[echo wagon]

Yes

It's definitely injective if it's strictly monotone

So if you change the codomain to be the image, then it's bijective

[shut shuttle]

Hmm I get it, thanks

[echo wagon]

Np

[muted steeple]

ezzz passed it with 80%

[stuck lark]

@echo wagon eh can always restrict codomain to image as you say, so it's not of much weight

[echo wagon]

Bitch, was that a test?

[muted steeple]

nah, don't worry.

[echo wagon]

[muted steeple]

i'm doing an online course.

self-study type thing

[echo wagon]

Lol

[stuck lark]

yeah so no big deal as long as one recalls an injection can be made bijective that way

[muted steeple]

do you guys intuitively understand logarithm laws or just memorise it?

[stuck lark]

both

[muted steeple]

dayum.

[echo wagon]

Both. Like when I was in high school, if I saw the rule, I could explain it. But I couldn't necessarily recall them all instantly without thinking about it

So I also memorized them

[viscid thistle]

Which laws

[unborn blade]

can someone help me with this problem? kind of confused how to go about tackling it

[jolly raven]

Even though this is gonna be kinda weird, let's just assume a diagonal is completely muddy

So then you have to decide if a diagonal is 3 kmh is less or more time than the horizontal and vertical distance combined at 5 mph

@unborn blade

I hope that's right

[viscid thistle]

How do you go about solving 4^x - 15 x 2^x = 16?

[sick steppe]

15(2^x) or 15x2^x?

[viscid thistle]

no brackets

oh

[echo wagon]

It's a bit confusing to use x as a variable and for multiplication

Use *

[viscid thistle]

yeah..

4^x - 15 * 2^x = 16

[echo wagon]

Notice that 4^x = (2^x)^2

So this is a quadratic equation in terms of 2^x

If you struggle to see it, let y = 2^x then y^2 = 4^x

[viscid thistle]

thnx!

[echo wagon]

Np

[dense pendant]

Does anyone know what important topics are in pre-calculus, pre-calculus H, or anything that will help for AP Calc BC? I will be self-studying precalc over the summer, so I dont really know what I will be expecting on how to prepare for AP Calc BC.

[placid ledge]

summation notation and polar forms (also maybe vectors I don't really remember)

[dense pendant]

i see

[tribal ravine]

You have some time for vectors (those will show up in Calc 3 but most courses have a reintroduction).

But yeah, summa notation is good for the end of BC (series) and polar forms is a huge part, as well.

[placid ledge]

w8 a minute

[placid ledge]

are you going from algebra 2 or trig directly to calculus BC?

[dense pendant]

alg2/trig H directly to AP Calc BC, pre-calc over the summer

so we did went over summations, statistics, probability if that makes any difference

[placid ledge]

oh

[dense pendant]

and trig, but not trig identities

[placid ledge]

Everything here except, Matrices, Probability and Combinatorics will probably be used https://www.khanacademy.org/math/precalculus. If you feel shaky on a topic then just review it.

[dense pendant]

oh ok thank you very much! interesting--

[placid ledge]

np

also

[dense pendant]

well this site offers a lot of content

[placid ledge]

If you start studying the basic concepts of calculus right now it'll really help later in the course

I did that the summer before calc and it made it a breeze

[dense pendant]

ooo-- i see

thank you so much though!

[tribal ravine]

+1 to what keto said

it really helped when i took calc a few years back

theres a lot of weird and new concepts introduced to you very quickly

good to get your own understanding of it beforehand

[dense pendant]

o-O ok will do. i suppose i can just start off memorizing the rules then--

[muted steeple]

can someone explain what's going on here?

wtf man, why do i figure this stuff out as soon as i post it

lmao

actually i still don't get the last equality

0.996A

nvm, i figured it out also 😆

[manic widget]

Hi everyone
Could someone explain to me how to tackle this question?

[glacial epoch]

Not really sure but based on multiple choice and other data it would be C

[edgy blaze]

can anyone explain reference angles? what do you subtract pi or 2pi?

[hallow thunder]

https://en.wikipedia.org/wiki/Derivative_(mathematics)

[unborn blade]

hi can someone help explain this to me

[edgy blaze]

f'(g(x)) * g'(x)

[rough carbon]

So with the chain rule, the derivative of what’s inside the function comes outside, multiplied by the derivative of the outside function

2x differentiates to 2

Sin(u) differentiates to cos(u)

Hence 2cos2x

[unborn blade]

oh that makes sense

thanks

[prisma saffron]

i was wondering if anyone could help me get started on any of these trig bc functions. especially the ones with multiple identities. i’m not sure how to start the multiple identity ones

[edgy blaze]

1. you can start by factoring cot

the cos(2x) = sin(x) use an identity to turn that cos(2x) into one with just sin

[prisma saffron]

how would you get it to the nearest degree though

[vagrant herald]

@manic basin got the answer ?

I can help if not solved

[manic basin]

I really didn’t mate I’m too confused
The answer ( checked from the back ) is a=1 and b € R

But I don’t know how

[vagrant herald]

@manic basin

[restive bridge]

https://i.imgur.com/ecv8RBr.png

What's killing me is the second condition. I get that 2^k+1 is the equivalent of 2* 2^k. But they're losing me on how they got the other side of the >

[stuck lark]

@restive bridge if x>y and z>0 then xz>yz

[restive bridge]

I don't follow.

How is 2 * k the equivalent of k + 1?

[uncut mulch]

its not

they're applying the assumption that 2^k > k

and from that
2 * 2^k > 2*k

[willow bear]

@restive bridge multiplying both sides of 2^k > k by 2 gives you 2*2^k > 2*k

does that make more sense to you?

[restive bridge]

I get it now. TY people

[muted steeple]

did i setup this problem correctly?

<@&286206848099549185>

I worked out the k in $Ae^{tk}$ by $5 = 10e^{30k} \therefore \frac{1}{2}=e^{30k} \therefore \ln \left( \frac{1}{2}\right) = 30k \therefore k = \frac{\ln \left( \frac{1}{2}\right)}{30}$

[obsidian monolith]

Vocal

[austere void]

Well, I'm not really a chemist but

[muted steeple]

then $10\left(e^{\frac{t \cdot \ln \left(\frac{1}{2}\right)}{30}}\right) = 10\left(e^{\ln\left(\frac{1}{2}\right)}\right)^{\frac{t}{30}}$

im not either

[austere void]

I think you're making the solution way too difficult

[obsidian monolith]

Vocal

[muted steeple]

i'm following along the course

that's how this was taught

[austere void]

Look, by definition this means that the amount of caesium halves every thirty years, right

[muted steeple]

yes.

[austere void]

Therefore you get 1/4 the amount after 60 years, 1/8 after 90, etc.

After 30k years you're left with $\frac{1}{2^k}$ of your original amount, which you want to be equal to $\frac{1}{10}$

[obsidian monolith]

NickPro

[austere void]

@muted steeple right?

[muted steeple]

yeah

[austere void]

Just solve for k and then find the number of years which is 30k

Are you familiar with how logs work?

[muted steeple]

well hopefully, because this chapter is on logs.

probably not if i'm unable to do this problem.

[austere void]

Take a look

You want to solve $$\frac{1}{2^k} = \frac{1}{10}$$
As the fractions are equal and the numberators are equal, you can set the denominators equal to each other:
$$2^k = 10$$

[obsidian monolith]

NickPro

[muted steeple]

okay

so

[austere void]

This can be solved by definition of the logarithm

[muted steeple]

$k \cdot \ln(2) = \ln(10) \therefore k = \frac{ln(10)}{ln(2)}$ ?

[obsidian monolith]

Vocal

[austere void]

Yes

[muted steeple]

okay, that's a new way of looking at the problem

thanks.

[austere void]

No problem

[muted steeple]

wait

actually

[austere void]

That's an alternate way to express $log_2 10$

[obsidian monolith]

NickPro

[austere void]

The number that literally says "what is the power to which you have to raise 2 if you want to get 10"

or "2 to the what is 10"

[muted steeple]

i'm trying to figure out how $\frac{\ln(\frac{1}{2})}{30} = \frac{ln(10)}{ln(2)}$

[obsidian monolith]

Vocal

[austere void]

It isn't

[muted steeple]

hmm then i need to figure out how my original solve went awry

[austere void]

What did the 5 in your original equation stand for?

[muted steeple]

half of 10

i.e. half-life

cuz $5 = 10e^{tk}$ when k = 30

[obsidian monolith]

Vocal

[austere void]

Well then we were using the letter k for different purposes

[muted steeple]

really?

how come?

[austere void]

I used it as the variable for the number of 30-year cycles you have to wait

[muted steeple]

yeah

[austere void]

You used it as the constant meaning the halflife of caesium

[muted steeple]

but it's a product of those cycles correct?

okay, but the number of cycles is a constant no?

for 30 year cycles

i'm so confused right now

[austere void]

No, that's the answer you're looking for

[muted steeple]

yes, that's what i'm trying to find first.

in order to model the problem

the k

constant k

then i can plug t and figure out the answer for any t

[austere void]

Then k isn't equal to 30

[muted steeple]

sorry t=30

my bad

oh darn..

$5 = 10e^{tk}$ when t = 30

[obsidian monolith]

Vocal

[austere void]

Yes, now you have correctly found your k to be $\frac{ln(\frac{1}{2})}{30}$ or $-\frac{ln(2)}{30}$

[obsidian monolith]

NickPro

[muted steeple]

or -?

[austere void]

That's just the property of the log

1/2 = 2^-1

[muted steeple]

oh

woiw

i didn't know this

and it makes ense

so it's hidden behind this one right?

[austere void]

Yes

[muted steeple]

but if that's the case how does it equal to what we worked out together?

[austere void]

What I calculated to be k was actually t in your system of variables

[muted steeple]

ahh, but then what was the k in that case?

[austere void]

$\frac{ln(10)}{ln(2)}$ as far as I remember

[obsidian monolith]

NickPro

[muted steeple]

yeah but that's time as you mentioned.

oh

[austere void]

your k was correct

[muted steeple]

so now i just get the product?

99 years

[austere void]

Yeah around that

[muted steeple]

$\frac{\ln(10)}{\ln(2)} \cdot 30$

right?

[obsidian monolith]

Vocal

[austere void]

Yeah

[muted steeple]

okay, it's actually a lot cleaner to solve for k and t your way

you skip a lot of ugly stuff

thanks.

[austere void]

No problem

[muted steeple]

РУН

HEY

mind explaining one more thing about logs?

why are we always use base e log?

[sick steppe]

choice

[muted steeple]

looking at logarithm laws, you can use any base, as long as it's the same...

i see

[sick steppe]

and ln just plays nicer in math

[willow bear]

the importance of the natural log over all the others is a little hard to explain without calculus

[austere void]

There is a cute fact about the natural log that relates it with primes

But that's by far not the only reason

[willow bear]

i mean

that takes way more mathematical experience to get to

[austere void]

It does, yeah

[austere void]

@muted steeple On a very high level, e^x is the only function which equals its own rate of change, and for this reason it has lots of applications in physics

In a lot of applications of math, using a natural log is a lot cleaner than any other base because you wouldn't want to always multiply things by some factor you don't even want there

Since ln(x) is the inverse to e^x

Again, it's not easy to explain because most of the stuff that relates to this comes from calculus, which you are apparently still yet to learn

[austere void]

It's also easier to use some specific base for calculations, since you might need to compute, say $\log_2{3}$ or $\log_4{7}$

[obsidian monolith]

NickPro

[austere void]

And it just turns out that natural log is the easiest to find using a computer

@willow bear Does the latter count as an explanation without calculus

[willow bear]

i guess?

like, idk

this opens the door to the question of "how do computers even calculate logs"

[austere void]

I mean, yeah

But this is still better than nothing

[muted steeple]

i'll take it as a gospel for now.

calc is not too far away from my grasp

trig is the next topic.

[austere void]

Trig is pretty easy, but get ready for equations which can take pages to solve

[muted steeple]

idk what i want to do after calc.

i'll probably look at some machine learning content and see what sort of math i might make use of.

[austere void]

Calculus is usually the last topic at high school

For machine learning, you would need something called "linear algebra"

And also calculus

[muted steeple]

i know about lin alg, a little bit of it at least.

[austere void]

Check out 3Blue1Brown, he has an amazing playlist in both of these

[muted steeple]

are partial differential equations part of calc curriculum, or is it later down the line?

[austere void]

Nah diffeq are later

They are significantly more difficult

[muted steeple]

well fuck.

i guess i'll need to learn basic proof techniques too.

[austere void]

At my school everything was messed up when it came to the order of topics

We had trig, calculus, and then exponentials and logarithms the following year

[muted steeple]

wow, that's strange.

[austere void]

The most illogical thing ever

And then another part of calculus

[muted steeple]

what's the general pre-req for partial diff eqs?

[austere void]

Normally it goes like this

It's calculus, specifically derivatives and integrals

Then you have to learn multivariable functions, partial derivatives and ordinary differential equations

[muted steeple]

" multivariable functions, partial derivatives and ordinary differential equations"
are these big topics on their own?

[austere void]

The latter is

Multivariable function is just a function of several variables, like $f(x, y) = x^2 - y$

[obsidian monolith]

NickPro

[austere void]

Like, that's it

[muted steeple]

so you have 2 independent axis?

[austere void]

Partial derivative isn't all that different from a regular one

@muted steeple In math it's not always the best idea to think about the function as a graph

[muted steeple]

i see.

[austere void]

Just think of it as something that takes multiple numbers and returns one

[muted steeple]

but it should still be able to be plotted on a graph no?

[austere void]

not always

[muted steeple]

provided it's not too many variables?

[austere void]

Then it could... in theory

[muted steeple]

hopefully trig doesn't make me wanna quit lol

[stuck lark]

a function of n vars has a n+1 dimension graph, eg a function of 1 var has a 2d graph, that of 2 vars has a 3d graph

[austere void]

I mean, we sorta have a way to show the complex functions without using a 4-dimensional space

But normally yeah, if you graph it the usual way, that's how it works

@muted steeple Anyway, however, just as anywhere in math, make sure you understand where things come from

[muted steeple]

as in?

so far my introduction to maths was definitions and rules.

[austere void]

Math isn't a bunch of formulas to memorise, it's all filled with reasoning and connections

[stuck lark]

a function of 3 vars has a 4d graph. we can't really visualize it but we can still get a sense of it by taking contour plots, ie plotting the set of points that are mapped to a particular value

for any more vars it's impractical to try visualizing a graph and we stick to imagining outputs as just numbers instead of heights along another axis

[austere void]

Thank you, brilliant explanation

[stuck lark]

some intuition can still be recovered by imagining outputs as a physical thing like temperature varying over any number of spatial coordinates

[austere void]

Well, if it isn't 1, you can just reduce the fraction

4/6 = 2/3

26/65 = 2/5

no problem

[muted steeple]

need a little bit help here

so it should be $\frac{1}{2^k}=\frac{1}{5}$

correct?

[obsidian monolith]

Vocal

[austere void]

Yes

Where k is the number of half-lives

[muted steeple]

which is $k=\frac{ln(5)}{ln(2)}$

[obsidian monolith]

Vocal

[muted steeple]

okay, since i have k

i need now to find t

[austere void]

so the result you get is $5730 \cdot \log_2{5}$

[obsidian monolith]

NickPro

[austere void]

Yeah that

[muted steeple]

hmm, i was trying to find it differently after that step.

$\frac{1}{10} = \frac{1}{5}\cdot e^{t \cdot \frac{\ln(5)}{\ln(2)}}$

[obsidian monolith]

Vocal

[muted steeple]

that's what i was trying to do

where 1/10 is 1/2 of 1/5

didn't seem to give sensible answer.

[austere void]

Indeed, since there is no reason to divide your 1/5 by 2

[muted steeple]

ohh

it should've just stayed 1/2?

[austere void]

1/5*

[muted steeple]

hmmm that doesn't quite register to me.

[austere void]

But have in mind that ln(5) / ln(2) is t, not k

[muted steeple]

t is time

k is coefficient

constant

i need to find years

where years are t

[austere void]

Okay, t is $5730\cdot\frac{ln(5)}{ln(2)}$

[obsidian monolith]

NickPro

[austere void]

Look, by solving the equation 2^x = 5 you have found the number of half-lives

[muted steeple]

yep

i'm trying to figure this out algebraically, step by step.

it does make intuitive sense.

what you're saying

i'm just trying to see the processes that lead to that above.

the initial amount was 100%

$\frac{1}{5} = e^{t \cdot \frac{\ln(5)}{\ln(2)}}$ ?

[obsidian monolith]

Vocal

[austere void]

I just used the wrong variable name at the beginning

[muted steeple]

cuz A = 1?

[austere void]

If we rewrite this a little

We know that after 5730 years the amount is 1/2 of what it was

[muted steeple]

1/5

oh

no

[austere void]

No, that's one half-life

[muted steeple]

yeah

[austere void]

So the number of these half-lives is t/5730, where t is the number of years

Right?

[muted steeple]

yes.

[austere void]

Therefore to find $t$ we need to solve $2^{\frac{t}{5730}} = 5$

[obsidian monolith]

NickPro

[austere void]

Because that fraction is the number of half-lives

[muted steeple]

yep

[austere void]

So $t = 5730\cdot\frac{\ln(5)}{\ln(2)}$

[obsidian monolith]

NickPro

[muted steeple]

okay that makes a lot of sense

man, i'm just thrown off because i was taught to model the problems like $Ae^{tk}$

[obsidian monolith]

Vocal

[muted steeple]

and work from there

[austere void]

Well, let's create a general formula for k if you need

[muted steeple]

the way i had it figured out that the initial value is is 100%

[austere void]

Let T equal the half-life of whatever substance you need

[muted steeple]

hence A = 1

$\frac{1}{2} = 1 \cdot e^{5730 \cdot k}$

[obsidian monolith]

Vocal

[austere void]

so $Ae^{Tk} = \frac{A}{2}$

[obsidian monolith]

NickPro

[muted steeple]

okay, so you keep the A variable in?

[austere void]

It cancels out

[muted steeple]

okay.

ohh, that makes more sense than what i did i guess.

but T = 5730 right in the above equation?

[austere void]

Yes

[muted steeple]

$5730k \cdot \ln(A) = \ln\left(\frac{A}{2}\right)$

[obsidian monolith]

Vocal

[austere void]

So $e^{Tk} = \frac{1}{2} \implies Tk = \ln(\frac{1}{2}) \implies k = \frac{\ln(\frac{1}{2})}{T}$

[muted steeple]

yeah

that's what i modelled right?

$\frac{1}{2} = 1 \cdot e^{5730 \cdot k}$

here

[obsidian monolith]

Vocal

[muted steeple]

A = 1

[obsidian monolith]

NickPro

[austere void]

That's it

You can also use ln(1/2) = -ln(2)

The log property I mentioned before

@muted steeple Does that make sense?

[muted steeple]

okay

this makes so much sense

can you remind me the benefits of that property?

damn this is some kewl stuff

[austere void]

That's the same property you have been using all along

I just took $\ln(2^{-1})$

[obsidian monolith]

NickPro

[muted steeple]

yeah, ik but i don't see the value of ever putting it back to $-1 \cdot ln(2)$

[obsidian monolith]

Vocal

[muted steeple]

but i guess you get a more complete general equation

[austere void]

If you get some experience with working with logs, you will find it useful to remember numbers like ln(2) or ln(3)

Logs of fractions are rarely used and they can be simplified

Though at this point it doesn't matter

[zealous horizon]

Does anyone have notes or a resource that I can use to answer something like this.

We did I notes on a lot of this but we did it take notes for questions 5-9

[frosty rover]

there is a rule to convert sin2x term into sinx

@zealous horizon

for 8

or you can convert sinx's into cosx's and get this

there are called half angle theorems or something i think if translated

[sick steppe]

those are double angles

[zealous horizon]

I have found out how to do most of them does anyone k ow how to do 5

[sick steppe]

recognize a compound angle identity

[zealous horizon]

I see the cos(s-t) and the sin(s-t) I just don’t know what to do after you use the identities

[sick steppe]

no, the whole thing can be reduced into 1 trig expression with a compound angle

[zealous horizon]

Does the whole thing used cos(s+t)

[sick steppe]

what do you even mean?

[zealous horizon]

Like if s is s-t and t is t

[sick steppe]

that's just confusing for no needed reason

just write the final answer

[zealous horizon]

the answer cos s

[sick steppe]

yes

[zealous horizon]

Ayy thanks

[onyx fern]

Anyone know how to find rectangular polar coordinates?

[willow bear]

@shrewd loom you wanted help with precalculus, is that still the case?

[shrewd loom]

thanks

question is 28xs^5

over 5t^5u3

over 7rs^2

over 10t^3u

[willow bear]

uh

that's a lot of "over"s... is it a nested fraction or something?

do you have a picture of it maybe?

@shrewd loom ?

[shrewd loom]

ugh yea

[willow bear]

okay

idk why you put the x there but whatever

let me just transcribe it for my own convenience

[shrewd loom]

could u call

and explain

[willow bear]

no, i can't vc right now.

$\frac{28s^5}{5t^5u^3} \bigg/ \frac{7rs^2}{10t^3u}$

[shrewd loom]

ok

[obsidian monolith]

Ann

[willow bear]

so this is your thing

do you know how to divide fractions? Y/N

[shrewd loom]

y

[willow bear]

okay

this is a problem which requires you to divide two fractions.

are you able to do that? Y/N

[shrewd loom]

y

thanks

i got it

[willow bear]

you're welcome

[muted steeple]

how does this happen?

[willow bear]

tan(θ) = sin(θ)/cos(θ)

[muted steeple]

that should reduce to a/b?

[willow bear]

what?

tan(θ) = b/a here, if you're looking at the diagram

[muted steeple]

oh, i was imagining cos(90-theta)/sin(90-theta)

nvm.

ok

[muted steeple]

how was this length determined?

[willow bear]

which length?

the h/2 you highlighted?

[muted steeple]

yep.

[willow bear]

put two copies of this triangle together and they make one big triangle whose angles are all 60°

[muted steeple]

ooh nice.

senkju.

[viscid thistle]

@willow bear need help

[willow bear]

???

[jolly raven]

@viscid thistle the equation of the midline is always the vertical transformation for a sinusoidal function

The very last number outside of the cosine is the vertical transformation

[willow bear]

they are muted.

[jolly raven]

Ah

Interesting

[grizzled blade]

hey could anyone help me with this?

we're supposed to factor out the x to find it

im a bit confused on what to do with e.

and the process in of itself.

[lapis sphinx]

U mean solve for y?

[grizzled blade]

yes y

sorry

we have to expand

the equation

[lapis sphinx]

Combine like terms first

[grizzled blade]

2e^y = 12

[lapis sphinx]

Constants

Yep

[grizzled blade]

after that.

how do we turn that into logarithimic form.

[lapis sphinx]

Then get rid of the two

[grizzled blade]

so e^y = 6

[lapis sphinx]

Yes

Then

[flat basin]

lmk when u guys are done i have a precalc question

[lapis sphinx]

Then take log of both sides

[grizzled blade]

so yloge = log6

[lapis sphinx]

It would be

Log base e of 6 = y

Or in other words ln 6 = y

Have u studied the natural logarithm?

[grizzled blade]

yes, Ine = 1

correct?

[lapis sphinx]

Yes, and log with base of “e” is same thing as “ln”

[grizzled blade]

i see

so

the final equatino is

y = log6?

[lapis sphinx]

Log base e of 6

Or ln6

Not just log6

[grizzled blade]

ok

yes

that makes sense

how about this one?

i have all the answers, its just a little confusing to catch on to the process

[lapis sphinx]

So can u rewrite the left side

Into exponent form

[grizzled blade]

to x/2log10?

[lapis sphinx]

Nah

Dw abt log yet

[grizzled blade]

💀

aight

[lapis sphinx]

Just rewrite left side

[grizzled blade]

so 10^x/2

[lapis sphinx]

Nah, here I will give u an example

[grizzled blade]

alright

[lapis sphinx]

What if I had

[grizzled blade]

ah shit

10^-x

[lapis sphinx]

Sqrt 10, how would u write that in exponential

[lapis sphinx]

That’s 1/(10^x)

[grizzled blade]

i thought sqrt 10 would be 10/2

?

in exponential

form

[lapis sphinx]

10^ (1/2)

[grizzled blade]

10 ^ 1/2

i see

so if the power is x

it would be

10 ^ 1/x

[lapis sphinx]

Correct

Nice

[grizzled blade]

ok so where do we go from there

we do 1/xlog 10

?

[lapis sphinx]

So now we have 10 ^ (1/x) = 9

Then we take the log

What is the base of the exponent?

[grizzled blade]

10

[lapis sphinx]

So that is the base of our log

[grizzled blade]

so it would be that

?

[lapis sphinx]

Nah we can’t bring down the 1/x to the left like that

[grizzled blade]

damn

so what would we do

log - xlog?

[lapis sphinx]

So now we have 10 ^ (1/x) = 9

That is exponential form

Using that format I just posted, can u change to logarithmic

?

Or try defining the variables first

[grizzled blade]

hmm

so

n = 9, b = 10 and a = 1/x

[lapis sphinx]

Correcttttt

Nice

[grizzled blade]

how does this help us?

[lapis sphinx]

Now put those numbers into the logarithmic form

What do u end up with

[grizzled blade]

1/xlog9?

that isn't right is it.. lol

[lapis sphinx]

Write it into the log form (the left side of the image I posted)

The variables u defined

Just bring the numbers into there

[grizzled blade]

log9 = 1/x (10^1/x) = 9

[lapis sphinx]

What u wrote first

Log 9 is 1/x

[grizzled blade]

i see

[lapis sphinx]

Now solve for x

That’s ur answer

If we had any other base other than 10, u would have needed to write “base n”

But with 10 u don’t need to

Tag me if u need smth

[grizzled blade]

uhhh

i got 0

which isn't right

how would i solve for x?

Log 9 = 1/x

i multiplied both sides by x to get

xlog9 = 1

then i turned 1 into log form

so xlog9 = log

but i dont think thats right

@lapis sphinx

[lapis sphinx]

So u were right at xlog9 = 1

Now isolate x

How should we do that

[grizzled blade]

x = 1/log9

i factor out the x

[lapis sphinx]

Correct

There’s ur answer

[grizzled blade]

oh

tf

its right

ok i got this really challenging one next so ill lyk if i need help

[lapis sphinx]

Try to commit to memory that format I gave above

[grizzled blade]

yEAH

*yeah

i will

thank you

[lapis sphinx]

Yea no prob

[grizzled blade]

ah fuck

im confused again

@lapis sphinx

this one is like completely different from what i've seen

first thing i did was, i turned it into exponential from

e^x-1/2 = 11

from there i did

x-1/2 In = In 11

wait one second i think i figured it out

[lapis sphinx]

This is right

[grizzled blade]

ah shit

i was off the answer by 1

[lapis sphinx]

U don’t put an ln or a log on both sides

Only on 1 side

[grizzled blade]

i think i figured out why i got it wrong

[lapis sphinx]

Remember the format

[grizzled blade]

yeah

[lapis sphinx]

U didn’t solve for x but x-1 I’m guessing

[grizzled blade]

hmm

wait hold on

lemmie send you my work

the answer is 5/79

**5.79

do you know where i went wrong

is it readable for you lol

@lapis sphinx

[lapis sphinx]

(2ln 11 ) + 1 is answer

U forgot the plus 1

[grizzled blade]

yeah

where

would it

come form

when i add the In?

the -1In?

i figured it was that

[lapis sphinx]

I don’t think so

[grizzled blade]

wait

i only factored x

i should've factord

xIn

correct?

[lapis sphinx]

Lemme write it down

[grizzled blade]

alright

[lapis sphinx]

Remember don’t put the ln on both sides

[grizzled blade]

what the

your handwriting

is so neat

you used your formular correct?

*formula

https://cdn.discordapp.com/attachments/363224154469826562/839305037826097233/image0.png

this

[lapis sphinx]

Yes

B = e

[grizzled blade]

yeah thats helpful

[lapis sphinx]

X-1/2 is a

N is 11

Commit it to memory asap cuz it’ll make life much easier

[grizzled blade]

yeah clearly

[lapis sphinx]

I gtg, pce

[grizzled blade]

thank you for all the help

[lapis sphinx]

Fosho

[valid estuary]

yes

multiplication is just repeated addition

so like 2 * 4 is saying add 4 two times, 4 + 4 = 8, 2 * 4 = 8, same thing with your example

[willow bear]

yes, 2x means 2*x. in algebra if two things are next to each other it tends to mean multiplication

[valid estuary]

Could someone help me out with this problem?

36 / pi doesn’t seem to be the answer

[uncut mulch]

why did you cross out the term with dr/dt?

[valid estuary]

sorta just guessed that the radius would be constant at that time so the derivative is 0

but that's the problem isn't it

I don't know how else to find dr/dt

[uncut mulch]

consider expressing r and h in terms of each other using similar triangles and the given height and radius

and then volume in terms of a single variable

[valid estuary]

ohh I gotchu

h = 4r

and then you can plug it in

thanks

[wanton horizon]

does anyone know how to find polar coordinates?

[rough carbon]

Converting from Cartesian to polar ?

[wanton horizon]

yea that and to rectangular

[rough carbon]

Sure, which question do you need

Help with?

[wanton horizon]

From this to rectangular

[rough carbon]

By ‘rectangular’ do you mean (x,y)

[wanton horizon]

yea i think so

[sick steppe]

Draw (3,pi/2) on a polar plane

[wanton horizon]

ok

[obsidian monolith]

Shen

[fleet yew]

Correct.

[unborn blade]

Hi, can someone tell me where I'm screwing up?

The question says to find derivative using log laws

I labeled my steps with numbers on the right side if someone can let me know which steps I'm messing up

[willow bear]

(3)

ln(a)+ln(b) is not ln(a+b)

@unborn blade

[unborn blade]

oof

i see that now

[willow bear]

what follows after is unsalvageable really

[unborn blade]

i see

yeah my brain for some reason was reading ln as a constant or something

rather than a function

[unborn blade]

am I on the right track with this?

[past meadow]

you are correct.

you can factor it further if you want, but you are correct.

[unborn blade]

great, thank youu

sorry, I have one more

I'm fairly certain I did this wrong

judging by the answer key

just not sure where I'm going wrong

seeing as my answer is way off of the actual answer, i'm guessing i'm doing something drastically wrong

which is weird considering I did the previous question correctly

[lament schooner]

@somber jewel have you heard of special right triangles before?

[somber jewel]

i have not

[lament schooner]

what about the pythagorean theorem?

[somber jewel]

yes

can you please help me solve that?

[lament schooner]

can you vc?

[somber jewel]

ye

[somber jewel]

'

[lament schooner]

https://www.youtube.com/watch?v=nVTtSE5nv7c

[tranquil ruin]

Line a perpendicular from Jay to (Luis-Bob)'s road then you can use Pythago

[rough carbon]

Or alternatively, you can work out interior angle JLB

Which is 180-45 bc angles on a straight line are 180 deg

Then use cosine rule to find lJBl

Remember that’s your a

The A is the 135

And b and c are the 10 and 8 given

[viscid thistle]

Is there some software I can use/install to compute Taylor series without actually doing tem by hand? Because doing taylor series by hand, even to the 5 term is not fun at all

[timber tree]

ok

what is it

[timber tree]

probably

[viscid thistle]

Hello

Okay

So

[timber tree]

type out ur full thoughts

and ping when done

ok

so u sound quiet

[viscid thistle]

I am

Not

I have

Pictures

Typing the question in the format of the question on the quiz was proven to be difficult

@timber tree

[timber tree]

wait

u want explanation?

or u want me to do them

because im not sure if i am allowed to do either ngl

[viscid thistle]

I just need to know if I did them correctly

[stuck lark]

@viscid thistle this is a quiz and you're taking it now?

@viscid thistle answer

[viscid thistle]

No I’ve already submitted it but my teacher does not grade quickly or give me any feedback on my questions and I NEED to know which ones I got wrong or right to prepare for the actual test

I’m ab to graduate and I need to submit the test after this quiz ASAP

[stuck lark]

so that's a practice quiz?

[viscid thistle]

It is a practice, but I will figure it out

I did not read the rules prior to asking

That’s my bad

[stuck lark]

yes, asking for help on tests etc is academic dishonesty and bannable

so don't do that on the real test

[viscid thistle]

I would never, I only asked because it was a quiz

[stuck lark]

practice, right? not graded

[viscid thistle]

Yeah

[unborn blade]

is the answer just "no min/max"?

i got this far

[distant valley]

why would u say that

[unborn blade]

@distant valley is that to me?

basically if we try to solve the resulting quadratic

you would get complex roots

so im guessing we wouldnt be able to establish any real intervals

so no real min/max

honestly im not sure

[distant valley]

there arent any crit values so

you should use this approach

[sick steppe]

$0=2\cos(2x)+\sin(x)-4$ is correct

[obsidian monolith]

moshill1

[sick steppe]

$0=2(1-2\sin^2(x))+\sin(x)-4$

[obsidian monolith]

moshill1

[sick steppe]

$0=-4\sin^2(x)+\sin(x)-2$

[obsidian monolith]

moshill1

[sick steppe]

then just solve that quadratic

[unborn blade]

oh

so i guess i simplified wrong

i got -4sin^2x-sinx-2

which gave complex roots

[sick steppe]

yeah the sin(x) shouldnt change

[unborn blade]

yeah makes much more sense now

thank youu

[sick steppe]

$0=4\sin^2(x)-\sin(x)+2$

[obsidian monolith]

moshill1

[somber jewel]

think you can help me with this one?

[short adder]

The solution I'm getting for texits question is imaginary roots

[unborn blade]

yeah it's gonna be complex roots

[viscid thistle]

I'm not sure how I would criticize it. I don't see why it wouldn't be correct

[sick steppe]

consider 3 points which are colinear

[viscid thistle]

The line segments wouldn't then connect all three points

[sick steppe]

why wouldnt they?

[viscid thistle]

nevermind

[rigid trail]

Hello @everyone I'm Azariah and I am horrible at math (SMH) lol that's a way to start. I am an Cyber Security major at my college and I am taking Precal this summer. I joined this group because instead of just going through this class I actual want to learn it. Will anyone be willing to help me out? I am asking in advance My class starts May 25. You will have to be patient with me 😅 👍

[sick steppe]

#❓how-to-get-help

[viscid thistle]

I'm certain that the converse is false, but is this statement false? I think it is

[sick steppe]

The statement written is true, but the converse isnt

[viscid thistle]

It's true if QPM are the same point?

Assuming that they are different points, I think it is false

[sick steppe]

my guess is they assume P Q and M are distinct

[viscid thistle]

Could you explain how could this statement be true, then?

If they are distinct, then I don't see how that could be the case

[sick steppe]

Ok well draw a 0 degree angle

you'll see that P Q and M lie on the same line

[viscid thistle]

could you please draw a 0 degree angle? for me a 0 degree angle is the angle between one and the same point

[sick steppe]

angles make a V shape right?

so 0 degrees is just the 2 lines from the central point being ontop of each other

[viscid thistle]

i guess

[sick steppe]

so if I have a 0 deg angle, the points that form it must be colinear

but if I have 3 colinear points, the angle doesnt have to be 0, it could be 180

[viscid thistle]

I get it

[harsh stratus]

do you know how the focus and directrix are related to a parabola?

[astral lark]

can anyone help me with this one, it has me stuck

[grave tartan]

$\lim_{x\to\frac{\pi}{4}}{\frac{1-\sin{(2x)}}{(\cos(x)-\sin(x))(x-\frac{\pi}{4})}}$

[obsidian monolith]

AyeWaddup

[grave tartan]

I can’t seem to remove the annoyance that (x-pi/4) is bringing in this limit

[fringe raft]

Use the limit definition of LHL and RHL

[grave tartan]

I’m sorry could you elaborate

How would splitting the limit work out

[mighty pier]

Notice that as $x\to\pi/4, x-\pi/4\to0$ It should be easier if you make that substitution

[obsidian monolith]

error 404

[grave tartan]

Oh yeah sorry forgot to say

I got help and this question is resolved

[tight ore]

Pls sent the solution @grave tartan

[willow bear]

we do not give out answers here

[scenic spear]

Ok so im doing some practice before finals, but I dont really know how to find part b

plugging in 0 and 1 for t and dividing them

it didnt work

how would i set this up?

[willow bear]

@scenic spear what you are saying is that the population more than doubles every year

increasing by 107.573% would mean that the population would go from 35000 to 72650 in a single year, which it doesn't

[scenic spear]

ok, if I plugged in 1 for part a can i divide from year 1/ year 0 to get my answer?

so 37650.56/35000

for 107.573 i multiplied it by 100

dividing 36650/35000 would be 1.0757

[willow bear]

that's the scaling factor

but you want the percentage change

you would agree that if a quantity was multiplied by, say, 1.2 then it grew by 20%, not by 120%. right?

[scenic spear]

yes

[willow bear]

yeah so subtract 1

[scenic spear]

1 minus the 1.0757?

[willow bear]

no

1.07573 - 1

[scenic spear]

srry i typed that wrong

would that give me the % population increase?

.0756 didnt work

[willow bear]

.-.

okay so like

idk if it's a mental block or just a misunderstanding or what

let's say i started out with $5000 and now i have $520.. by what percentage did my savings grow, and how did you work it out?

[obsidian monolith]

Ann

[scenic spear]

It decreased by 89 percent

[willow bear]

okay so i intended to have 5240 in there and not 520

but fine

how did you figure this out

[scenic spear]

v2-v1
--------*100
v1

i forgot abt this formula but idk if its used in this context

[willow bear]

that way of writing fractions hurts me on a spiritual level

anyway, percentage change = (new - old)/old * 100% yes

now apply it to your thing

[scenic spear]

idk how to implement fractions lol

in discord *

ah i see

so i got 7.573

%

[willow bear]

just use a slash and proper parentheses.

[viscid thistle]

does the logarithm power rule work when the argument is x+5 and not just x?

[willow bear]

what do you mean

are you asking whether $\log( (x+5)^p) = p \log(x+5)$? yes, of course that's true.

[obsidian monolith]

Ann

[viscid thistle]

2*ln(x) = ln(x^2) right?

[willow bear]

yes

[viscid thistle]

so what if it is 2*ln(x+5)

[willow bear]

you can write 2 ln(x+5) as ln( (x+5)^2 ) if you want

[viscid thistle]

is it ln(x^2 + 5^2)?

[willow bear]

no

[viscid thistle]

oh

so the argument is (x+5)^2

[willow bear]

if you wish to put it that way

[viscid thistle]

This is my question but I assume my second line is wrong

[echo wagon]

👀

[viscid thistle]

what have i done 😂

[echo wagon]

(a+b)^2 ≠ a^2 + b^2

[viscid thistle]

oh you right

thnx

[echo wagon]

Np

[analog rose]

how should i make further progress? i gotta solve using induction

[sick steppe]

simplify what you got

[analog rose]

am i supposed to foil everything?

that sounds like a giant mess

[sick steppe]

Cant really read your working clearly and also cba to actually do the question rn

but there's likely a trick for simplifying the num

[analog rose]

is this better

[frail moat]

You mind if I try to complete it for you?

[analog rose]

sure go ahead

[frail moat]

@analog rose

Sorry if the picture isn't that good. I don't have a good camera @analog rose

[analog rose]

wait

[frail moat]

And this is mathematical induction. I don't know if this belongs in #precalculus

[analog rose]

i can see it

@frail moat im confused on why you stopped at teh final step

[frail moat]

What final step?

The solution is complete @analog rose

[analog rose]

how do you know its complete?

[frail moat]

Ohh. That. So listen

Now substitute (n) for (k+1). If you get the original equation then you can come to the conclusion that the solution is completed @analog rose

Go on. Substitutes (n) for (k+1) @analog rose

I'm sure you'll get the original term

[analog rose]

ohhh ok that makes sense

thanks a lot man

[frail moat]

I'm glad that I was of any help bro 🙂

[frail moat]

Hey @analog rose
How come this text is different in color from the rest of the texts?
Apologies for my ignorance but I don't use this app much

[analog rose]

don't apologize lol

but I'm not sure what your asking

ohh wait thats because i pinged you

and when you get pinged the text appears yellow you

*to you

[frail moat]

What's ping? And why would someone do that?

[analog rose]

you just pinged me by saying "@analog rose"

when you ping someone they get notified

[frail moat]

Like this? "@analog rose"

Does my text looks different in color to you?

[analog rose]

yep

[frail moat]

Discord is just 🔥

[magic totem]

question : is it true that through permutation of digits of a number it is impossible to find the same modulo if the quotient is remains the same?

[ocean temple]

Guys I need help

<@&286206848099549185>

[jagged glade]

Let's continue in #precalculus shall we?@ocean temple

,rotate

[jagged glade]

Wow，these are hard

[ocean temple]

Ok

[jagged glade]

As I see, Nyann has propose a solving method for question 3 on question channel 0

[ocean temple]

Yeah saw that

I am trying on it

[jagged glade]

Ah, that's good

[ocean temple]

Yup I have done question no. 3

Can you you give some clue for others

[jagged glade]

I'm working on those too

[ocean temple]

Oh ok

[jagged glade]

So, for question 4
We need to make 6{x}²-5{x}+1≤0

Then we will have to range for {x}

[ocean temple]

Like do we have to use restricted domain ?