🤔
#precalculus
viscid thistle
aw what
viscid thistle
can someone help me with that question
dense terrace
define invariant points
viscid thistle
points that stay the same on both graphs
dense terrace
oh
viscid thistle
so any point on the line y=x is an invariant?
dense terrace
theres only one point
dense terrace
draw the line y = x out and it only interseccs one point
calm whale
is it not 2 🤔
dense terrace
i mean two points sorry
viscid thistle
oh i see iy
calm whale
🤦
dense terrace
sorry im baka
viscid thistle
yes thank you
calm whale
don't worry about it 😃
dense zealot
@dense terrace yap ur a big baka but it's okay ~ @calm whale is too <3
dense zealot
Lol
dense terrace
aha
dense terrace
x = cos 145°
viscid thistle
oh
dense terrace
y = sin 145°
calm whale
what the fuck is a terminal arm
viscid thistle
haha ye
calm whale
what
dense terrace
what
calm whale
what
calm whale
WHAT
dense terrace
@calm whale lets go out
calm whale
. . .
dense terrace
went out
dense zealot
by judgement of bleh
I place a decree on you of
Death
(did I use that word right)
calm whale
theres nothing racist about sohcahtoa 🤔
calm whale
stop memeing
calm whale
gnight
severe verge
good night!
dense zealot
Lol tru
viscid thistle
can someone help again
I understand the b and c value
I just dont know how to get the A and D
thought the A = (max value - min value) / 2
elder smelt
how is this wrong?
like for one of the problems they did x/6 instead of pi/6
wanna try that but I only got 1 try left
k
fuck me
it's in terms of t
not x
LOL
illiterate
ripe fog
I'm trying to find the value of f(x) for function 3x^2+2x-4 where x = x+1 ... I plug x-1 into x, but never arrive at the answer in the book which is 3x^2+8x+1 ... I don't see how they got the 8x. Could someone help me find where my algebra falls short? My solution was 3x^2+2x+1
ripe fog
yes, sorry @clever inlet
ripe fog
3(x+1)^2+2(x+1)-4 ... I'm guessing I failed to factor the first term here?
I can never remember how you handle something like 3(x+n)^2 ... do you first distribute, then apply the exponent?
ripe fog
ah, let me try that
clever inlet
first one makes more sense i think
ripe fog
They both make sense, you just combine like terms all in one step in the first one. That worked though, handling that first term screwed me, thanks a lot 😃
clever inlet
np
viscid thistle
hello
can someone help me with this
I know how to get the a value
dont know how to get the B
i know period = 2pi/b
so b = 2pi/period
but whats the period
ripe fog
@viscid thistle It looks like period represents the number of months. (between 5 and 12).
nimble oxide
So, what I've learned so far when finding zeroes, multiplicity, you usually just solve each part by setting f(x)=0.. but for this problem wouldn't each zero be (0,0)? ... due to the fact if you set each one to 0 it would be 0/[the number]?
true vigil
you have a zero at x = 0 with multiplicity 1
you need to factorize it to get the other zeroes
nimble oxide
ohhh yeah you're right.. duh!
😂
thanks
factor out the 50 yah-duh-yahduh .. yeah i gotchaa
dawn loom
I'm having difficulty with understanding phythagorean illerations
Not sure how to pronounce/spell it, but it's in Pre-Calc
dense zealot
:o
dawn loom
@dense zealot its pythagorean identites
willow bear
sec(x)tan(x) - sin(x)
not trig
nimble oxide
clever inlet
^
dense terrace
=tex x + y = 200 \
y = 200 - x \
xy = 200x - x^2 \
\frac{d}{dx}(200x - x^2) = 200 - 2x \
x = 100
granite stirrupBOT
dense terrace
the value of x will give a stationary point in the curve 200x - x^2 which gives the maximum area of the whatever fence
and clearly
== 200(100) - 100^2
granite stirrupBOT
Error: 200 is not a function
dense terrace
== 200*100 - 100^2
granite stirrupBOT
10000
nimble oxide
Exactly! Ok, at least I know I'm not dumb!
I've literally been trying to figure out the answer forever!
nimble oxide
Yeah, I chose 1000.. but that's a huge typo
I was literally stuck on that problem for 30 minutes... I got a 100 on the quiz.. and I slammed down my laptop.. Im so mad 😐 i did tha problem over and over again.. for a typo qq
dense terrace
^
slate hill
its a non real number
mental maple
they're implicitly saying that $$f : \mathbb{R} \to \mathbb{R}$$
granite stirrupBOT
slate hill
idk what that means
mental maple
if $$16-15x < 0$$, then the output is not a real number
granite stirrupBOT
mental maple
$$\mathbb{R}$$ means real numbers
granite stirrupBOT
mental maple
So the domain would be $$16-15x \geq 0$$
slate hill
granite stirrupBOT
mental maple
What does that mean for x?
slate hill
i dont know
mental maple
ahve you learned how to solve inequalities?
slate hill
oh didnt say you were
need to get this done first
no we have not done inequalities yes
mental maple
=tex 16 - 15x \geq 0 \implies 16 \geq 15x
granite stirrupBOT
mental maple
this is adding 15x to both sides
then, dividing by 15
=tex 16/15 \geq x \implies x \leq 16/15
granite stirrupBOT
mental maple
it's just like solving an equation
slate hill
why is this on a "piecewise function homework"
granite stirrupBOT
mental maple
this is not a piecewise function
it is a thing about domains, just like the last two
mental maple
no problem m8
dense terrace
i didnt even help lol
prisma epoch
Let S be a sphere of radius r covered with bumps. All of these bumps are identical and are the projection of
bump(x) = \frac{r sin(δx)}{5}dx
Knowing that
δ*\frac{r}{4} = 90x
All bump’s heights meet at the center of the sphere.
What is the general formula for the volume?
<@&286206848099549185>
silver ibex
$$Let S be a sphere of radius r covered with bumps. All of these bumps are identical and are the projection of
bump(x) = \frac{r sin(δx)}{5}dx
Knowing that
δ*\frac{r}{4} = 90x
All bump’s heights meet at the center of the sphere.
What is the general formula for the volume?$$
granite stirrupBOT
Rendering failed. Check your code. You can edit your existing message if needed.
silver ibex
Damnit
prisma epoch
Lol
thick raptor
Facepalm
prisma epoch
Let S be a sphere of radius r covered with bumps. All of these bumps are identical and are the projection of
$$bump(x) = \frac{r sin(δx)}{5}dx$$
Knowing that
$$δ*\frac{r}{4} = 90x$$
All bump’s heights meet at the center of the sphere.
What is the general formula for the volume?
calm thicket
1-800-you-thought
calm thicket
;p
granite stirrupBOT
Rendering failed. Check your code. You can edit your existing message if needed.
calm thicket
:P
prisma epoch
Whatever you guys get it...
Right?
calm thicket
No, if you don't get it to work right we'll never be able to understand you.
dun dun dun
That being said, this math is well above me, best of luck
granite stirrupBOT
prisma epoch
THANKS MAN
prisma epoch
No one?
worthy oar
I'm not a helper but what's up
prisma epoch
Whathus
prisma epoch
Let S be a sphere of radius r covered with bumps. All of these bumps are identical and are the projection of \ $$\operatorname{bump}(x)=\frac{r\sin(\delta x)}5{\rm d}x$$ \ Knowing that \ $$\delta \frac r4=90x$$ \ All bump's heights meet at the center of the sphere. \ What is the general formula for the volume?
granite stirrupBOT
worthy oar
Hoo boy
prisma epoch
What?
worthy oar
i have no clue how to do this
worthy oar
What grade are you in?
prisma epoch
Im 14 years old
worthy oar
damn
prisma epoch
Yeah
worthy oar
I dont think i remember enough of precalc to do this
lemme see if i can find something to help you
worthy oar
Sure
prisma epoch
So i find the volume of a bump
By finding “y=(bump(x))”’s area through derivation and integration.
Then
I need to find how many bumps there is
I find the length of one bump, and from that i find the area of the curve of the base of a bump. I divide the area of the regular sphere by that, and i round down.
That’s the number of bumps
By finding I apply disk method on the area of bump. I take that, and i substract the volume of a paraboloid which is the inclination of the base of a bump. That’s the volume of a bump
I apply disk method on the area of bump. I take that, and i substract the volume of a paraboloid which is the inclination of the base of a bump. That’s the volume of a bump
severe verge
calc is spooky
worthy oar
this is precal apparently
fringe stream
what's the question?
worthy oar
severe verge
precal doesn't cover integration i thought
grim yarrow
bump functions????
fringe stream
it doesn't lol.
prisma epoch
Really?
worthy oar
yeah its a calculus thing apparently
severe verge
different bump XD
prisma epoch
Im so confused. Because we don’t have precalc or calc in canada
fringe stream
yes... we do.
severe verge
at my school we had "advanced functions" and "calculus and vectors"
prisma epoch
But there’s not the same division
fringe stream
what's a projection?
prisma epoch
You take two 2d images and make a 3d one with them i believe
vital tartan
yeah exactly it's called advanced functions and calculus and vectors ^^^^
there's definitely calc. in canada, lol
clever inlet
Well, it doesn't really make sense to have a negative amount of units produced in this case
In terms of asymptotes
Try x =
slate hill
why are the asympototes not correct
clever inlet
And y =
slate hill
ah
clever inlet
Actually y=, then x=
clever inlet
Yes
clever inlet
Well it must be positive right?
slate hill
oh so 0
clever inlet
Yes
Cause it doesn't really make sense here to have negative units produced
In this context
clever inlet
Np
unique bolt
Hey I was recommended to take this course as precalculus https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/syllabus/
However it seems to be a calculus 1 course and I had already taken a different one though without homework
So am I supposed to retake a course until I become competent or what?
I didn't feel like the basic concepts were too difficult, it's just in the exercises of some things that it fails to deliver
Because I have to link the specific question of the exercise to the larger body of concept
true vigil
it's calculus not precalculus
do the homework
so you see where you have gaps
and fill them
ripe fog
Questions channel is getting flooded with questions so I'll ask here.
Complete the square of each quadratic expression then graph each function by shifting. For the expression f(x) = x^2+x+1. From that I get x^2+x=-1 --> (x+1/2)^2 = -1/2 --> (x + 1/2)^2 + 1/2 = 0. The solution says the answer is (x + 1/2) ^2 + 3/4. Where did I mess up?
grizzled hull
Good question! Let's break it down step by step. (If the LaTeX bot lets me....)
No?
I can do it myself, then.
The kicker is in $$\frac{1}{4}-\frac{1}{4}$$.
granite stirrupBOT
Rendering failed. Check your code. You can edit your existing message if needed.
grizzled hull
In our case, that makes $$\frac{1}{4}$$ as the element to add and subtract.
Does this make sense, @ripe fog ?
granite stirrupBOT
ripe fog
looking now
grizzled hull
It seems the LaTeX bot and I get into fights. I'm glad I can post my own images.
granite stirrupBOT
granite stirrupBOT
ripe fog
how do you split lines?
grizzled hull
\\
granite stirrupBOT
ripe fog
$$x^2+x +1$$\
$$x^2=-1$$\
$$x^2+x+1/4 = -1*1/4$$
granite stirrupBOT
granite stirrupBOT
grizzled hull
You'll have to write \\\\ for Discord to understand.
ripe fog
$$x^2+x +1$$\
$$x^2=-1$$\
$$x^2+x+\frac{1}{4} = -1\frac{1}{4}$$
granite stirrupBOT
ripe fog
second expression should say x^2+x ffs haha
ripe fog
so I halve the coefficient of 1x and get 1/2, then square it to get the third term (1/2)^2 = +1/4
ohhhhh I see
I need to add the 1/4 to the other side, not multiply it hahah darn.
granite stirrupBOT
ripe fog
Oops, damnit
$$x^2+x +1$$ \
$$x^2+x=-1$$ \
$$x^2+x+\frac{1}{4} = -1+\frac{1}{4}$$ \
$$(x+\frac{1}{2})^2 + \frac{3}{4} = 0$$
granite stirrupBOT
grizzled hull
I'm going to be terribly persnickety for a moment. The second line does not follow from the first at all.
ripe fog
What do you mean?
grizzled hull
If you want that to be true, you need $$x^2+x+1=0$$ as the first line.
granite stirrupBOT
grizzled hull
Otherwise you're telling me that a general function which has no input has a defined output unrelated to the variables.
ripe fog
yeah yeah... persnickety indeed XD
ripe fog
so then I just take the $$x^2$$ graph, shift it to the left 1/2 and up 3/4
ripe fog
@pine kindle It wasn't asking to solve for x, but yeah that is the solution for sure
haha, never hurts to see it a few different ways xD
My mind was blown when I finally realized exactly what a quadratic equation is, and solving \ $$ax^2+bx+c=0$$ \ by completing the square. I always wondered why they have you repeatedely do these arcane quadratic equations in algebra. "Quadratic" sounds so intimidating to the uninitiated :P.
granite stirrupBOT
ripe fog
I mostly get stumped on assumed prerequisite knowledge of a certain notation when reading math texts. I'm self-learning so it took me a while to inventory where my knowledge gaps lay.
true vigil
it's about finding the right books
if you don't have the prereqs something's wrong with your choices
ripe fog
Yeah, I'm just going through my precalculus book since it touches on trig functions, set notation, logarithms, and the unit circle. I've actually found some older books from like pre-1960 were really good. Lyman Kells books are pretty concise and well written.
true vigil
really? old books tend to be unbearable for me
but for precalculus any book should do
as long as the exposition is clear
ripe fog
@viscid thistle depends on the book... some are terrible. Jacobson Basic Algebra? Forget it 😛
The colorful textbooks of today trigger my adhd... back then it was just black text on an off-white page. Nothing fights for your attention except the content.
ripe fog
@true vigil Yeah I realized that after I purchased it. But I haven't seen half the notation used in that book yet.
true vigil
you purchased it before knowing what it was?
ripe fog
@true vigil I did some research on "good algebra books" and "basic algebra" sounded just right. I didn't even know abstract algebra was a thing at the time.
true vigil
I see
in math books, people don't explicitly say "let's define this to be that"
they just go "(notation) is (that)"
ripe fog
this book doesn't even do that, at least from the first few pages of reading
true vigil
but it's to be expected that it's all new notation for you, it's higher mathematics
it does
it defines some set operations
also defines the combinations symbol
ripe fog
actually it doesn't look as bad since I've done a lot of learning since. The union and belongs to symbols are at least recognizable
true vigil
it should be okay, if you're interested in higher mathematics is looks like an ok place to start
but right now, when you're learning precalculus, it's probably not the time
ripe fog
haha yeah... it's something I'll come back to when the time is right
ripe fog
Running into some more trouble with one of these:
$$2x^2 - 12x + 19 = 0 $$ \
$$2x^2 - 12x = - 19 $$ \
$$x^2 - 6x = - \frac{19}{2} $$ \
$$x^2 - 6x + 9 = \frac{18}{2}$$\
$$(x - 3)^2 = \frac{18}{2}$$\
The books solution was $$2(x - 3)^2 + 1 = 0$$ \
But I have no idea how they arrived at that.
granite stirrupBOT
indigo topaz
-19/2 + 9 =/= 18/2
patent beacon
@ripe fog
When you add 9 to both sides, where does -19/2 go?
ripe fog
@patent beacon $$ - \frac{19}{2} + \frac{18}{2} = -1/2 I see... I subtracted the 9. Ugh.
Subtracted the 9 in my first answer, and forgot to add it on the second attempt haha.
dense zealot
Shouldn't this go in #prealg-and-algebra
ripe fog
$$
-3x^2 - 12x - 17 = 0 \
-3x^2 - 12x = 17 \
\frac{-3x^2}{-3} - \frac{12x}{-3} = - \frac{17}{3} \
x^2 + 4x + 4 = - \frac{5}{3} \
(x + 2)^2 = -\frac{5}{3} \
(x+2)^2 + \frac{5}{3} = 0 \
3(x+2)^2 + 5 = 0
$$
The book's solution is $$ -3(x+2)^2 - 5 $$ . I've done the problem out 3 times on paper and I keep getting the same thing 😛
granite stirrupBOT
Rendering failed. Check your code. You can edit your existing message if needed.
ripe fog
ohhhh haha
Oh I see what I did... should have got rid of the fraction on the right by multiplying both sides by -3 first
then subtract the 5
haha 😄
@pine kindle The graph for theirs points downward, I just did each of them on the graphing calculator and my y values are opposite theirs
started with f(x) = x^2
The questions want you to complete the square of the quadratic, then graph using transformations
yeah, it's f(x) = -3x^2 - 12x - 17
please tell me the book is wrong
yah or a(x+stuff)^2+other stuff form
hmm ok
but the general form of the quadratic is ax^2 + bx + c = 0 , does setting it to y change anything?
oh i think I see
let me give it a try
patent beacon
A quadratic equation is of the form
y = ax² + bx + c
"Solving a quadratic", or finding the roots of the quadratic is the solution for
ax² + bx + c = 0
ripe fog
I got it... this explains why I've been having trouble with these all day haha. Ok so this is what I got:
$$ - \frac{3x^2}{-3} - \frac{12x}{-3} = \frac{y}{-3} + \frac{17}{-3} $$ \
$$ (x+2)^2 = \frac{y + 5}{-3} $$ \
$$ -3(x+2)^2 - 5 = y $$
woot!
thanks for clearing that up fellas 😃
granite stirrupBOT
indigo python
Currently working on find a sector of a circle, I have a basic formula set up; however I'm having a issues on finding R
What would be the proper procedure to find it?
Would it be okay to just get x by it self to find out the variable?
ripe fog
$$ \frac{x^2}{6} = 2 $$ \
$$ x^2 = 12 $$ \
$$ x = 2\sqrt(3) $$ \
I think that's right
granite stirrupBOT
indigo python
It is, thank you
So simply using algebra to get that missing variable is okay?
ripe fog
yeah, because you already have the area... now you're solving for another unknown
prisma epoch
Yeah
indigo python
Thanks guys
still yew
Why $$\cos(60)=\frac{\sqrt{3}}{2}$$
granite stirrupBOT
clever inlet
Do you mean sin(60°)?
hasty wadi
alright
i need help boyos
im on last attempt
im tryna find cosine and secent
yes
severe verge
sin²+cos²=1
hasty wadi
i tried that
severe verge
(-6/7)²+cos²=1
hasty wadi
but im still fucking up for some reason
severe verge
36/49+cos²=1
rain totem
draw a triangle @hasty wadi
hasty wadi
i did
severe verge
cos²=(49-36)/49
severe verge
i guess it should be negative
fringe stream
tangent is positive and cosecant is negative, so it should be in the third quadrant.
dense terrace
uwu draw a triangle
severe verge
ya
hasty wadi
if this is wrong, tide pod challenge is gonna be uploaded to youtube tonight
lets see.........
thank the fucking lord
were right boyos
fringe stream
😄
hasty wadi
thank you 😃
fringe stream
😄
granite stirrupBOT
rocky plinth
I got sin14x/cos10x, but I feel like it can be simplified in a better way.
hasty wadi
I USED PYTHAGOREAN IDENTITIES, I CANT FIND COSINE
THIS QUESTION IS BULLSHIT
oh yes
thank you
lmao
uhhh let's see
it's still wrong
this question is straight up garbagio
yes
isn't that what i put?
hasty wadi
im confused when you look at it from the graphing perspective?
fringe stream
cot(x) > 0 and sin(x) < 0, therefore, IIIrd quadrant.
fringe stream
yes, in the third quadrant, only tangent is > 0.
i find the quadrant approach more simpler. 😶
sotto isn't wrong either though.
hasty wadi
im so lost
like, i understand how to find them algebraically but idk why it's negative
fringe stream
okay, do you know CAST?
hasty wadi
all students take calculus?
fringe stream
what...?
hasty wadi
yes
fringe stream
now.
look at your given values.
it's given that cot(x) > 0 and sin(x) < 0.
you know that cot(x) > 0 in 1st and 3rd quadrants.
hasty wadi
yes
hasty wadi
ahhhh
fringe stream
now, all values except tangent/cotangent will be negative in the third quadrant.
pseudo bronze
man that 10 minute wait time was killing me. I saw the question and I was dying to type lol
hasty wadi
so therefore all of the trig functions (besides tangent) have to be negative because it's in the third quadrant
i get it
pseudo bronze
wait what is "all students take calculus" ???
fringe stream
that's the order.
pseudo bronze
ahh
i just figure it out logically based off the info given
with trig identities
fringe stream
all functions in I
sin/csc in II
tan/cot in III
cos/sec in IV
fringe stream
i can try.
hasty wadi
it's right 😄
pseudo bronze
it should be simple. thanks!
hasty wadi
thank you erryone 😃
hasty wadi
alright idk why this is wrong
oh im fucking stupid
i have to use theta
shiettttttttt
webassign stop fucking with me
earnest nymph
And the question is?
clever inlet
you would still have 1/sec(θ) though
hoary yoke
no
limpid plover
I tried using pythahgoras theroem to get the hypotenuse
hoary yoke
trig is trig
limpid plover
but it didn't work
hoary yoke
wait what?
limpid plover
I also tried the a^3 + b^3 identity which didn't work either
limpid plover
You there?
granite stirrupBOT
limpid plover
You might be thinking too hard, This is grade 10
limpid plover
Hmmmmmm
past veldt
What's the question?
hoary yoke
If tanA = a/b, What is the value of sinA/cos^8A +cosA/sin^8A
tbh i could expand it differently
:/
limpid plover
I never thought there would be someone below my level.........
Everyone here seems to be way above my level
Level as in qualifications
hoary yoke
i havent started trig in school yet
limpid plover
I thought so
There is also a question in #geometry-and-trigonometry that is easy, but I can't get it
I will be right back
past veldt
You can find an identity for sec^7 in terms of tan, and one for cosec^7 in terms of tan
hoary yoke
adding is same as multiplying for those 2
;\
i remember proving that identity
=tex \frac {a^7+3a^5b^2+3a^3b^4+ab^6}{b^8} + \frac {b^7+3b^5a^2+3b^3a^4+ba^6}{a^8}
granite stirrupBOT
hybrid charm
Bye
true vigil
cant see
somber trellis
did you take that picture using a 2002 nokia 7650
true vigil
take a screenshot and not a photo
true vigil
how did you arrive at those angles?
somber trellis
your angle is meant to be positive
true vigil
^
somber trellis
👍
somber trellis
same method as before
viscid thistle
I tried to transform it backward
somber trellis
this one's easiest
somber trellis
equate the real and imaginary parts
real part is zero, so cos(theta) should be zero
rest is easy
true vigil
what angle are you putting
true vigil
draw the vector -5i and look at the angle, what should it be?
viscid thistle
Or 180
viscid thistle
So it would be at quad 3 or 4 right?
true vigil
between them yeah
viscid thistle
Cuz it has no number b4 it. That means its 0
But i still dont know what to put for the answers
@true vigilur there still? Sorry for pinging
true vigil
can I see?
viscid thistle
Like this right?
true vigil
why are there two arrows
viscid thistle
?
2 arrows?
Cuz idk which quad it lies on
Between 3 and 4
So i drew for both
And connect with 0
viscid thistle
O i drew from the outside
true vigil
so the angle is
viscid thistle
180?
Straight linr
*line
Its not 90
Why this easy question triggers me so much ugh
true vigil
dont mind the C points
the angle is measured from the x axis
counterclockwise, like that
true vigil
yeah what were you doing in the others?
viscid thistle
I calculated them
Cuz im pretty bad when i have to draw it out and measure
I see. U taught better than my teacher
So now
From that 270?
U will figure out the number infront?
U switch 270 to the unit circle thingy right?
Wait no it doesnt work that way
viscid thistle
Ye
true vigil
the number in front you need to take sqrt(a^2 + b^2)
try to do the others by drawing the vector on the plane like I did
and relating the angle to the one you know how to calculate
it will be instructive
true vigil
cool
hybrid charm
dense zealot
Hai
hybrid charm
Hi
granite stirrupBOT
frozen osprey
oh is that equation your problem?
hoary yoke
hai
frozen osprey
heeello
frozen osprey
Hm... I'm not sure what it means to compute this by parts. Do you know about the common kinds of triangles?
frozen osprey
okay, are you allowed to simply use the fact that sin(pi/4) and cos(pi/4) = sqrt(2)/2?
or do you guys nto know this yet
hoary yoke
oh i know that
im just solving online problem
and I dont understand what to do but not compute by parts
frozen osprey
Hm well I guess it's worth noting that you can use a trig identity here to simply
if you factor our sin(pi/4) from the expression on the left, what do you get?
frozen osprey
!!! wow thank you
dense terrace
maybe toa cah soh
hoary yoke
no
dense terrace
maybe compound angles
dense terrace
dis owo
prisma epoch
Indeed
viscid thistle
Ok so my Algebra 3 class is super slow and I want to start learning precalc. What is the fastest way for me to get started?
true vigil
you could get like SAT preparation materials for the math section
they tend to be good and to the point
princeton review has great ones
viscid thistle
Neato
Now. What free sources are there? Anything like an online workbook or list of formulas?
true vigil
you can get textbooks for personal study for free in this day and age
dense zealot
Use aops
It's really good
The understanding it gives u of precalc is wayyy better than anything school could do
teaches u a lot of other stuff that'll come in handy later
calm thicket
=tex \sin^4(x)+\cos^4(x)=1-2(\sin^2(x))\cos^2(x)
granite stirrupBOT
calm thicket
Supposed to show whether it's true
granite stirrupBOT
granite stirrupBOT
calm thicket
So there's a difference of 1?
Because it's 1-2sin^2etc
no I mean it's false because there's a diff of 1
because sin(x)^4etc = 1
granite stirrupBOT
calm thicket
Ohh
okai
so ye
👍
ohfuck now there's solve for x ones D:
gonna see what I can do on my own tho
=tex \cos(5x)+\cos(x)=0\\cos(5x)=-\cos(x)\5x=-x\x=0
But solution is pi/6 or pi/4
patent beacon
What do you do to go from line 2 to line 3?
true vigil
gg
patent beacon
I don't know how you'd get an easy solution for this, without breaking down that cos(5x)
And that doesn't sound like fun
patent beacon
@pine kindle
Smart. I always forget about those
calm thicket
=tex \cos(5x)+\cos(x)=0\2\cos\left(\frac{5x+x}{2}\right)\cos\left(\frac{5x-x}{2}\right)=0
granite stirrupBOT
calm thicket
🤷
true vigil
what
granite stirrupBOT
dense zealot
No
patent beacon
There's another solution you have to consider
granite stirrupBOT
granite stirrupBOT
Support the bot on Patreon: https://www.patreon.com/dxsmiley
dense zealot
Ya
patent beacon
Ahh, so 5x = x + π and 5x = -x + π are both solutions
Perhaps not the only solutions in 0 < x < 2π
gritty pewter
dense terrace
yes
v = rw
v is the linear speed, r is the radius which is the 16-inch
find w <- this is the angular velocity
blazing raven
Synthetic Multiplication:
(x - 3)(x^2 - 2x + 5)
Write the polynomial to be multiplied at the bottom With the remainder 0. Fill in the whole second row. How? Multiply the bottom row by 3 to get the second row. Then subtract up.
3 \
----------------
1 -2 5 | 0
3 \
0 3 -6 15
----------------
1 -2 5 | 0
3 \ 1 -5 11 -15
0 3 -6 15
----------------
1 -2 5 | 0
Answer: 1 -5 11 -15 --> x^3 - 5x^2 + 11x - 15
thick raptor
lol, nice
blazing raven
😃
calm thicket
Let xy = a and xz = b, how can I find the minimum possible value of zy?
(Assume all values are positive if it matters)
calm thicket
but how tho :0
calm thicket
ohh clever
frank dock
How do I factor 6(x^2-4x+4)^2 + (x^2-4x+4)-1 so that it will equal x = - 5 and x = 1
blazing raven
um, do you see this pattern there: 6a^2 + a - 1?
frank dock
Yeah i think so
The textbook suggested that I set aside what was in the brackets and write it as 6r^2+r-1
and put the variables back in afterwards
but I got stuck at 6r^2+r-1 and don't know how to factor it from there
hexed ermine
@blazing raven Nice! How did you find it? Interesting concept :D
blazing raven
sorry, so we let r = x^2 -4x + 4.
Or r = (x - 2)^2. But if we look at the original expression it is equal to 6r^2 + r - 1 = (3r - 1)(2r + 1)
hexed ermine
(Synthetic Multiplication)
blazing raven
wolfram says your problem does Not equal -5 and 1 at the end.
So there must be a typo somewhere
(I also got to the same point)
anyway I gotta go.
frank dock
Sorry, looking at the wrong answer key the answer for this question was (2x^2-8x+9)(3x^2-12x+11)
stumped
So how do I go from 6(x^2-4x+4)^2 + (x^2-4x+4)-1 to (2x^2-8x+9)(3x^2-12x+11)
clever inlet
@frank dock ok
So you saw the substitution a one a two did?
He/she let r = (x-2)^2
And factored like a normal quadratic to get (3r - 1)(2r + 1)
You just need to sub (x-2)^2 back into r now
frank dock
ooh ok
thank you
very much
I have one last question can't quite wrap my head around, but why does the x intercepts for x^2-9x+8 being x = -1 and x -8 when graphed become positive
clever inlet
Hmm
Did you factor
(x-8)(x-1)?
Cause if so
The X intercepts are not -8 and -1.
They are 8 and 1
Find x intercepts by solving each factor = 0
x - 8 = 0
x = 8
clever inlet
np
granite stirrupBOT
smoky lance
when I multiply by the conjugate to rationalize the top,
do i distribute the 5 and the sqrt?
still yew
It should look like $$\frac{5-\sqrt{x ^ 2-2}}{13}$$
granite stirrupBOT
granite stirrupBOT
tired cedar
how'd u get rid of the cot^2 lol
hasty wadi
i did the inverse of tangent but did the reciprocal of cot
tired cedar
what are the interval restrictions
hasty wadi
there isn't any
tired cedar
2pi*
hasty wadi
from what i know
tired cedar
oh wait it nvm
hasty wadi
i don't even know what interval restrictions mean lmao
like im totally confused on how im wrong
tired cedar
where do u get pi/3 from
tired cedar
idk how to put the answer in your format