precalculus

[simple helm]

It is

[gaunt mason]

Or something like that. We don't use these in europe

[simple helm]

Oh reslly

:(

So you don’t know what it is

[gaunt mason]

I mean I can google it.

[simple helm]

I believe it’s the opposite of secant

[drowsy spear]

@gaunt mason instead of using the maximums for examples of the starting and ending points where they repeat she would have refer to the midline

[gaunt mason]

@simple helm we dont use secant and cosecant 🙂

[drowsy spear]

Its would be 1 and she would use that

[gaunt mason]

We use tan, cotan, sin, cos 🙂

[simple helm]

Ah okay i gotcha

We use all of them

[gaunt mason]

you dont use that. I think even max and min are not guaranteed.

In sin and cos it is. But in general you need to visually conclude what is the period.

[simple helm]

Do you understand this one?

[gaunt mason]

Yes

Regarding your first image and this, the y-offset, that is the free number on the right is irrelevant fot the asymptotes

Because it shifts the graph vertically.

[simple helm]

Yeah

[gaunt mason]

So to find the asymptotes you need to observe the y=2csc(1/2theta - 90)

And y=tan(2theta -180)

Now, the constant in front of the trig. function is also irrelevant.

As it stretches the graph vertically

So the first equation y=csc(1/2theta-90) is what you need

[drowsy spear]

@gaunt mason would you help me with the other questions on that worksheet above?

[simple helm]

Yeah

[gaunt mason]

@drowsy spear What you don't know, type it.

@simple helm Now. tan is not defined at PI/2, 3PI/2, 5PI/2, ....

And also in negative, -PI/2, -3PI/2, ...

[simple helm]

Yeah

[gaunt mason]

Which can be written commonly as PI/2 + n*PI , where n is any integer number.

[drowsy spear]

Alright well I need the find a sin or cos function that models the graph function

[gaunt mason]

So you have that 2theta-180 must not be equal to PI/2+n*PI

@simple helm

You can put the equation 2theta-180 = PI/2 + n*PI the values of theta that satisfy this are the ones that are not allowed, thus are the asymptotes

2theta - PI = PI/2 + n*PI

2theta = PI/2 + n*PI + PI the last two terms can be collapsed into a single n*PI.

So the solution is:
theta=PI/4 + n*PI/2

In degrees theta = 45° + n*90°

(also a note, it is better to call this n as k, as n is mostly used for positive integers, but here it can be any integer)

SImilar procedure for the csc.

@drowsy spear Do you have any idea how to do it?

[drowsy spear]

No clue unfortunately

[gaunt mason]

It asks you to find a sin function to match the graph, you have no idea what could you examine first?

The task asked you to find the period, it asked you to find the midline, it asked you find the amplitudes, it asks you to find the matching sin function.

Does this turn on the lightbulb

[drowsy spear]

Ah ok so I would then try and use them in the equation right

[gaunt mason]

yes, gj, try to think a bit harder next time on the question and previous subquestions 🙂

the thinking and focus on the problem will make you improve in math

[drowsy spear]

Yes thats something I've been drifting away from... 😔

[gaunt mason]

I see that you know all the answers, you just need to focus 10% more than you currently do 😄

[drowsy spear]

🙂 thats what everyone says

Lol

[gaunt mason]

I know it is hard when it is mandatory 😐

I hate mandatory tasks 😐

but, math is fun in general 🙂

[drowsy spear]

100% agree

[viscid thistle]

^^

[strong ermine]

🌝

[neon mural]

oiler

[strong ermine]

yes thats me

[viscid thistle]

Hey guys so I need help in this problem

[strong ermine]

send it

[round quest]

how do i do this

[astral mantle]

i suggest you draw it

[grim meadow]

Topic: systems of equations

Idk how to find x or y

I've been going back to this problem for the past 3days

[viscid thistle]

4y^2-16y=4(y-2)^2-16

[grim meadow]

Is this it?

[viscid thistle]

?

omg that was just a hint lol

[grim meadow]

Oh lmao

Idk how to do it

I've done 1 problem

And it was cause y was easy to get

We're expected to learn it by ourselves

9x^2+18x=9(x+1)^2+9

4y^2-16y=4(y-2)^2-16

Idk where to go from here

[uncut mulch]

consider the first equation in your system

note the presence of the (y-2)^2

[grim meadow]

I tried to find y

[uncut mulch]

don't try and get y like that

consider the presence of $4y^2 - 16y$ in your second equation, \
the hint:
$$4y^2 - 16y = 4\red{(y-2)^2}-16 $$
and the first equation in your system:
$$\red{(y-2)^2} = 9(x-2)$$

[obsidian monolith]

ℝamonov

[uncut mulch]

and do some substitution to eliminate one of your variables, in this case y would be elimintated

leaving you with a quadratic equation in x,

which you should be able to solve,

and plug them back in to either equation in your system to get their respective y values

[simple helm]

Hey

@uncut mulch could you help me with a problem?

[sick steppe]

,rotate

[smoky pagoda]

do you just wanna know if you're right?

[simple helm]

I started it

I need a bit of help graphing it

[smoky pagoda]

first of all there are asymptotes

since the transformed function, tan(x), has infinitely many asymptotes

[simple helm]

Ah i gotcha

Aren’t they at oo

Pi*

[sick steppe]

Regular asymptotes of tangent occur at odd multiples of pi/2

iirc

[smoky pagoda]

^

[simple helm]

I gotcha

So how would I go upon graphing it

[leaden thorn]

can someone explain to me how this is wrong

[bronze star]

You want to check the last interval, the minimum isn’t at x=1

[leaden thorn]

its at a decimal place right?

[sick steppe]

yes

[leaden thorn]

oh my lord

skipping that problem

[sick steppe]

...

why?

[bronze star]

You can estimate the minimum

[sick steppe]

if you're asked to eyeball, it's obvious what the value is

[leaden thorn]

i tried eyeballing it but it said it was wrong

should be 0.5 right?

[bronze star]

Try using the segment that is shown on the graph?
Like (-4, -2) and (0.5, 2) ?

[leaden thorn]

i tried that already but it was wrong

its fine i can come back to it

[ripe adder]

If a question is asking me

A would be

(1/2(e)^(4x-4))(x)

right?

[rapid lance]

yes, i think so

[proper moss]

hi guys i need help with maths

2 questions of maths

[gaunt mason]

Post one

[proper moss]

o.o

[vast ravine]

Can someone help me, i've got 4(x-3)^2 / 25 + 16(y-2)^2 / 49 = 1, ellipeses -- i'm trying to find the vertices

i see the center is 3,2

normally if the x -3 didn't have coeffecients it'd be the squared denominator - the x and y values

but with the coeffecients i'm confused what to do

[proper moss]

anyone alive to help?

[vast ravine]

@proper moss http://www.hanlonmath.com/pdfFiles/29.SyntheticSubstitution.pdf

[proper moss]

what

[vast ravine]

just plug in -3 for x

and then 4 for x

normally you would follow the instructions on that pdf

but they are only asking for the answer

of this ellipse

[gaunt mason]

What is the formula for the foci

[vast ravine]

i'm trying to find the vertex

not the foci sorry

i see sqrt(25)= 5

so i'd take 4/5+3(x coordinate)

to get (1/2 , 2)

for one point

but i can't figure out the other

i'm thinking my logic is wrong

[gaunt mason]

What is the vertex of the ellipse?

[vast ravine]

derp

the center is (3, 2)

[gaunt mason]

ok

[radiant notch]

wtf is a least degree

[sick steppe]

context?

[vast ravine]

still confused if anyone can help with this

[sick steppe]

@bitter basin 2 is in quad4

-0.57 is an angle in quad4

you wrote an angle in quadrant 1...

@bitter basin

you answered ~33 deg, which is in quadrant 1

the angle in radians is -0.57 which is between -pi/2 and 0, so it's in quadrant 4

You got 2 wrong, 1 is fine

since 1.461 < pi/2, and 83.7 is acute

[bitter basin]

oh

really

[sick steppe]

yes, 1 and 3 are right

the fact you got 3 right makes me confused why you got 2 wrong

yes

you have -0.57

[bitter basin]

ye

[sick steppe]

how are you getting a positive answer out?

[bitter basin]

ohh

[sick steppe]

$-0.57 \times \frac{180}{\pi}$

[obsidian monolith]

moshill1

[sick steppe]

yes

ok

"round to 3 decimal places"

81.8 isnt 3 DP

81.81818181....

you just plug it into a calculator..

yes

900/11 ~ 81.818

what do you get if you do 1/3?

yeah

cause most calculators handle repeated decimals

so idk how you only get 81.81

the bar means repeated

you just said you know and dont know the repeater

[gaunt mason]

@topaz obsidian Don't crosspost on multiple channesl.

[topaz obsidian]

can i have some help

[gaunt mason]

Have you made a sketch of the problem?

[topaz obsidian]

no

[gaunt mason]

why?

[golden dome]

Can someone help me do rram on this

[gaunt mason]

whats rram

[robust nest]

struggling with my review lol, already forgot. Can someone give me hints on where to start with them

[robust nest]

any help?

[gaunt mason]

Do you know 17?

[robust nest]

not even 17 lol

i forgot a lot

[gaunt mason]

Any ideas on 17,

something that catches your eye

[robust nest]

that 3^2 is 9

thats about it

could try log

[gaunt mason]

good job!

[robust nest]

ok so since I now know that

hmm

[gaunt mason]

You should write 9 as (3^2)

[robust nest]

ah makes sense

let me do that

[gaunt mason]

There is a rule that if you have (a^b)^c that is actually a^(b*c)

[robust nest]

Ok so now I did that I have 3^(x+8) = 3^(4x^2)

[gaunt mason]

https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties

[robust nest]

since they have the same base should i set them equal to each other?

x+8 = 4x^2

[gaunt mason]

Wait

how did you got 8 and all this

stop for a second

[robust nest]

kk

[gaunt mason]

oh, its 8, I thought it is 2

on the left

[robust nest]

yeah its 8

so once i have that subtract x+8 and move it to the other side?

[gaunt mason]

This ^2 on the right came out of nowhere

[robust nest]

4x^2 - x- 8?

hm

[gaunt mason]

Its just 4x on the right side of equation

(3^2)^2x

[robust nest]

now im confused lol

[gaunt mason]

3^(2*2x)

[robust nest]

oh you multiply it

[gaunt mason]

yes

[robust nest]

hm

kk

[gaunt mason]

Check the wikipedia link

there you have all the identities

which have to be remembered, or have a table with them next to you

There are these nice tables with formulas to buy

But there are basically only 3 identities for exponents... so not that hard to remember

[robust nest]

Oh thanks

This is helpful

Okay how about question 18

[gaunt mason]

In 17 have you got the solution

Now the important part is to plug the solution back in the question, for verification!

You always have to verify your work if possible

Then you are sure you did it correctly

[robust nest]

Yes I got 8/3 and plugged it back in

it worked

thank you!

So how do I work out #18

[gaunt mason]

Hmmm

Again you need to check the identities for the logarithms

https://en.wikipedia.org/wiki/Logarithm#Logarithmic_identities

There are several ways to go, but you see that you have "x" on the left and right, in both cases in the logarithm

And checking the identities, do they give you any ideas?

[robust nest]

hmm im looking through it rn

im looking at the cahnge of base right now but im very confused

[gaunt mason]

Hehe, you chose the most difficult rule first.

Above that there are 4 other rules in the table

[robust nest]

OH WAit

i think i got it

one second

ok so i got to log base 2 (x+3/x-1) = 1

[gaunt mason]

good job

this is one way, and a good way. there are others, but this is good 🙂

Now you can solve this simply by "resolving the logarithm"

[robust nest]

Got x=5

[gaunt mason]

Great, how?

[robust nest]

so after solving that previous stuff I used my knowledge of logarithms and figured out that 2^1 = (x+3/x-1)

2^1 is just 2

so 2 = (x+3/x-1)

[gaunt mason]

Yes, great.

[robust nest]

and then I multiplied x-1

[gaunt mason]

The rest is trivial

[robust nest]

yup

[gaunt mason]

OK, you can use a trick also

[robust nest]

Alright

[gaunt mason]

log base 2 (x+3/x-1) = 1

The trick is to catch that log_2(2) = 1

So log_2(x+3/x-1) = log_2(2)

Always log_n(n) = 1

[robust nest]

ahh

u used a substitute makes sense

[gaunt mason]

So then you have left and right with the same base so the inner thing is the same

Can you apply that knowledge (the trick) to the original quesiton also, immediately?

[robust nest]

Let me try

hmmm

[gaunt mason]

It is not needed.

Let me show, but your solution is better.

[robust nest]

Alright go ahead

[gaunt mason]

So the question is:
log_2(x-1) + 1 = log_2(x+3)

[robust nest]

yup

[gaunt mason]

log_2(x-1) + log_2(2) = log_2(x+3)

[robust nest]

ohhh

[gaunt mason]

You use the rule from wiki, but for +, not for -

[robust nest]

so that would be log_2 (2x-2)

on the left?

[gaunt mason]

You get log_2[(x-1)*2] = log_2(x+3)

yes

[robust nest]

makes sense

[gaunt mason]

Again, bases on the left and right are the same

So you have 2x-2 = x+3

But your original solution requires less tricks so it is the best.

Now, the last thing, what I said before.

Always -> verification

[robust nest]

yup just verified it

it works

[gaunt mason]

cool, class dismissed

[robust nest]

thanks

[thorn moat]

im not understanding how the answer to the 3rd question is calculated. The 1st and 2nd i finaly get not but not the 3rd.

[uncut mulch]

if you can get the 2nd you should have no issue with the 3rd

[thorn moat]

i understand that the max profit is my y cord of my vertex and the number to maximize the profit my the x cord of my vertex. what am i missing?

[uncut mulch]

if you got the 2nd one, that should've been the x-coord of your vertex

[viscid thistle]

hello

i was wondering if i could get help understand linear combination

[sick steppe]

Linear combination means adding things that have scalar multiplication

$V = c_1v_1 + c_2v_2 + ... + c_nv_n$

[obsidian monolith]

moshill1

[viscid thistle]

yeah i looked into that

but tyhe way hes explained it here

just a sec i bring jamboard

@sick steppe

[sick steppe]

not sure what's confusing you then

Ccos(D) is one of the scalars, same with Csin(D)

[viscid thistle]

how does it work in the jamboard and like the standard definition

i dont know how it all works, the text on that whiteboard is like mayan scripture to me

[sick steppe]

Ok so the first line is the definition of a linear combination

Acos(x) + Bsin(x)

[viscid thistle]

wait wait wait wait

how does c1v1+c2v2 compare to acos(x)+bsin(x)

[sick steppe]

c are the scalars

v are the things

You learn linear combinations w/ vectors so I just wrote out the way I learned linear combinations lol

c1 = a
c2=b
v1=cos
v2=sin

[viscid thistle]

i see

okay

that part is a bit clear now

what next

[sick steppe]

Not entirely sure on the 2nd line, but my guess is showing that squaring a linear combination isnt a linear combination

But the g(x) statement is basically showing that: Any horizontal translation and vertical dilation on a sinusoidal function is equivalent to a linear combination of "basic" sinusoids

namely Ccos(x-D) can be expressed in the form: Acos(x) + Bsin(x)

[viscid thistle]

anyone here?

wait

huh

https://tenor.com/view/hold-up-jbtzxclsv-wayment-no-way-for-real-gif-16110940

[sick steppe]

Ccos(x-D) = C[cosxcosD+sinxsinD] = (CcosD)cosx + (CsinD)sinx

in this case A = CcosD and B = CsinD

[viscid thistle]

I see

like i see the similiarity

but whats the point

of this

i fail to understand its relevance/function

and how it works, I haven't seen a problem to date

and my asshat of a teacher took a topic out of the textbook so i cant even reference the textbook

[sick steppe]

Fourier is your man

[viscid thistle]

who

[sick steppe]

Fourier basically used this idea to represent any function (even discontinuous) as linear combinations of sines and cosines

Mainly this comes up when dealing with waves (duh)

For example, your voice is a sound wave, but dealing with the exact wave your voice makes is hell

it's much easier to deal with a broken down sound wave made of basic sinusoids

[viscid thistle]

oh splitting one function into components

i see

[sick steppe]

But it's also useful going backwards for solving trig (the process is called auxillary angle)

[viscid thistle]

huh

[sick steppe]

3sin(x) + 4cos(x) = 7 has no neat way to solving it

so you can combine the trig expressions into 1

[viscid thistle]

ah

how so

like can you walk me through it

[sick steppe]

$a\cos{x} + b\sin{x} = A\cos{(x-D)} \ A = \sqrt{a^2+b^2} D = \arctan{(b/a)}$

[lunar axle]

can someone plz help??

[viscid thistle]

wouldnt A = sqrt(9-16)

[obsidian monolith]

moshill1

[sick steppe]

meant to be a plus

[viscid thistle]

ok so its 5 cos(x-arctan(4/3)

[sick steppe]

yeah

the combined expression is ugly yeah, but it's much easier to solve

[viscid thistle]

idk what I did wrong

[hallow bison]

To find the difference in height of these 2 balloons I would use cosine of the 2 triangles angles and then solve for the heights and subtract them to get 266.6140266 as the difference of the height of the 2 balloons right?

[sick steppe]

yes @hallow bison

[hallow bison]

so 266.6140266 is final answer right?

[sick steppe]

cosine is negative in that domain

[viscid thistle]

oh
figures

[sick steppe]

sin2u = 2sinucosu

[viscid thistle]

ty

@sick steppe

is that it

is that all there is too it?

[sick steppe]

That I'm aware of yes

[lunar axle]

can someone plz check if these are correct for me

[barren trellis]

could someone help me with my understanding how to solve these quadratic equations?

I am in the mathematics call

[sick steppe]

just post the question...?

[barren trellis]

ok the question is t^4+2t^3-3t^2=0

[uncut mulch]

that isn't a quadratic equation

[barren trellis]

how do I make it quadratic

or rather how do I solve it

[sick steppe]

common factor then use quadratic techniques

[barren trellis]

ok I'll give it a go

so is it once the leading coefficient is 1 it is quadratic

[sick steppe]

no

do you know what a quadratic is?

[barren trellis]

yea sort of

[sick steppe]

define it

[barren trellis]

the equation

ax^2+bx+c=0

[sick steppe]

yes

so where did you get that the co-efficient has to be 1...?

[barren trellis]

i'm not sure I have always thought the leading coefficient had to be one to use the quadratic formula

[sick steppe]

no

[barren trellis]

at least all my problems that I have had so far have been that way

[sick steppe]

anyway, what's the common factor in the question?

[barren trellis]

it would be t^2 right?

[sick steppe]

yes

[barren trellis]

so would I rewrite the equation as t^2(t^2-3=0)

[sick steppe]

no, cause that's horrible notation

and you lost a term

[barren trellis]

so how would I rewrite.

[sick steppe]

properly..

[barren trellis]

yes please

[sick steppe]

you common factor t^2

$t^2(\text{quadratic})=0$

[obsidian monolith]

moshill1

[barren trellis]

um

would it be t^2 (t^2+2t^3-3t^2)

[sick steppe]

no

cause that's not a common factor

[barren trellis]

you said that t^2 is the common factory right

factor*

[sick steppe]

Yes

[barren trellis]

ok

so you would "Subtract" t^2 from everything

right

[sick steppe]

Subtract?

[barren trellis]

well take out

[sick steppe]

You divide

Factoring is dividing

[barren trellis]

Im sorry its not clicking yet

one last try

[sick steppe]

Common factor means factor out of everything

[barren trellis]

t^2(t^2+2-3)

[sick steppe]

2t^3

Factor t^2 out

[barren trellis]

it would be 2t^1

which would be

2t

[sick steppe]

Yes

[barren trellis]

so we are at

t^2(t^2+2t-3)

[sick steppe]

Yes

[barren trellis]

thank you

so now how would I further complete this

would I do zero product property

[sick steppe]

Yes

[barren trellis]

Ok thanks you so much, I i'm pretty sure I've got it now. That was really frustrating for some reason

[gloomy mortar]

So uh. I obviously did something wrong because I can’t take a square root of a negative soooooo...could someone help me out here? What did I do wrong

Also don’t understand these two questions

<@&286206848099549185>

[drowsy karma]

How do I determine the nature of a root without solving or graphing?

One of the questions is

$-2(x+3)^2+40=0

[willow bear]

are you talking about quadratic equations exclusively?

are they all in completed-square form?

[drowsy karma]

Its not in completed square form

some are

and yes quadratic equations

[willow bear]

well theres this thing called the discriminant

which you might find handy

especially for equations in expanded form

[drowsy karma]

Yeah we learned about that today

[willow bear]

ok

[drowsy karma]

I forgot what it was though

lmfao

[willow bear]

the discriminant of ax^2 + bx + c is given by D = b^2 - 4ac

[drowsy karma]

thats part of the quadratic formula right

[willow bear]

it's what's under the root in the QF

[drowsy karma]

So how do I use the discriminant

Like plug in abc?

I'm guessing I have to make the equation in standard form first

[willow bear]

you don't have to

if the equation is in completed-square form it's much easier to analyze

but if it is in expanded form

then you find the value of the discriminant

and its sign will tell you how many roots there will be

D > 0 means two roots, D = 0 means one (doubled) root, D < 0 means no (real) roots

[drowsy karma]

Its not in completed square

It's factored

How do I use the bot

Isn't it $

[willow bear]

it's FACTORED?

[drowsy karma]

fuck

um

[willow bear]

also, it's dollars at the start AND at the end for math mode

[drowsy karma]

idk what its called

$-2(x+3)^2+40=0$

[obsidian monolith]

Stewie

[drowsy karma]

that

[willow bear]

that's completed square

[drowsy karma]

okay

wouldn't i have to expand that

[willow bear]

no

anyway with an equation in completed-square form $a(x+h)^2 + k = 0$ you can rewrite that as $(x+h)^2 = -\frac{k}{a}$

[obsidian monolith]

Ann

[willow bear]

so the sign of -k/a is what interests you now

[drowsy karma]

so -40

[willow bear]

no, -k/a is not -40 in your case.

[drowsy karma]

then waht is it

wait

-20

?

[willow bear]

no, -k/a is not -20 either in your case.

[drowsy karma]

im so confused

so what would it be

[willow bear]

your k is 40 and your a is -2

(-40)/(-2) = ?

[drowsy karma]

20

right

[willow bear]

was that so hard

[drowsy karma]

yes

okay so its 20

bigger than 0

real and unequal

what about like

$x^2+10+3x=0$

[obsidian monolith]

Stewie

[willow bear]

this is in expanded form

[drowsy karma]

so make it a completed square?

or is there a different way

[willow bear]

you don't need to

you have the discriminant, which works on eqs in expanded form

[drowsy karma]

ah ok

so if its completed square i could use -k/a but if its expanded i could just use the discriminant

[willow bear]

i mean

yeah

[minor rapids]

Hey! Could anyone solve this for me please? Thx ;)

[willow bear]

no

this is not a "do things for you" server

[minor rapids]

Well maybe someone knows rhe answer

[neon mural]

well yes

but why should we tell you

this server is for help

not

[minor rapids]

To help me?

[neon mural]

for

someone to do it for you

ah yes

help and doing things for something aren't the same thing though

[minor rapids]

Well i dont know how to solve it

[neon mural]

why not?

[minor rapids]

Because idk

I just dont understand it

[neon mural]

alright sure

uhh

do you know how to simplify the left term?

[minor rapids]

Not rly tbh lol, i know only some

[neon mural]

alright what happens when you cube a cube root?

[minor rapids]

Im really bad at this kind of thing

I dont know those terms sorry

Im in high school

[neon mural]

mate

so am i

alright

do you know what a square root is?

[minor rapids]

No

Maybe in dutch but idk

[neon mural]

ah maybe that's why

$\sqrt{2}^2=?$

[obsidian monolith]

lofi hip hop radio

[neon mural]

can you solve this?

[minor rapids]

Oh root

Thats root 2x2

[neon mural]

can you give me a numerical answer

like

is it

1

2

3

4

?

[minor rapids]

Oh ye

[neon mural]

yep

so what is it?

[minor rapids]

Root 4?

[neon mural]

which is?

[minor rapids]

Idk, no round number i think

[neon mural]

alright

[minor rapids]

Wait

Thats 2

[neon mural]

yep nice

so

[minor rapids]

Bruh lol

[neon mural]

$\sqrt{2}^2=2$

[obsidian monolith]

lofi hip hop radio

[neon mural]

so similarly

[minor rapids]

The root and ^ thing solve eachother

[neon mural]

$\sqrt[3]{x}^3$

what would that be

ugh

wait

[obsidian monolith]

lofi hip hop radio

[neon mural]

can you solve this?

[minor rapids]

3x

[neon mural]

why 3x?

[minor rapids]

Idk lol

Wait

[neon mural]

didn't you say the root and the ^ cancel out?

[minor rapids]

3 root x^3

[neon mural]

which is....

[minor rapids]

Root x^9

[neon mural]

not quite

[minor rapids]

Which is 3

O

[neon mural]

do you remember this?

[minor rapids]

Ye

[neon mural]

so apply that to the question i gaev you

[minor rapids]

9?

3x3

[neon mural]

$\sqrt[3]{x}^3$

[obsidian monolith]

lofi hip hop radio

[neon mural]

if the cube root and the cube cancel out

what is left?

[minor rapids]

3x

[neon mural]

no

try again

like

the cube root

and the cube

[minor rapids]

X^3

[neon mural]

cancel each other out

so there's nothing left

except for the x

[minor rapids]

Just x?

[neon mural]

just x

remember how when i asked you

$\sqrt{2}^2=2$

this question?

[obsidian monolith]

lofi hip hop radio

[neon mural]

do you see how the root and the square cancel each other out?

[minor rapids]

Ye

Is that for any square

[neon mural]

so that's the same, no matter what the power is

[minor rapids]

Ah ok

[neon mural]

so for example i could write

$\sqrt[10000]{x}^{10000}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

and that would be?

[minor rapids]

X

[neon mural]

right

because the powers are equal

alright now

[minor rapids]

Ye

[neon mural]

moving on

$\left\sqrt[3]{\frac{x}{2(2x^2)}}\right^3$

[minor rapids]

Uh

[neon mural]

one sec

alright

alright

so using what you've learnt

what do you think this becomes

[obsidian monolith]

lofi hip hop radio
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[neon mural]

ugh

but ohwell

what do you think this becomes

[minor rapids]

Everything except the power and root?

[neon mural]

nice!

so that's the first part of your question

now

second part

[minor rapids]

Epic

Does the first 3 also stay

[neon mural]

no

[minor rapids]

On the left

Ok

[neon mural]

the root and the three cancel out

because it's a CUBE root

and it's CUBED

so they cancel out

$\frac{1}{x}^{-1}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

would you know what this becomes?

[minor rapids]

I have to go in a couple minutes brb, will be back in 2 hrs after thag

[neon mural]

ah well

good luck

i'm not available in 2 hrs lmao

[minor rapids]

I still have 2 min lol

Euf

[neon mural]

ure

sure

$\frac{1}{x}^{-1}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

do you know this

[minor rapids]

Idk

[neon mural]

alright so

when x is raised to a negative power, like this, $x^{-1}$, it becomes flipped

[minor rapids]

I guess its just x?

[obsidian monolith]

lofi hip hop radio

[neon mural]

so like this

$x^{-1}=\frac{1}{x}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

so

using the same logic, if i do this: $\frac{1}{x}^{-1}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

what does it become?

[minor rapids]

X^1

[neon mural]

yep

so now

if i

$\frac{4}{x}^{-1}$

do this

[obsidian monolith]

lofi hip hop radio

[neon mural]

what does it become

[minor rapids]

4x^1

[neon mural]

no

not quote

not quite

so what the ^-1 does

it makes the fraction flipped

so

do you agree that $\frac{x}{1}=x$?

[obsidian monolith]

lofi hip hop radio

[minor rapids]

Yes

[hard wasp]

hey guys so i got a question with binomials if someone can help, basically..

[neon mural]

@hard wasp make sure you read #❓how-to-get-help

$\frac{x}{1}^{-1}=x^{-1}$

[hard wasp]

oh

[obsidian monolith]

lofi hip hop radio

[neon mural]

would you agree?

[minor rapids]

Yes

[neon mural]

so

$\frac{1}{x}=x^{-1}$?

[obsidian monolith]

lofi hip hop radio

[minor rapids]

Uh

I dunno

[neon mural]

well just know it works like that, yeah?

[minor rapids]

Yes

[neon mural]

so basically what ^-1 does is that it flips the fraction

so

$\frac{4}{x}^{-1}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

is what

remember to flip the fraction!

[minor rapids]

Ok

So 4x?

[neon mural]

so what does it become

nope

[minor rapids]

X/4

[neon mural]

you havent flipped it

exactly

so now you have

[minor rapids]

I gtg man, thx for the help!

[neon mural]

ah well

it

it's quite trivial

so i trust you can do the resst

[viscid thistle]

$(\frac{a}{b})^{-n} = (\frac{b}{a})^n$

[obsidian monolith]

Shen

[gaunt mason]

Whats your attempt

[viscid thistle]

You're welcome @bitter basin

[unborn yoke]

does anyone know how to solve a quesiton like this?

i saw it on a test the other day as extra credit and never figured it ou

did some googling and found a similar question but the answer is so weird i couldnt hear a word the guy was saying

basically saying

cos(21) + cos(22) + cos(23)... + cos(290) =

[gaunt mason]

Oh

How it came that bad in the image.... @unborn yoke

Well, cos(n) + cos(180-n) = 0 for every n.

So the terms from 21 till 180-21=159 annihilate.

[unborn yoke]

omg

i love you

thank

i didnt think about that

[gaunt mason]

Also cos(180+n) + cos(360-n) = 0

[unborn yoke]

yep

[faint cipher]

Judt realized I can't do a 3x3 matrix determinant to save my life

[gaunt mason]

But that still leaves a lot of terms open 😐

[unborn yoke]

yeah

[gaunt mason]

It leaves the terms from cos 160 till cos (180+69) = cos 249 alive.

[unborn yoke]

cos 160 til cos 249?

[viscid thistle]

Angle β=7/2π lies in quadrant 3 in standard position. What is the measure of the principal angle?
is it 7pi/2?

[viscid thistle]

yes

[thick python]

Hi, completely lost. Could someone explain if this needs to be put in standard form and if this is a horizontal stretch by 3 and a horizontal compression by 4. thank you

[patent beacon]

@thick python

Note you will need to adjust your form a little bit, to get it to look like this

[gaunt mason]

@patent beacon seems like a mistake for a>1

its a vertical expansion

[patent beacon]

Oop, that's what I get for grabbing something off Google

[robust nest]

need some help with this question and another one

[patent beacon]

Make a right angled triangle that satisfies tanθ = 7/24

That is, opp = 7, adj = 24

What's cosθ of that same triangle? That is, what's adj/hyp?

[robust nest]

Hm real quick question

Where would I place theta

[patent beacon]

Either angle. You just want any right triangle that satisfies tanθ = 7/24

[robust nest]

kk one second

I got 24/25, however the key says -24/25

[patent beacon]

Ye ye. This is where the tan > 0, sin < 0 part comes in

Note that what we did with the triangles assumes we're in Q1

But we're not. What quadrant are we in?

[robust nest]

hmm

Q3?

[patent beacon]

You got it!

In Q3, cosθ is actually negative. Meaning, we want the negative of the answer we got in Q1

[robust nest]

Ahhhh

That makes sense

Thanks!

I have one more question

I had the equation for this a while ago. Now we are doing a review I cant find my notes anymore lol. Any idea of what it is or where I can find the equation

[patent beacon]

You've got the circle equation
L = rθ

Take the derivative in terms of t:
v = rω

@robust nest

So 1.8 rads/s

[robust nest]

Alright thanks!

[quick aspen]

ANyone canm help me on precalc test tommorow?

<@&286206848099549185>

[fleet yew]

No.

[north lagoon]

Like, if you ask your doubts now and not when you're taking your test, there would be some help coming.

[viscid thistle]

I HOPE that's what they meant

[neon mural]

asking for help on a test is a bannable offence

[astral mantle]

im suprised how many people dont think cheating on a test is wrong

[uncut mulch]

they probably do, they just don't care

[blissful ridge]

I could go on rant on how "our education system emphasizes on scoring good grades rather than actual learning" but I won't

[lament fiber]

how do I even check if the coordinates are rational?(please ping me when you answer, if it isn’t much trouble)

this is the only thing I can think of as something that can concretely be said about this
where do I go from here?

can I say that if we consider x and y as being rational then the left hand side is rational while the right hand side is irrational(product of rational and irrational numbers) which isn’t possible(or am I wrong here, I think I am)

but that can’t be said because r can be irrational?

[lament fiber]

please help <@&286206848099549185>

#precalculus message

[lament fiber]

anyone?

[lament fiber]

pls help

[viscid thistle]

@lament fiberl

I can try to help

[lament fiber]

thank you

[viscid thistle]

so what it wants you to do

is form a circle around the point given

[lament fiber]

yes

I was able to get that

but how do I find how many points on the circumference have rational coordinates

all the answers I could find online assume that r is rational

[quick aspen]

stay mad

[viscid thistle]

I think you should make the radius 1 and use the unit circle

[lament fiber]

I could, but what if r is irrational

either way

I got rational=irrational in the end

which isn't right?

[viscid thistle]

https://www.toppr.com/ask/question/let-n-be-the-number-of-points-having-rational-coordinates-equidistant-from-the-point-0sqrt/ @lament fiber sorry I couldn't be much help, but I did find an explanation online

Use parametric form

[lament fiber]

I found that explanation

it assumes r is rational
what if it's irrational
can that even be a thing

[sinful maple]

Hello, is this the appropriate channel to ask about function tangent curves or tangent things in particular.

[viscid thistle]

yes

[sinful maple]

OK, so I have f(x)=x^2 -4x +3. I need to find the point of the slope of f in which the tangent is parallel to the line e with the equation y=2x.

Basically I'm lost and no matter the things I tried to learn about tangent line stuff just don't make quite a lot of sense to me.

[viscid thistle]

start off with making a line parallel to y=2x. Have you done that yet?

[sinful maple]

Uhhh, I never solved using a graph. Whatever we do, it's all theory and stuff if that makes sense.

[patent beacon]

We say two lines are parallel if they have the same slope. What's the slope of y = 2x? What slope must your line be?

It never hurts to have a graph of both curves as well

[sinful maple]

Sorry, I have no idea. I'm directly translating the exercise. It says "You are given a function f(x)=x^2 -4x +3, x belongs to R"
i) Find the point of the slope of f in which the tangent is parallel to the line (e) with the equation y=2x.

[patent beacon]

So a line comes in the form:
y = mx + b
Where m = slope
And b = y-intercept

The line they gave you is
y = 2x + 0
Can you identify m for this line?

[sinful maple]

I don't know. I think you have to find the derivative of f(x) first and then the derivative of y so the solution would be f'(x)=2.

I truly am not sure.

Sorry.

[echo wagon]

y = 2x + 0 and y = mx + c. What number takes the spot of the m in the first equation?

Because m is slope

You really shouldn't be touching calculus if you don't know the slope of a straight line, because that's kind of what the derivative is all about

[sinful maple]

OK then, thank you for your time.

[patent beacon]

Mind you, you did get it correct. The slope of the line is indeed 2

[echo wagon]

I did not mean this in a discouraging way, but in a "you should recap straight lines before continuing to learn calculus" way

[patent beacon]

Just, using calc on that is like using a sledgehammer to break a cracker

Either way, you now just need to find a point on your curve that has a slope of 2

[echo wagon]

@sinful maple Kaynex is still helping, if you want it. Don't leave because of what I said

[sinful maple]

I know, I'm just thinking.

[viscid thistle]

I highly suggest graphing each equation

[sinful maple]

I do not know how to make a graph.

I was never taught.

I mean not like x'x and y'y with O in the cross section.

I mean making the line.

[patent beacon]

,w graph y = 2x and y = xx - 4x + 3

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[sinful maple]

It's actualy x^2 or it doesn't make a difference?

[patent beacon]

I put xx because I am lazy, haha

That's the same as x²

[sinful maple]

Ok, so I found the point and it's 0.

[sinful maple]

All right.

[patent beacon]

So the line is easy enough, you have
y = 2x + 0

Which means you have a slope of 2, and a y-intercept of 0. That is, (0,0) and (1,2) are on the line and you just extend it forever in both directions

[sinful maple]

OK, thank you.

[patent beacon]

The point on the quadratic that is parallel to that line is not x = 0.

[lament fiber]

@patent beacon if you don’t mind, can you please help me

#precalculus message

with this

[patent beacon]

@lament fiber
I think that the answer you were presented with is the intended answer, though you have noticed the proof isn't correct

The "right" proof is too complicated

[lament fiber]

I got rational=irrational if r is assumed to be rational based on how I rearranged it

so did I do something wrong?

because that answer was indeed the intended answer

[patent beacon]

The truth is that r² is not always rational, and that (√3)y is not always irrational

The reasoning is too messy to make sense of here

[lament fiber]

oof

thanks for the help

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:irrational-numbers/x2f8bb11595b61c86:sums-and-products-of-rational-and-irrational-numbers/v/proof-that-rational-times-irrational-is-irrational

I went by this and concluded that the y* sqrt(3) part is always irrational

[lunar axle]

someone plz help

[lament fiber]

what have you tried out?

||solve the quadratic first by replacing the trig function with some variable t||

[viscid thistle]

can someone show me how to split this into two fractions?

$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

[obsidian monolith]

HoboSas

[viscid thistle]

yeah but like I want to bring out the exponent n

[pastel cloud]

log?

[viscid thistle]

$\sum_{n=0}^{N} \frac{1+2^n}{3^{n-1}} = \sum_{n=0}^{N}\frac{1}{3^{n-1}} +2\left( \frac23 \right) ^{n-1}$

[obsidian monolith]

HoboSas

[viscid thistle]

hm thank you I think I get this now

[lone robin]

can you guys help me?

[scarlet oak]

Check out the unit circle

I don't have one on hand, but Google it and you should find a labeled one

[lone robin]

this??

[scarlet oak]

Yep

[lone robin]

how can I apply it? i really dont know what to do. I just need guide

[scarlet oak]

How much do you know of radians? Like πradians is 180 degrees

2πradians is 360

1/2 πradians is 90 degrees

Then the ordered pairs like (1,0) states how far it is from the center

It's a unit circle, so from the center of the circle to the edge is one unit

https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/v/unit-circle-definition-of-trig-functions-1&ved=2ahUKEwjpnMe4jMTtAhUBy1kKHZKODJwQ28sGMAB6BAgBEAc&usg=AOvVaw1JaYnKbY6YnrDsbHVCahGB

This is a vid from khan, it may explain it better than I can though text

[lone robin]

OMG THANK YOU VERY MUCH

ill try to understand it

[scarlet oak]

Welcome :) lmk if you need more clarification later on and I'll help best I can

[lone robin]

OK THANK YOU VERY MUCHi

[lone robin]

@scarlet oak is radian = arc length?

[scarlet oak]

Iirc the arc length is just the length of that specific part of the circle

So I believe it depends on the angle, I'm refreshing rq

[copper vigil]

what is the fundamental theorem of precalculus?

[fresh marsh]

Never. Du. Dwug.

[bold knoll]

really embarrassing but
anyone have any resources on why absolute value equations are solved the way they are? i know to solve them but dont get why it works

oh and i mean the eqs with variables both in abs and outside, and eqs with multiple separate expressions in the abs (like abs(x) + abs(x+1) = 2-x^2)

[lapis sphinx]

why is it embarassing

if anything its admirable

[bold knoll]

I was doing multivar calc the other day and now alg got me lol

[fleet yew]

@bold knoll over the reals?

[bold knoll]

yeah

[fleet yew]

Are you familiar with the sign function

[bold knoll]

dont think so

[fleet yew]

Look it up

[bold knoll]

you have another name for it?

[fleet yew]

The sign function

[bold knoll]

oh ok. yeah it looks ok

piecewise function with -1 1

what about it

[fleet yew]

Its the sign of the number

[bold knoll]

what does it have to do with absolute value?

[fleet yew]

$\frac{x}{sgn(x)} = |x|$

[obsidian monolith]

AMD

[bold knoll]

arent i just replacing my problem with abs with a sgn?

i would have to restrict sgn(x) to some interval?

[fleet yew]

sgnx has only two values

[bold knoll]

oh, just use both pos and neg sgn(x)

[fleet yew]

+-1

abs(x) = a

x/sgn(x) = a

x = a*sgnx = +-a

[strong ermine]

is this a quiz?

[bold knoll]

abs(x) + abs(x + 1) = 2 - x^2
so for something like the above, im forced to check all four cases of neg and pos?

[strong ermine]

$s=r \theta$

[obsidian monolith]

Euler's Identity

[strong ermine]

so yeah youre good

[fleet yew]

@bold knoll there is only one variable. So there are three cases. x=+, x=-, x=0

[strong ermine]

but simplify

[sick steppe]

@bitter basin are you going to ask if every question is right?

[bold knoll]

btw im working on it @fleet yew thank you

[dull peak]

hey can someone help

i got the root x-1 but thats about it

[sick steppe]

@dull peak does Vieta ring a bell?

[brave blaze]

A trip to another city is 180
miles. In miles per hour to the nearest tenth, how fast does one have to drive to average 60
mph on a three-hour trip during which one stops for 10
minutes?

[dull peak]

?

[brave blaze]

can someone help me with dat

ello anyone?

[sick steppe]

yes or no

[dull peak]

no

wait is the answer not 1