precalculus

[peak path]

soz

[inland igloo]

@peak path umm you tryna solve for me or...

[peak path]

@inland igloo yes i am solving it

@inland igloo im stuck too so maybe someone else can help? but ill ping you when i get it

[inland igloo]

@peak path Alright

[peak path]

do u get it ?

@inland igloo

[inland igloo]

what did you do to get it?

the answer

[peak path]

multiplied both sides by csc(x) + cot (x)

but im not very good at writing proofs, im not sure if thats considered a proof cuz i think you'll need LHS and RHS ?

[inland igloo]

I just started learning that unit today

so its kinda a challenge for me

[peak path]

its okay :)

yea proofs are annoying and they are always worth so many marks in tests

[slender river]

write one yourself

not trying to be rude

lol

i encourage you to try to justify this to yourself

yes

[peak path]

like -(quadratic formula for -f(x)) = quadratic formula for f(x)
@viscid thistle this is not always true

are you familiar with odd and even functions ?

ahaha i know what u mean

x = -b +- root ....

yes they do

-f(x) is f(x) translated in the x-axis

the roots will be the same

[restive stirrup]

Hey guys! Help me with this problem 🥺

[willow bear]

what have you tried so far and where are you stuck?

[restive stirrup]

Actually I already know the answer. It’s just that I’m having trouble with the solutions.

[viscid thistle]

That's exactly what Ann asked

Where are you stuck

Which part

[willow bear]

nah @viscid thistle don't you know that the best way to get help is to be vague and not provide any useful info until the helper magically knows exactly what the issue is and solves it for you? /s

[viscid thistle]

[willow bear]

[viscid thistle]

[latent siren]

Still nothin on this?

Idk how someone expects help then doesn't even say what they need help with

[sand harbor]

magic

[charred frigate]

for intervals of inc and dec do you include the actual local max/min?

for example would you say sin is increasing on (-pi/2,pi/2) or [-pi/2,pi/2]

[sick steppe]

Not included since the slope is 0

so it's neither increase not decreasing

[charred frigate]

tyty

[charred frigate]

Does anyone know of any precalc problem sets online? I don't know of any but someone was asking

[tame bluff]

Most textbooks are probably fine

[wicked hearth]

I’m confused with where I went wrong when plugging in the values into quadratic formula

[uncut mulch]

why do you think you got something wrong in what's posted?

your positive case just isn't completely simplified

[wicked hearth]

I thought I already simplified it can I cancel those twos?

@uncut mulch

[uncut mulch]

divide numerator and denominator by 2

yes

[wicked hearth]

I’m just a little confused because this is what my teachers answer key has

[uncut mulch]

the teacher is wrong

they're the one that didn't apply the QF properly

[wicked hearth]

Oh okay thanks that’s why I thought I was wrong

What about the negative part of my answer?

[uncut mulch]

neagtive case is incorrect

you miswrote the QF

[wicked hearth]

Oh lol I see it

Thanks @uncut mulch

[viscid thistle]

@restive stirrup
Find equation of ellipse where, 15 is the radius and 13 is the height
x^2 + y^2 = 169
(13x/15)^2 + y^2 = 169
Given that the bus is 10 feet high, and just barely almost touching the ceiling. We can just find what X is when Y is 10

(13x/15)^2 + 100 = 169
(13x/15)^2 = 69
13x/15 = sqrt(69)
13x = 15 * sqrt(69)
x = (15/13) * sqrt(69)

The two X's are ~ 9.585 and -9.585
Now find the distance between these two X's
~9.585 * 2

Answer: The bus has to be below ~19.17 feet wide to be able to fit through the tunnel.

if you don't understand something like how i got the equation of the ellipse, or why i found the distance between the two x's you can ask me

[restive stirrup]

It’s all good. I already know the answer. Thanks tho for the help. :))

[cobalt carbon]

Let Q = (-2, 3, 4). Then the distance between Q and the plane through the points A = (0,1,1), B = (1,1,0) and C = (1,0,3) is
d/sqrt(11) for some value of d. What's d? How do you do this type of problem?

[willow bear]

do you know how to find the distance between a point and a plane?

[cobalt carbon]

I have to use the normal vector?

unit normal vector?

@willow bear

[willow bear]

i asked you a simple yes-or-no question

[cobalt carbon]

err

no then

[willow bear]

i expected one of three answers:

1. yes, i do know how to find the distance between a point and a plane.
2. no, i do not know how to find the distance between a point and a plane.
3. i think i know a method but i'm not 100% sure, could you look over it for me and tell me if it works?

[cobalt carbon]

No, I dot not @willow bear

[willow bear]

ok, then look up how to do it and come back to me.

[cobalt carbon]

@willow bear i searched and then solved myself

[willow bear]

bravo

[steel tulip]

Is it true that the complement of the union of sets A and B is the same as the intersection of the complement of A and the complement of B? So $ (A \cup B)^{c} = A^{c} \cap B^{c} $?

[viscid thistle]

yes

[obsidian monolith]

Tiessie:

[viscid thistle]

you can see why by drawing a Venn's diagram

this is De Morgan's law btw

[steel tulip]

That's what I did but I dont know if it's an actual proof

[viscid thistle]

can i see it

[steel tulip]

One sec let me draw it again I kinda drew over it

The bar over the set indicates complement

[grave tundra]

Any idea how to simplify this further?

[viscid thistle]

@steel tulip whoops, i lowkey forgot, yeah no that's not a valid proof, that's just a visual of it

need help with the proof?

[steel tulip]

uh sure i have no idea really where to begin

[viscid thistle]

Alright

Let $Y=(A\cup B)^c$ and $Z=A^c\cap B^c \ $ For example, let x be an arbitrary element of Y, hence $\ x\in Y$ implies $x\in (A\cup B)^c\ $ $x\notin (A\cup B)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Can you try to continue from here? @steel tulip

[steel tulip]

Because $ x \notin A \cup B, x \notin A$ and & x \notin B$ Thus x must be an element of $ A^{c} \cap B^{c}$ ?

I feel like that's still missing a step

[viscid thistle]

Because $ x \notin A \cup B, x \notin A$ and & $x \notin B$ Thus x must be an element of $ A^{c} \cap B^{c}$ ?

[obsidian monolith]

Al𝟛dium:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

(quoting what you said)

Where do you feel it's missing a step

[steel tulip]

I don't see how the conclusion that x must be an element of Ac intersect Bc is strictly true

Thats only really obvious to me with a diagram

[viscid thistle]

Do you not see the step
$\ x\notin A$ and $x\notin B\ $
$\implies x\in A^c \cap B^c$?

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

(just being sure of which step you are referring to)

[steel tulip]

That's what I'm referring to, that step is only obvious to me with a diagram

And since you said the diagram was only a visual, I'm not 100% certain that that step is necessarily true

[viscid thistle]

Indeed it is a visual to clarify stuff

Maybe if i break that step into 2 you see it clearer

Let me write it

$\ x\notin A$ and $x\notin B\ $ $\ \implies x\in A^c$ and $x\in B^c\ $
$\implies x\in A^c \cap B^c$

[obsidian monolith]

Al𝟛dium:

[steel tulip]

Ah yes of course that makes sense

[viscid thistle]

I mean the diagram is a good way too in order to understand it, you can use it for steps too ofc

I didn't say the opposite

Ah yes of course that makes sense
I'm glad to hear

[loud marsh]

i know this is simple af but i forgot how to

[viscid thistle]

so

$y=4^{x}

[loud marsh]

what does $ mean

[viscid thistle]

it is for writing TeX equations

[loud marsh]

ohh okok

[viscid thistle]

dk why it doesn't work

$ y=4^{x}

welp

[loud marsh]

welp

[viscid thistle]

y=4^{x}

anyway

so the inverse of the equation should be starting with x=

[willow bear]

you need to put dollars on both sides

[viscid thistle]

ah, tysm

[willow bear]

dollars start AND end math mode $y = 4^x$ like this

[obsidian monolith]

Ann:

[viscid thistle]

so, since you know that log is how much time you need to multiply 4 to get y, $y=4^{x} \implies x=\log_{4}{y}$

[obsidian monolith]

DrifAssault:

[viscid thistle]

as for b

you know that 6 power to y is x

so, you get $y=\log_{6}{x} \implies x=6^y$

[obsidian monolith]

DrifAssault:

[viscid thistle]

@loud marsh here is your explainations, I hope that I explain it well enough

[loud marsh]

life saver

thanks

[viscid thistle]

np

[wicked hearth]

I just wanna double check my answer

[sudden bay]

can someone provide me with a simple guide about how to factor quadratics

Assume i know nothing ( i dont know anything )

[quaint mason]

how do i find c and d from this graph??

[sick steppe]

c is just go up from 5, then across to the vertical axis

d is basically eyeball the point of inflection, which is usually in the middle of logistic curves

[quaint mason]

u mean up from 6? so estimate it a bit?

ah gotcha but question for d

how am i suppose to round it to one decimal place? its just worded weird @sick steppe

[sick steppe]

Yeah whatever the number is it's small lol

[quaint mason]

thats what im thinking

[sick steppe]

and the graph basically exists between 0 and 15

so you expect concavity to change around 7.5

[quaint mason]

ah got it

[sick steppe]

But the way to verify is to put a dot on the graph at 7.5, cover one half of the graph and see if that is 1 concavity, then repeat for the other side

[quaint mason]

sorry i was just confused

yeah

[sick steppe]

so like cover the left side you get down, cover right you get up

[quaint mason]

its inflection point is at the half mark

where it goes from concave up to down

[sick steppe]

yeah

[quaint mason]

sorry i was just confused a bit and thought i had to do some calculations

ty :)

[sick steppe]

no worries

[quaint mason]

alrighty gotcha tyvm bro @sick steppe

[gilded brook]

I'm having a tough time figuring out this last part

It = -pi/4 but the answer needs to be in 0<=theta<2pi so I add pi to it to get 3pi/4

But clearly it's wrong

[echo wagon]

Why add pi?

[sick steppe]

-pi/4 is right, but 6-6i is in quad 4 not 2

so you want a full revolution, not half

[echo wagon]

Arguments for the same point differ by multiples of 2π

[sick steppe]

yep

[gilded brook]

Oh and pi is half right

[sick steppe]

so -pi/4 + 2pi

[gilded brook]

So I just needed to do -pi/4+2pi

[sick steppe]

yeah

[gilded brook]

Forgot pi is just 180

[sick steppe]

you're right w/ 3pi/4 if the complex number were -6+6i

[gilded brook]

circly speaking

[worn violet]

hey could someone help me with this calculus problem

Give an example of a polynomial of degree 5 with roots -2 and −i.

[cobalt carbon]

@worn violet well, we know one of the factors of degree could be (x + 2). What factor could result in a root of -i?

[slate venture]

How do you find the 0's of this equation?

(x^3) + (11x^2) + 39x + 29

[uncut mulch]

try applying rat root first

also not an equation

[viscid thistle]

guys how would you do part c?

im thinking of sutracting the cases where no men, one man and two men sit together from the total number but i am struggling to mathamatically form the number of cases when one man sits between the boys and 2 men sit between the boys

[wide ocean]

anyone know range?

[clear sundial]

@wide ocean range represents possible y values for the function

keep in mind closed whole means the function has a y value for the corresponding x value at the point

[wide ocean]

@clear sundial oh i see.. not sure how to present it

[clear sundial]

you can use interval notation

[wide ocean]

since there's a gap.. stumped

ohh, like how?

[clear sundial]

[value, value) U [value, value]

have you learned that notation?

[bitter basin]

correct

?

[burnt sonnet]

How do you have a negative mass on the top one? 😮

@bitter basin

What equation did you set up for that one?

[bitter basin]

A= Pe^kt?

[burnt sonnet]

what is A,P,k?

[bitter basin]

i forgot

lol

[burnt sonnet]

HMm

Not good xD

But I tihnk of it as N(t)

the mass as a function of time

and I know that it is exponecial decay

$N(t)=N_0 \cdot e^{-\lambda \cdot t}$

[obsidian monolith]

egocarpo:

[burnt sonnet]

where N_0 is the starting quantity

then lambda is ln(2)/T_{1/2}

where T_{1/2} is the half life

rewrite it to

$N(t)=N_0 \cdot 2^{\frac{- t}{T_{\frac{1}{2}}}$

[obsidian monolith]

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

[bitter basin]

hmm

[burnt sonnet]

no familiar?

we can try and put in t=T_1/2

$N(T_{\frac{1}{2}})=N_0 \cdot 2^{-\frac{T_{\frac{1}{2}}}{T{\frac{1}{2}}}}=N_0*2^{-1}=\frac{N_0}{2}$

well if N is A is it okay then?

Just a name

[obsidian monolith]

egocarpo:

[bitter basin]

idk

[burnt sonnet]

WEll in my formula if I plug in T_1/2 I get half of the starting amount which is what I want

so you know that N(1000 yrs)=1.5grams

but if you accept my formula

you also know that $N(1000 yrs)=N_0 \cdot 2^{-\frac{1000 yrs}{T{\frac{1}{2}}}}$

[bitter basin]

yea

[obsidian monolith]

egocarpo:

[burnt sonnet]

You have the half time

so the only unkown is N_0 that is the starting amount

[bitter basin]

yea

[burnt sonnet]

$1.5=N_0 \cdot 2^{-\frac{1000 yrs}{1599yrs}}}$

[obsidian monolith]

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

[burnt sonnet]

solve for N_0

2^-... is just a number do it on a calculator

because it decays

the amount should reduce with time so the exponent must be negative

for the right part?

It must be bigger than 1.5 grams

yean N_0

should be bigger than 1.5

[bitter basin]

oh

[burnt sonnet]

1.5 is the mass after 1000 years

so the starting mass should be more than 1.5 since it reduces over time

[bitter basin]

hmmmmm

[burnt sonnet]

well then you messed up the 2^-..

I get 4.5 something

on the right side

[bitter basin]

yeah

[burnt sonnet]

$1.5=N_0 \cdot 2^{-\frac{1000 yrs}{1599yrs}}}$

[obsidian monolith]

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

[bitter basin]

yeah i did

[burnt sonnet]

$N_0=1.5\cdot2^{\frac{1000 yrs}{1599yrs}}}$

[obsidian monolith]

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

[burnt sonnet]

the minus sign dissapears

THE MINUS SIGN xD

[bitter basin]

oh

[burnt sonnet]

you have issed it

missed it*

[bitter basin]

lol

[burnt sonnet]

hihi

yeah I think so

I might have misscalculated

It is negative before you move it over the 1.5 sid

e

[bitter basin]

yep

[burnt sonnet]

But most important

do you understand how I was able to say that 0.9 was wrong?

this is super imoportant understanding that you need to have

ok Good

awesome

Yw

hmm

why do you think it is e^0.05?

Is it the same model here?

You can use that but r*t is not 0.05) in that case

I pref to look at these as $y(x)=C*a^x$

[obsidian monolith]

egocarpo:

[burnt sonnet]

where C is the starting amount since

$y(0)=C*a^0=C$

[obsidian monolith]

egocarpo:

[burnt sonnet]

then a is the growth rate

well it grows with 0.05

so the growth rate is 1+0.05=1.05

is this familiar

[bitter basin]

oh yes

[burnt sonnet]

so do you see what C is?

[bitter basin]

compound

interest

[burnt sonnet]

C is the starting amount

[bitter basin]

o

ye

[patent lance]

so i dont think this is a precalc question but its in my precalc class so im asking here

this question

feels so simple

[burnt sonnet]

the constant infront of the expoential

[patent lance]

but i got it wrong

twice

and im so confused

[burnt sonnet]

Frosty we are doing a question here

[bitter basin]

ok

lo.

l

[burnt sonnet]

either wait or go to questions alpha, beta, ..

Are you able to find C blueTurbo?

yes good

C is 9000 dollars

[bitter basin]

ye

[burnt sonnet]

the growth rate is 1.05=a

so $y(x)=9000*(1.05)^x$

[obsidian monolith]

egocarpo:

[burnt sonnet]

okay so what was the question again

I am asking you xD

[bitter basin]

lol

[burnt sonnet]

So you understand what we are looking for

yes good

[bitter basin]

so two?

[burnt sonnet]

well two what?

[bitter basin]

years

[burnt sonnet]

NO!

yes

That is what the amount should be equal to yes

can we call the time when it this happens $x^*$

[obsidian monolith]

egocarpo:

[burnt sonnet]

yeah

I wanna put it like this

[bitter basin]

yes

[burnt sonnet]

$y(x^)=2\cdot9000dollars=9000\cdot(1.05)^{x^}$

[bitter basin]

got it

[obsidian monolith]

egocarpo:

[burnt sonnet]

ok cool

$2=1.05^{x^*}$

[obsidian monolith]

egocarpo:

[burnt sonnet]

seemes like it could be something like that yes

yes!

agreed!

[bitter basin]

ok

[burnt sonnet]

I don't know the numbers I just know the operations to do 😄

Is this one clear?

[bitter basin]

still helps

yes

[burnt sonnet]

awesome

hmm

OK I think I have time for it

ok what is the problem?

well

t is given aswell right?

hmm

no 😦

you find out that the population is 363000 at year 2006

the starting year is 2000

[bitter basin]

yes

[burnt sonnet]

ok

[bitter basin]

hmm

[burnt sonnet]

then you see how you find k?

[bitter basin]

no

[burnt sonnet]

😮

ok

$363 = 289.81\cdot e^{(k\cdot6)}$

well just remove 3 0's

[bitter basin]

ok

[burnt sonnet]

this equation is ok!

then

[bitter basin]

ok

[obsidian monolith]

egocarpo:

[burnt sonnet]

what is next step to get k

or k*

[bitter basin]

divide

[burnt sonnet]

yes

$\frac{363}{289.81} =e^{k(6)}$

[bitter basin]

oh yes

i had that written

[obsidian monolith]

egocarpo:

[burnt sonnet]

ok what is the next step

maybe some nice logarithm

[latent siren]

@patent lance A hint is to find a line perpendicular to the one given that passes through the point given then find the point where they cross

[bitter basin]

hmm

log

[burnt sonnet]

you can use any but log base e is the nicest

😄

since ln(e)=1

[bitter basin]

yeah

[burnt sonnet]

$\ln(\frac{363}{289.81}) =\ln(e^{k(6)})=k\cdot 6 \cdot \ln(e)=k\cdot 6$

[obsidian monolith]

egocarpo:

[bitter basin]

ok

[burnt sonnet]

do you see how you get k now?

[bitter basin]

yes

[burnt sonnet]

ok sick

then use the k and the correct t to find the population in 2017

[patent lance]

@latent siren ty that helped

[burnt sonnet]

or what ever year it was

what is 6k?

[bitter basin]

i mean

[burnt sonnet]

you need to find k in the top equation

no there is only one k

k=ln(363/289,81)/6

one number one solution

me 2

so that is the k

[bitter basin]

ok

[burnt sonnet]

now we have the function for the population x years after 2000

[bitter basin]

yea

[burnt sonnet]

A(x)=289.81*e^(0.0375x)

now compute this for x=17 🙂

what is th

548 what?

people?

[bitter basin]

yes

[burnt sonnet]

no

yes good!

[bitter basin]

so

[burnt sonnet]

548.247 thousand people

maybe you have to round down or up since

people are usally whole 😂

[bitter basin]

lol

[burnt sonnet]

or well

it is thousant people so

[bitter basin]

oh

[burnt sonnet]

548 247

[bitter basin]

yep

[burnt sonnet]

yeah you have to be careful!

did you round off k?

it says without rounding the value of k

I get 548 522

if you take

k=ln(383/289.81)/6

and plug in this ln expression as k

😮

I am not sure what is wrong on that one tho

because this investment thing is pretty straight forward I think xD

[bitter basin]

ye

[burnt sonnet]

. ,?

[bitter basin]

,.

[burnt sonnet]

yeah which one did you use?

have that one worked before?

okay

have that worked before

or did they want comma before?

ok

try comma instead of period maybe?

yes i get it to 14.207 aswell

[bitter basin]

ok

[burnt sonnet]

you take ln(2)/ln(1.05)

hmm

[burnt sonnet]

is the last one correct atleast?

548 522

ok cool

yeah I got the same 😛

what is compount intresst different from normal intrest?

Could we have the formula wrong?

how do you mean? 🙂

You could try it but I don't think that is how it works 😛

after a year you will have

y(1)=9000*e^1.05

it have not increased with 5%

it have increased with like a factor 3

but try your thing

xD

^_^

It will get worse 😂

well

I think it just decimal wierd thing or something

I am certain of the method we used .P

Yeah I don't know what to tell you 😛

[bitter basin]

alright

[burnt sonnet]

yw

[cobalt storm]

Can someone explain to me how #11 is wrong?

[latent siren]

Explain to me how you got your answer

[cobalt storm]

I made a function f(x) = ax(x+5)(x-3)(x-10)

Then put in f(5)

And I got an a value of -0.009

From there I fixed the equation to be f(x) = -0.009x(x+5)(x-3)(x-10)

Then inputted f(1)

And got a negative value of -0.972 as my answer for f(1)

So that would mean the answer was true

Bc f(1) is negative

But then I got it wrong apparently

[latent siren]

@cobalt storm

[cobalt storm]

I don’t get it, when I plugged in 1 to my equation I got a negative

[crystal osprey]

@latent siren That graph I guess makes sense if you build it off the ambiguity of the wording of the question

@cobalt storm You're given 3 x intercepts to a quartic function correct?

hold up gonna eat dinner first before explanation

[cobalt storm]

Yeah

The function I came up with was f(x) = -0.009(x+5)(x-3)(x-10)

And it works cuz I checked with f(5) and I get an output of 4.5

I also put it in a graphing calculator and f(1) is negative

[crystal osprey]

That is not a quartic function.

@cobalt storm

Also sorry for the wait

(x+5)(x-3)(x-10) produces to a polynomial degree of 3.

Unless if you include -0.009x as a factor

The question is, why -0.009x particularly?

The question stated 3 x intercepts. That doesn't mean there isn't a 4th.

The 4th x intercept doesn't have to be at 0 either. It could literally be anywhere as long as it satisfies the condition that f(5) = 4.5

KoolAidWannaBe gave an example of this

[cobalt storm]

I did -0.009x as a way to make it quart in

Quartic*

[crystal osprey]

The question is also ambiguous as to not state if there's ONLY 3 unique x intercepts. The important giveaway here, is that it should apply to all instances. Is it true that f(1) will always end up as negative no matter the circumstance? Even one exception is enough to prove it false.

@cobalt storm right but you're stating that one of the x intercepts are at 0

[cobalt storm]

Oh I get the idea the principle that it has to equal negative based on the idea that there are only 3 x int

[crystal osprey]

Not necessarily

[cobalt storm]

Wait but 0 isnt an x intercept is it?

[crystal osprey]

it doesn't matter whether there are only 3 unique x intercepts nor if there's 4 unique x intercepts. Only confirmed information you have are that you are given the x coordinates of 3 of the 4.

@cobalt storm 0.009x is a factor.

[cobalt storm]

So this isn’t really a question that involves actually substituting in numbers and determining the y value for f(1)?

It’s more of just a theory based question?

[crystal osprey]

Yeah, I'd say so

If you review the laws of polynomial functions enough you can do this mentally

here is your graph for reference

[wide ocean]

'find domain and sketch graph'

this is so WIERD. are there three separate equations? @frail bloom

[abstract ore]

@wide ocean It's a piecewise function

[velvet granite]

Exactly

[patent beacon]

Rofl who did you just ping

[wide ocean]

Mb, what is a piecewise function mean?

[abstract ore]

It means that there are different equations for the line depending on what x is

[crystal osprey]

You probably cannot do this if you do not learn about it

[patent beacon]

When x≤1, then f(x) = -1

[wide ocean]

Ohhh so it's an irregular graph

I see I see

[patent beacon]

And when |x| < 1 (fancy way of writing -1 < x < 1) then f(x) = 3x + 2

[crystal osprey]

It's a singular function that can be split into different expressions to represent each portion basically

[patent beacon]

So this is literally pieces of graphs being put into one

[velvet granite]

When given a < or > you get an open circle on your graph and when we get a >or equal to we get a closed circle on the graph just remember that

[patent beacon]

Piecewise

[crystal osprey]

Yup

Name checks out

[wide ocean]

Thanks everyone

Wait I'm still confused on the y= -1 when x is ≤ 1

like huhh

[velvet granite]

its a straight line on y=-1

but its only included from the x value 1 to negative infinity

[patent beacon]

,w graph -1

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Kaynex#7520

[wide ocean]

Oh, I see!

Until x=1

[patent beacon]

The function that is always -1

[crystal osprey]

yup

[wide ocean]

And 3x+2 ≤ 1, you isolate for x

[patent beacon]

Or, will always output -1 no matter the input

[wide ocean]

so X = -2/3

[patent beacon]

Oh boy

[wide ocean]

for positive values

[velvet granite]

no thats a slope equation

[wide ocean]

Oh right my bad

y = Mx + b

so it's a postive slope

[velvet granite]

Yes

[wide ocean]

and it intercepts at y=2

[velvet granite]

What I usually do to graph these is create a table starting from the number given

[wide ocean]

for 3x + 2 , when x is less than one

I see

[crystal osprey]

sounds about right

[wide ocean]

May I ask if the slope of 3 intersects with the first line

because the y intercept is 2? For all positive values of X

And when |x| < 1 (fancy way of writing -1 < x < 1) then f(x) = 3x + 2
@patent beacon Oh I seee

Now for the last part

7-2x

the last part, when x > or equal to 1

it doesn't touch the rest of the graph?

My end result 🙂

Thoughts?

[velvet granite]

well it starts at the x value 1 and goes to positive infinity so i guess yes

[azure hedge]

hello

anyone know a good site to practice geometry and vectos precalculus?

[wide ocean]

@velvet granite What do you think of my graph? ty

Does it seem right ? 🤔

Or should it still touch the point (0,7) ?

but it says while X > or equal to 1

🙂 If anyone knows this ? please

[velvet granite]

wait disregard what i just said

[wide ocean]

@velvet granite ah okay. wait do you think my graph is right ? any tweaking

[steady fulcrum]

@wide ocean looks good to me

oh wait

one thing

the domain is really small

[velvet granite]

looks good to me buddy

[steady fulcrum]

since the domain is x is between 0 and 1

the x value shouldn't be smaller or greater than 1

so cut off the graph

oh i'm sorry that's a separate domain on the bottom

if the domain is the one defined on top then it's perfect @wide ocean

[viscid thistle]

Hey, can someone help me on this? It’s a SSA triangle.

[peak path]

@viscid thistleisnt this a SAS triangle? I recommend you draw it out and you'll see it :)

[viscid thistle]

Yes I did, but I don’t know how to find the other sides. 😦

:^(

[peak path]

you dont need the other sides to find the area

[viscid thistle]

ohhh

[peak path]

are you familiar with finding area using the sine rule?

[viscid thistle]

No

[viscid thistle]

I’ve only worked on finding all the other sides and angles.

[peak path]

this is the formula for it

its actually possible to find all the sides

even though the question doesnt require it

[viscid thistle]

Oh

But how would I obtain SinC?

And SinB?

[peak path]

that formula is just an example

[viscid thistle]

Oops I meant b

Oh

[peak path]

in your case, it would be 1/2 ac sinB

the values are given in the question

[viscid thistle]

So you only use what you have?

[peak path]

the formula i sent can only be used if u know two sides and the included angle

in your case, since its a SAS triangle, u can use it

[viscid thistle]

Ohhhh that makes more sense

[peak path]

now u have to substitute the values correctly into the formula

[viscid thistle]

:^D

I thought you had to find them and substitute them into the formula

But it makes more sense now

[peak path]

basically the formula is A= 1/2 (side)(side) sin (included angle)

ahaha nah, the formula is based on the diagram next to it

[viscid thistle]

Okay

Thank you!

[peak path]

no problem! the sin rule and cos rule is very useful in finding the sides and angles too

i can send u the formula if youll like :)

[viscid thistle]

Sure! Thank you!

[peak path]

the sine rule is used when:

1. 1 side and 2 angles are given
2. 2 sides and 1 non-included angle are given

the cosine rule is used when:

1. 2 sides and the included angle is given
2. 3 sides are given

no problem !

[viscid thistle]

Thank you!! :’)

[lapis iris]

got this notation on an assignment

the question is asking about the dot product of the vectors

so I assume á and ñ are vectors (idk what’s up with the accents or wierdly sided coefficients lol)

[willow bear]

talk to your teacher and tell them the formatting is fucked up

[spice wadi]

yeah

just assume a' is x, n~ is y and try it

[warped dagger]

have i done anything wrong here?

[viscid thistle]

Your handwriting is lowkey messy

And you are addicted to "cancel" anything you can where for example at 2² you don't cancel the 2.

But seems fine so far from what i can manage to read @warped dagger

[warped dagger]

Thanks, my handwriting sucks

[calm whale]

these 4's will haunt me

[warped dagger]

not sure how to go about solving for H, square roots are scary

$24\sqrt3 cm^3 = 3\sqrt3 * H$

[obsidian monolith]

ddm4313:

[calm whale]

I would just not write cm cubed on every line

to solve for H simply divide by 3sqrt(3)

as the roots cancel out, you'd get 24/3=8

[warped dagger]

thanks, appreciate it.

[calm whale]

rip latex but nps, I think it can get really confusing if you leave the metric units in the eq-n

[copper vigil]

can someone explain matrix multiplication to me

i don't get it

[viscid thistle]

@copper vigil search 3b1b matrix multiplication :)

[patent beacon]

A matrix can be thought of like a function that acts on a vector. Any linear operator can be instead written as a matrix.

In that sense, matrix multiplication is equivalent to function composition. That is,
A(Bx) = (AB)x

@copper vigil

[leaden stratus]

Can you help me with this? I need to restrict the domain of these functions to make them invertible. Guide me, please. Thanks!

Anyone? Tag me if you reply

[somber yew]

@leaden stratus Do you know about the horizontal line test for testing if a function is one-one(injective)?

Apparently you aren't active atm so I'll proceed with the explanation.

[leaden stratus]

I'm here

[somber yew]

Basically, for a function to be one-one(injective), no horizontal line should intersect it more than once.

Oh good.

[leaden stratus]

No, I don't know about it

[somber yew]

Basically, for a function to be one-one(injective), no horizontal line should intersect it more than once.

[leaden stratus]

Hm ok

[somber yew]

So if you draw a horizontal line through those 4 graphs, can you tell me which ones are one-one and which ones are not?

Remember, if your line can possibly intersect it at 2 or more places, it isn't one-one for the given domain at the very least.

[leaden stratus]

I don't know where to draw this horizontal line 🤔

[somber yew]

Mmm nvm

Do you know when a function is said to be invertible?

[leaden stratus]

When it's both injective and surjective

[somber yew]

Correct. When is a function injective?

[leaden stratus]

When every element of the codomain is the image of at most one element of its domain

I know the definitions of injective, surjective and bijection

[somber yew]

Correct. Look at graph (c) for a moment. Does it satisfy injectivity?

[leaden stratus]

No

Because it is y = x^2 (I know this one)

So y can have two images

[somber yew]

Correct. So you need to restrict the domain in such a way that half of the portion is cut out.

Do you know how you can do that?

[leaden stratus]

Only the positive values?

From 0 to + inf.?

[somber yew]

Yes. Either only the positive values, or only the negative values.

You can make the function invertible for either, but not both simultaneously.

[leaden stratus]

Ok

[somber yew]

So either [0,+infty) or (-infty,0], but not both.(don't take their union!)

[leaden stratus]

The graph (a) is injective, right? But not surjective

[somber yew]

Yes, it is not surjective, but has the range/codomain been written out explicitly in your original question?

Because none of these functions(except possibly (b)) is surjective with R as the range

[leaden stratus]

No, no range written

[somber yew]

Hmm, then I suggest you just assume the codomain=range, i.e., the function is surjective(mention this explicitly)

[leaden stratus]

For example, the (a) domain what could be?

[somber yew]

The function in (a) is injective for all real numbers.

Do you see any y values being repeated?

[leaden stratus]

Yes, starting from some x values, the y remains constant

[somber yew]

For future reference, this is the horizontal line test. Makes sense?

Well, that's a bit tricky- y isn't becoming constant here but actually increasing very, very slowly.

You could interpret this as the graph of arctan(x) with x in (-infty,+infty)

[leaden stratus]

Ah yeah, it slowly increases tending to the asymptote @somber yew

[somber yew]

Correct.

So every x value corresponds to a unique y value, making (a) injective.

And since we have conveniently assumed codomain=range, (a) becomes invertible for all real x.

[copper breach]

Can someone please help me with this problem and give me an explanation of how to do this? Thanks so much!

[leaden stratus]

So the correct domain for it to be invertible is D: {-infinity, +infinity}?

[somber yew]

@copper breach Channel in use. Try one of the question channels which isn't occupied.

Yes.

[copper breach]

sorry

[somber yew]

Avoid that notation though lmao, it's supposed to be (-infinity,+infinity)

{a,b} represents two elements of a set, not an interval.

[leaden stratus]

While (b) is some logarithmic function?

[somber yew]

Yeah, it seems invertible right from the looks of it.

[leaden stratus]

No, wait

It's exponential

[somber yew]

For (c) you know what to do, and (d) is more or less the same.

Nah I guess that graph pertains to -ln(x)

,w plot y=-ln(x)

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by TedNowKaczynski#1729

[somber yew]

[sick seal]

would you mind if i interrupt rq for a quick yes or no question?

[leaden stratus]

Yes, go ahead

[somber yew]

Uh sure

[sick seal]

is the thing flawed wrong or do i need to go back

[somber yew]

Mmmm a linear function has constant slope

So you see, chart (B) gives you a constant slope, while in chart (A) the slope keeps increasing.

Other than that, you can just plot these points quickly to check if they're linear or exponential.

Try it. The correct answer(don't see until you've tried) is : ||B is linear, A is exponential.||

[sick seal]

yea thats what i selected... but so the promped is messed up

thanks i thought i was goin crazy

[somber yew]

Hahah, no issues. Talk to your instructor about it.

@leaden stratus

[leaden stratus]

Hey

Well

[somber yew]

Can you now get (d)?

[leaden stratus]

The (b) is -ln(x) and tends to get close to y-axis

[somber yew]

Yes, keeps growing without bound for values of x close to 0.

The function is clearly injective

[leaden stratus]

So the domain for it to be invertible would be from 0 (excluded) to +infinite?

[somber yew]

Correct(that is actually the domain over which the function makes sense, since we don't define log for negative values)

[leaden stratus]

(d) is an even function

[somber yew]

Ah, you could have played the even card sooner XD

(c) is even too.

[leaden stratus]

Lol, you're right

True

[somber yew]

Regardless, there's an additional catch with (d)

Even when you consider only positive values/only negative values of x, the hill in first and second quadrant seems to be repeating y values.

So you need to define your domain in a way such that you only include one part of the hill.

[leaden stratus]

But it's not injective then

And neither surjective

[somber yew]

Well, that is why you need to restrict the domain, no?

(And we're taking surjectivity for granted)

[leaden stratus]

Yep

[somber yew]

Although in this particular case, things are a bit messed up since we need to know the coordinates of the top of the hill, i.e., the function maxima.

[leaden stratus]

So only negative or positive values?

[somber yew]

Nah, as I said, that doesn't work due to those two hills.

You need to know the function maxima to describe the domain sensibly.

[leaden stratus]

Ok

[somber yew]

Otherwise, just say the function maxima occurs at the points (a,b) and (-a,b). Then the domain can be one of these four: [0,a], or [a, +infinity), or (-infinity, -a], or [-a,0].

Makes sense?

[leaden stratus]

Yes

Makes sense

[somber yew]

Well, anything else left?

[leaden stratus]

Not atm

Thanks

[somber yew]

No worries.

[leaden stratus]

Here again @somber yew with another problem.
"Ingrid and Manuel's farm sells orange jam. The productivity q in kg of an orange tree in function of its age t in years is represented by the parabolic graph in the figure; the number of jam jars produced is described by the function n=4/25 * q.
Write the function that expresses the model of jars obtained by a tree in function of its age.
What function expresses the model of jars produced from 300 trees planted in the same moment in function of their age?

For the first question the answer is n(t) = -(1/100)t^2 + 2/5t
For the second question the answer is n(t) = - 3t^2+120t

[somber yew]

My brain's literally dead at this point mate, I'll have to pass. Someone else will help you out for sure.

[leaden stratus]

Oh ok, thanks

No problem

[sick seal]

how do i calculate C?

ive tried everything i could think of sofar

[sick seal]

except for being smart

i figured it out

[edgy blaze]

question for half angle, how do you know if it's + or -? I thought it would be based on quadrant. But when my Sin(a) was negative, the sin(a/2) said the negative version was wrong and positive was right

[cobalt storm]

I believe it’s false as I’m not given the whole function to verify, but I am given 2 x intercepts

[quick mirage]

if f(-3) = 0 and f(4) = 0, then anything in between must be either a strictly positive or negative number (because otherwise there would be another x-intercept). Since f(-2) is positive, then f(-1) must also be positive.

f(x) > 0, {x∈ℝ | x > -3, x < 4}

@cobalt storm

[cobalt storm]

Then would it possible to have a x int at 0,0?

For the rational?

[quick mirage]

It would have been stated if there was

|f(x)| > 0 with the above mentioned domain

[cobalt storm]

But in general is it possible for a rational function to have a x and y intercept at 0,0?

[quick mirage]

of course

y = x

[cobalt storm]

Doesn’t the parent function of 1/x not have it at 0 though

[quick mirage]

well some functions dont have intercepts at 0, 0 for sure

1/x doesnt have an intercept at all

but like polynomial equations

linear equation

absolute value functions

they all can contain x / y intercepts

[cobalt storm]

Thanks this helped a lot

Slowly starting to understand the theory behind all this stuff

[quick mirage]

np

another note: though 1/x doesnt have such an intercept, a transformed version of the function 100% can

[leaden stratus]

Can someone help me? I posted the question hours ago

[quick mirage]

What's your question?

[wide ocean]

17 c) please

Anybody know the steps. Make f(x) -> f(-x)

then what?

[patent beacon]

Pretty much just take a good look at f(-x)

If it's the same as f(x) then you've got an even function

If it is the negative, then you've got an odd function @wide ocean

[topaz obsidian]

looking for a math wizard please i have a couple questions left on my project for pre cal

[patent beacon]

Ye

[terse ravine]

How do I Isolate y^1

when I have 3y^1 = (-2x-3y/x-y^2)

@patent beacon can you help me?

[burnt sonnet]

oh you need y'?

the derivative of y

[terse ravine]

yeah I need it isolated..

[burnt sonnet]

oki!

I can help with that if you want

[terse ravine]

ok

[burnt sonnet]

So the first expression does not have a y' right?

[terse ravine]

no I did the work above

[burnt sonnet]

oh ok

[terse ravine]

and show where I got stuck

[burnt sonnet]

ok

so divide with 3

and you have y' isolated

[terse ravine]

yeah that is spooky :x

[burnt sonnet]

why?

3 is a constant

🙂

[terse ravine]

so divide 3y^1/3

[burnt sonnet]

What does that mean?

I am reading $3*y^{1/3 }$

[obsidian monolith]

egocarpo:

[burnt sonnet]

I am sure this is not what you mean

[terse ravine]

you mean this?

[burnt sonnet]

Yes!

but you can make it simpler

[terse ravine]

how because this is tricky and I'm learning still..

[burnt sonnet]

okey

I view it as multiplying with 1/3 on LHS and RHS

[terse ravine]

before we do that can you see if my algebra above is correct?

[burnt sonnet]

$3\frac{y'}{3}=\frac{1}{3}\cdot \frac{-2x-3y}{x-y^2}$

[obsidian monolith]

egocarpo:

[terse ravine]

This is implicit differentiation problem

[burnt sonnet]

yes

You can view it they way I put it in

I am lazy and don't wanna have to check xD

[terse ravine]

xD

[burnt sonnet]

but I will go through your algebra

It looks good!

$y'=\frac{1}{3}\cdot \frac{-2x-3y}{x-y^2}=\frac{-2x-3y}{3(x-y^2)}$

[obsidian monolith]

egocarpo:

[burnt sonnet]

do you see what I do here

[terse ravine]

ok

[burnt sonnet]

This is to make it easier to see how the division works

Do you see the difference in this and what you did?

Almost

[terse ravine]

darn

[burnt sonnet]

do you see where it is wrong?

[terse ravine]

oh

[burnt sonnet]

that doesnt mean anything to me first you say 3y' then you have an equation xD

The error is in RHS

you forgot a three on -y^2

the denominator is 3(x-y^2)=3x-3y^2

🙂

[terse ravine]

Oh of course!

[burnt sonnet]

Distibute the 3 on both 🙂

write out your y' expression for me now

[burnt sonnet]

yes

now you wanted y' at x=4 and y=-4

[terse ravine]

yes

[burnt sonnet]

Plug and play

[terse ravine]

will do

= 5/9

[burnt sonnet]

(-8+12)/3(4-16)

5 up top 😮

4/(3*12)

1/9

[terse ravine]

1/9

-2(4)-3(-4)/3(4)-3(-4)^2= 1/9

[burnt sonnet]

nice

do you wanna have a life hack?

[terse ravine]

its wrong hmm

[burnt sonnet]

is it?

[terse ravine]

yes

its negative -1/9

[burnt sonnet]

oh

yes

the bottom is negative

(-8+12)/3(4-16)
5 up top 😮
4/(3*(-12))
1/9

[terse ravine]

was my mistake

ty

[burnt sonnet]

4-16=-12

:d

[terse ravine]

c:

[burnt sonnet]

do you want the life hack for these questions?

[terse ravine]

yes

[burnt sonnet]

so after you have taken the derivative

you can plug in x=4 and y=-4

then you just have some numbers infront of y' and some constant terms

some people find this easier

2x+3y+3xy'-3y^2y'=0

at x=4 and y=-4

8+(-12)+(12)y'-3(-4)^2y'=0

-4+(12-3*16)y'=0

[fallow slate]

can someone help me with something

[burnt sonnet]

MIght be easier but it is good to know how to solve for y'

the way you did

I can try

is it a big question?

[fallow slate]

no it should be quick

its about rational functions

[burnt sonnet]

😮

spoky

shoot

[fallow slate]

ok these are two different rational functions

and their behaviors are very different

and i cant seem to figure out why

[burnt sonnet]

yes

[fallow slate]

like i cant wrap my head around it

[burnt sonnet]

hmm

[fallow slate]

see look

[burnt sonnet]

do you see the difference of them?

😂

yes

[fallow slate]

why does the one with x in the numerator go up in the middle

and the one with 1 in the numerator goes down in the middle

[burnt sonnet]

becase

x is negative there

so can we call the 2nd function f(x)

are you familiar with this?

[fallow slate]

yes

[burnt sonnet]

okay do you see the the first equation is f(x)*x

[fallow slate]

yes

[burnt sonnet]

ok

So for negative x values

[fallow slate]

its flipped?

[burnt sonnet]

it has a different sign

No I wouldn't say that

[fallow slate]

just that its going in the opposite direction

[burnt sonnet]

yes

you can say that

that explains why it goes to plus infinity instead of minus infity for x->3+

Can we focus on around x=5 first?

[fallow slate]

yes

[burnt sonnet]

What is different?

between the functions

sry

I meant 4

around x=4

Do you think that they are similar there?

[fallow slate]

oh well i mean the blue graph is steeper right

and theres the vertical asymptote

[burnt sonnet]

yes

it is larger

[fallow slate]

yeah they're pretty similar

i mean in terms of end behavior

[burnt sonnet]

maybe about a factor 4 larger

[fallow slate]

yeah

ohh

[burnt sonnet]

😄

around x=4

f(x)*x is 4 times larger than f(x)

since the top part is around 4

^_^

do you see this?

[fallow slate]

you're totally right!

oh i seem to get it now

[burnt sonnet]

okay cool

So there is only 1 thing that changes around x=4

we have 2 other intressting changes

[fallow slate]

is it the other asymptote?

[burnt sonnet]

yes that is one of them

[fallow slate]

at x= -3

[burnt sonnet]

x around -3