Youre looking at sec(4x+3)
#precalculus
shadow plaza
The solution that they gave is right only for k=4j where j is necessarily odd. That is what i am saying. You said it's valid only when k is odd
echo pelican
Youre solving for x not for the values on the unit circle that make sec undefined
Looking for specific values of x aaccording to this function
shadow plaza
Lol. No you're right. It is only valid when k is odd
Actually you know what, we're both doing it wrong
We should consider kπ+π/2 and kπ+3π/2 as the two cases to get the answer
lean tusk
I'm back - my bad, I didn't fully explain what the problem is asking for: here's everything it asks for (only looking at domain), also this is a different function this time
remote raptor
sick steppe
sin is opposite on hypotenuse
in this case y/r
sin(theta) = y only on the unit circle
@remote raptor ^
sick steppe
(R-C)(x) = R(x) - C(x)
arctic kestrel
oooooo
sick steppe
1/(1/cos) is cos
arctic kestrel
@sick steppe
i don't understand this?
What I understand is that i'm supposed to take 7 from g and replace it with f(g(7))
sick steppe
what's g(7)?
arctic kestrel
1
sick steppe
ok so what's f(g(7)) if g(7)=1
sick steppe
yeah
sick steppe
Just try the questions lol
sick steppe
no...
arctic kestrel
then divide y^2 by y
sick steppe
you just read off the center and radius
slate vessel
1/(1/cos) is cos
@sick steppe then why domain of both are different
sick steppe
cause 1/cos dne if cos(x)=0
sick steppe
but 1/(1/cos) = 1/sec = cos
slate vessel
Then why domain are different
1/sec will give same domain as 1/(1/cos)
which both giving different domain from cos
1/(1/cos=0)=1/infinity
which doesn't exist
viscid thistle
huh? $\frac{1}{\frac{1}{\cos(0)}}=\frac{1}{\frac{1}{1}}=1$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
i don't understand you at all
viscid thistle
what the fuck is that $\neq$, which 2 things you say have different domain?
obsidian monolithBOT
Al𝟛dium:
slate vessel
1/sec=1/(1/cos)
Domain will not contain odd multiples of pi/2
but
1/(1/cos)=cos
whose domain is R
this is where I am confusing
because it's the same expression but different domain
charred bridge
viscid thistle
^
we were told to make this shape using parametrics
i don't really know where to start
the rules are that the pattern should loop
meaning that t can be infinity and it'll still be a triangle
autumn marsh
@viscid thistle ask here
hearty robin
<@&286206848099549185>
viscid thistle
What ?
hearty robin
@viscid thistle
viscid thistle
?
willow bear
DE?
halcyon edge
Differential eq
willow bear
differentiate wrt x to get 1/a + y'/b = 0 and then y' = -b/a i guess
with initial condition y(0) = b
halcyon edge
Yep thanks
stuck lark
chain rule
halcyon edge
@stuck lark i see
viscid thistle
I need help with fundamental identities
viscid thistle
@viscid thistle still wondering? If so, do you have an specific problem you are stuck with
graceful dagger
Will anyone help me before my exams for calculus?
viscid thistle
@viscid thistle still wondering
can you help me?
do you have an specific problem you are stuck with?
none I just dont know how to solve like I am so confused
fossil crownBOT
viscid thistle
"fundamental identities" is very general
like this problem idk how to slove it
Do you have to prove it or solve for theta
What does your book tell you to do
prove it
like the rule it said I should start of with the complicated side which is the left right?
In 3 for example it does not really matter but generally yes
In 3, consider multiplying on num and den by something
will I be like needing this
to like reciprocal the identity
sand harbor
viscid thistle
"reciprocal the identity" is vague
lime bolt
you can multply both sides by cos*(1+sin) and u get the pythagorean identity
viscid thistle
.........
lime bolt
?
viscid thistle
The fact that you entered and said something that i wanted him to think in order to create intuition.
Sigh
yeah he shouldnt have said that
like its about learning not giving me the answer
lime bolt
hmm
viscid thistle
KANE, YOU WERE GIVEN THE LAST CHANCE.
It's NOT obvious for them
Kane
dont
thats not learning
viscid thistle
I'll leave this to a mod to decide over you.
Obvious is even worse than trivial
It's like changing with a synonym
lime bolt
ok ill note that for the future
viscid thistle
The thing is that you should've noted that in the previous 4 times.
lime bolt
ok
hasty wolf
hmm
@lime bolt
The point isn't "don't use this specific word". It's the overall idea of showing off that something is "easy" for you instead of being helpful.
lime bolt
i see
hasty wolf
So yeah. I did warn you.
sick steppe
at the x^2(x^2+1) line is where it goes wrong algebraically
viscid thistle
oh.....
sick steppe
ye
patent lance
Can anyone help me with this
Sorry for bad quality
Snipping tool like broke
Once I plug in 2+h for the equation I don't know what to do
echo pelican
You have to expand out the whole f(2+h)
And see if you can cancel terms from there
So e.g. the first term there would be
10*(2+h)^2
Expand that out using foil/distributive prooperty
And see what can be canceled out by other terms that appear
That would be my guess for what they want you to do
patent lance
Yeah gotcha
So you get (10h^2+36h+44)-44/h
Then you'd just take out the h yeah?
And get 10h+36
echo pelican
Yes if you did the steps correctly so far that works
patent lance
Alright neat ty
echo pelican
Based on how they wrote the answer i would assume you must be able to pull an h out of the top
patent lance
Yeah
This is gonna be a dumb question
But would you just do 6-7x>= 0?
To start off
Or would you distribute the 5
I got x<=6/7 and I'm not super sure if I did it right
And then after that I'm not sure how to write that as interval notation
echo pelican
No need to distribute the 5
The domain issues are all caused within the radical
So what you did is correct
You need all cases where the inside is greater than or equal to 0
echo pelican
Ok so all those pieces there describe one function
Look at the leftmost piece
It seems that when x is between -5 and -3
The way the function behaves is that it takes on a constant value of y=-3
You can describe either by inequality or interval which x values give the different y values
This one is called piecewise because the function behaves a certain way and then suddenly jumps to a different "piece" that behaves differently
echo pelican
The end point being dotted or open tells you if you include the endpoints
Yes exactly
patent lance
Yeah alright think I got it ty again
echo pelican
Yw
patent lance
It should be an open point right
But that'd interfere with the closed one with the other graph is my trouble
Unless I just did something else wrong as well which is possible
uncut mulch
open doesn't really interfere with closed
the issue is your graph isn't defined for x in (1,2]
echo pelican
The lower piece is closed at 1 whoch means it includes 1
The upper piece is open at 1 meaning that it doesnt quite reach 1
You can imagine that the open 1 on top is picking up exactly where the closed 1 came to a halt
patent lance
Oh ok
These would be right, right?
Or did I write inflection points wring5
Wrong
worthy summit
@everyone can someone remind me how to find asymptotes of a rational expression?
vertical
x^2-64/8x-64
void thorn
8x-64=0
pastel cloud
(x^2 - x) has to be less than or = 12
so (x^2 - x - 12) has to be less than or = 0
(x-4)(x+3) is less than or = 0
pastel cloud
i think b = 4 because 4^2 - 4 = 12 which is included in the domain
quick mirage
f * g = f(g(x)) = f(x^2 - x) = sqrt(12 - (x^2 - x))
= sqrt(-x^2 + x + 12)
= sqrt((-x+4)(x+3))
-x+4 >= 0
-x >= -4
x <= 4
x+3 >= 0
x >= -3
-3 <= x <= 4
Range = |-3, 4|
fiery wren
viscid thistle
first make a triangle with those points
find the y = mx + b equation that goes through b and c
then find the line perpendicular to that, that goes through A
then find the distance between A, and the intersection between the two lines
there's prob some formula for that actually
@naive prism
junior sable
Isn't 1/(square root of 2/2) 2/(square root of 2)? i don't get it
elder charm
uhh, no?
$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
but $\frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$
elder charm
proof: multiply numerator and denominator by $\sqrt{2}$ in both cases.
obsidian monolithBOT
ConfusedReptile:
junior sable
ok ty
pastel cloud
Any ideas on how to find m ?
I guess I know x+2 is a factor I could just use remainder theorem to find m ?
0 = 14(-2)^2 + m(-2) - 38
solve for m, and then use another method to find the rest of the factors?
14x^2 +9x-38 doesn't looks like it factors easily .. would I use polynomial division or quadratic formula after this?
I guess I could just graph it in my calculator to find the zeros
sick steppe
@pastel cloud you're right that 0 = 14(-2)^2 -2m -38, so you can get m easily
So you use the fact that the HA is y=-2/7 and the fact that the degree of the numerator = degree of the denominator
to find k ^
pastel cloud
oh I know how to get K, but I'm told to also find the other vertical asymptotes if there are any
sick steppe
The question you posted doesnt have anything to do with finding the other VA that I can see
sick steppe
Nope your pic is only find k and m
pastel cloud
it says in brackets (note there may be other vertical asymptotes)
pastel cloud
oh lmao
sick steppe
yeah
but you can find the other VA as 14x^2 + 9x -38 is factorable
since b^2-4ac > 0
sick steppe
oof
pastel cloud
ill be fine... i think
im comfortable with factor theorem and remainder theorem, reading graphs, knowing asymptotes, etc
u helped me the other day so thanks for that.
steel sequoia
Found the maximum on a graph but it gave me long decimal values. How do I instead see it as a fraction?
viscid thistle
Can i see your number
steel sequoia
For X I have -0.7320494 and Y is 2.5980762 so it's not rounding errors
I'm graphing 1/4(x-1)(x-4)(x+2)
And need the fractional values for the min and max
Unless they come out whole
sick steppe
@steel sequoia d/dx the function then quad formula is the only way you'll get exact answers
pastel cloud
Just finished my test... I think i did well
took longer than it should have, but I still had some spare time to go over my answers thoroughly.
nova ice
So I'm currently trying to precalculus. I wanted to know what you people thought would be the best sites and books to learn it
sick steppe
firstly: "Trying to precalculus"
2ndly: Khan Academy or Professor Leonard on Youtube
nova ice
Thank you
naive prism
im so confused on how you do this
i so far found the center of the circle and the radius and thats about it
idk where to really go from there
viscid thistle
What topic is this and how do I solve it ?
obsidian monolithBOT
moshill1:
viscid thistle
any hints?
uncut mulch
start with
z = k/t
use the info to determine k and hence the relation between z and t
viscid thistle
This is what I thought when I was doing this, what do you mean by the relationship between z and t?
uncut mulch
well you're starting with
z = k/t (from the inverse proportion)
you determined k to be 42
hence z = 42/t
obsidian monolithBOT
moshill1:
sick steppe
$S \propto pq \implies S = kpq$
obsidian monolithBOT
moshill1:
viscid thistle
Is there a formula that I should be following for these?
I don't really understand what they mean by, jointly, inversely, and directly.
sick steppe
If y inversely varies with x, $ y \propto \frac{1}{x}$. If y directly varies with x, $y \propto x$. Jointly means the x would be multiple variables.
obsidian monolithBOT
moshill1:
sick steppe
@viscid thistle little rule of thumb for you 🙂
viscid thistle
I'm starting to understand it a bit more now after doing a few questions
sick steppe
good
viscid thistle
Ex: y varies jointly with g
I think
y = kg, is this wrong?
that's direct variation, right?
viscid thistle
ok
I was doing this problem and thought
that was the way to answer it, and got stumped because I thought that's what the answer was
for this
sick steppe
ye
sick steppe
Ik wording is slightly weird
viscid thistle
yeah
sick steppe
cause we're use to "t squared"
viscid thistle
yep
sick steppe
but in proportionality you just phrase it that way
viscid thistle
I see
dawn karma
hey, I need help. lim sin(x) x--> (pi/2) = 0 ?
I was doing Khan Academy and this was one of the questions. I selected 0 and it marked as wrong
sick steppe
$ s \propto gt^2$
viscid thistle
my professor barely went over this, she just skimmed over it a bit so I'm slightly confused and need to do some touch up on it
obsidian monolithBOT
moshill1:
sick steppe
(which looks like distance but i digress)
@dawn karma sine starts at 0 and goes up to 1, then down to 0, to -1, up to 0
dawn karma
so the answer is 0?
full rapids
i’m a little confused about this, when i plug it into the calculator it starts adding the 4 into the answer but all the examples my teacher has given me are about only messing with the denominator and setting it equal to 0
my teacher never really spoke about finding the domain of equations where the radical is in the numerator 
uncut mulch
main things you need to be concerned about are
division by 0, sqrt of negatives, logs of non-positive numbers
full rapids
we’re doing fairly basic stuff as of now (and our course has skipped over a lot because of corona) so we haven’t worked with logs
other than that yes i’m just confused on the procedures i would use to solve this kind of problem
uncut mulch
well sqrt of negatives aren't defined over the reals,
so you'd also need to radicand: x-4 to be greater or equal to 0
full rapids
ahhh okay
lilac pier
@solid juniper is this a test
lilac pier
well for the first one it would help if you completed the squares
write it in standard form
lilac pier
= 1?
solid juniper
ohh yeah forgot about = 1
lilac pier
okay so you can write it as (x-2)^2 / (sqrt3)^2 + (y+3)^2 / 2^2 = 1
Now you know what the length of the major and minor axes are
lilac pier
$ \frac{(x-2)^{2}}{3} + \frac{(y+3)^{2}}{4} = \frac{(x-2)^{2}}{(\sqrt{3})^{2}} + \frac{(y+3)^{2}}{2^{2}}$
obsidian monolithBOT
Sup?:
solid juniper
wait i get the part on how to write it in that form but i tried to compute it and the result is dy/dx=-4x+8/3y+9. so i was wondering if thats the axis cause i thought it has to be a whole number
lapis sphinx
@solid juniper
solid juniper
yes?
lapis sphinx
How’d u get dy/dx ?
solid juniper
by doing (x-2)^2 / (sqrt3)^2 + (y+3)^2 / 2^2 = 1 on my calculator
lapis sphinx
Ur just trying to find the length of the major and minor axes right?
solid juniper
im actually just stuck at question 2 and 3
Ur just trying to find the length of the major and minor axes right?
@lapis sphinx im done now but i dont know if its right
lapis sphinx
Dy/dx is for derivatives; it’s calculus
solid juniper
2 and 3 are the only questions left
lapis sphinx
Alright I’ll take a look
solid juniper
(2,1) and (2,-5) major axis (2+sqrt3,-3) and (2-sqrt3,-3) minor axis
lapis sphinx
Alright
So for number 2
What was the equation u got from the information u had
U should’ve gotten this I believe
From there on, substitute in the values provided by the coordinate (3,3.2) into the equation for the ellipse
Then just solve for c
And ur done
U should’ve gotten c is 16 btw
Once u solve
Questions @solid juniper ?
Not sure abt number 3, haven’t done conics In a looooong time
solid juniper
so i would write in this from x^2/25+y^2/16?
lapis sphinx
=?
solid juniper
oh yeah = 1
solid juniper
ok
lapis sphinx
Do u know how I did it?
lapis sphinx
Idk. Haven’t done this in a while
Try to understand what I did for number 2
But I gtg
lapis sphinx
yea, np, just try to understand what I did lol
solid juniper
ok will do
plain turtle
i need help...
full pagoda
just post the question
full pagoda
a function cannot have multiple outputs for 1 input
so look at the given x and y values
if one x value corresponds to more than one y value, it is not a function
slate vessel
for each value of x there exists only one y for the relation to be a function
viscid thistle
while in the case of mod x there is no such function
spark veldt
lim Delta x 0 (x+ Delta x)^ 2 -2(x+ Delta x)+1-(x^ 2 -2x+1) Delta x
Help
lim Delta x 0 (x+ Delta x)^ 2 -2(x+ Delta x)+1-(x^ 2 -2x+1) Delta x
Sorry that’s hard to read
Nvm I got it
topaz obsidian
lapis sphinx
Isn’t it just graphing?
sick steppe
cause sqrt(-98) = sqrt(-1)sqrt(98)
cobalt storm
I don’t understand how this works
I don’t know the zeroes of the function so how can I create intervals
Would this be true bc 5 is in between 3 and 7
latent siren
5 being between 3 and 7 doesn't necessarily mean f(5) has to be between f(3) and f(7)
cobalt storm
Well x value wise if x values 3 and 7 are both producing positive y outputs wouldn’t 5 which is between them also produce a positive output
latent siren
Look at this function and take a look at f(-5) and f(0); f(-5) = 2 and f(0)=7 does that mean that f(-3) must be positive?
Sometimes it helps to try to visualize a contradiction
or a proof
cobalt storm
Oh so just because it is in between two x values that produce a positive output doesn’t necessarily mean it will for sure
Unless we actually see the whole function and can prove it
latent siren
yup 😄
cobalt storm
Thanks I appreciate it
latent siren
No problem, I'm very glad I could help! 🙂
sick seal
why is it not 1.3*140
astral mantle
try 140/.7
daring yarrow
waxen nymph
whats the answer
toxic karma
can some explain to me how to write 1x +2x +3x +4x in sigma notation?
half star
Note that all four factors in the sum have an $x$ and a positive integer between 1 and 4. So you can let $n=1,2,3,4$ and write it as
$$\sum_{n=1}^4 nx = 1x+2x+3x+4x$$
obsidian monolithBOT
derivada.schwarziana:
waxen nymph
can someone help me with this limit
nimble knoll
shut glade
i cannot figure out how this works
i cant get it to a probability and my math teacher is on vacation
so confused lol
honestly hate learning math in online school his "notes" were just setting the problem up and not doing anything
serene heath
you should be able to form 2 equations from the given info
shut glade
how so?
actually can anyone reccommend somewhere i can learn about these types of problems?
rich saddle
Got a probability question I'm having trouble wrapping my head around.
In a shipment of 30 total computers, 3 of which are defective, what's the probability of picking a sample of 6 computers with 2 defective computers in it?
Is 3C2/30C6 the correct answer? Or am I missing something?
It seems like that wouldn't scale well for larger amounts of defective computers. For instance, the chance of choosing 2 computers out of a set of 30 all defective should be 0, but that formula would give 30C2/30C6, which isn't 0.
patent beacon
rich saddle
How would one calculate that without having access to that calculator?
proud raven
sure
you can start with the easy one @viscid thistle
check out along the y axis, where x=0
if you plug in that x=0 to your equation
what do you get, and what does it have to equal?
uncut mulch
exponent laws.
power of 2/3 to each side
peak path
@loud marsh basically what @uncut mulch said, hope it helps :)
loud marsh
fr
peak path
yessir
valid lichen
Hey anyone mind helping me out with some graph stuff
its a curved graph, distance time
ive done the equation and gotten answer but idk how to put it into a graph
<@&286206848099549185> 👋
peak path
@valid lichen whats the equation?
valid lichen
So i figured out that its going linear from 0,0 to y=25.2 x=0.84
and then it starts curving
with -5.11
until it reaches x=6.72 y=113.4
idk how to put this into a graph
im stuck here idk what to do
Eh i honestly confuse myself now
peak path
with -5.11
@valid lichen is this the gradient?
valid lichen
So basically the problem was
car goes 30meter per second, and then it starts breaking after 0.84seconds with -5.11meter/second^2
ive worked out that it stops at 113.4meters
6.72seconds
peak path
and you're graphing the distance-time graph?
valid lichen
yes
peak path
how long did the car travel 30m/s for ?
peak path
and it stops breaking at 6.72s, 113.4 meter?
valid lichen
yep
it stands still at that moment
it breaks for 5.88seconds
and travels 88.2meters
whilst breaking
this maybe explains it better, i made a velocity time graph
peak path
is stracka area haha
peak path
which is 88.2
radiant ibex
could someone explain how they simplified this
i think i know what happened but i forget what its called when you multiply CONJUGATE
i think i just remembered
peak path
@valid lichen the distance time graph will look something like this
peak path
not that i know of
viscid thistle
@loud marsh
logx(27) = 3/2 is basically saying
x^(3/2) = 27.
we can further simplify this to
sqrt(x^3) = 27
square both sides
x^3 = 729
cuberoot both sides
x = 9
@viscid thistle imagine anything over 0 as infinity. (for this situation)
using this we know that at x = -3/2, the denominator is 0. and at x = 4 the denominator is also 0.
you then make a quadratic out of that.
need more help?
uncut mulch
🪵
quartz edge
what's the differential equation of 2x?
willow bear
the what?
obsidian monolithBOT
Svet L'octogone:
night fiber
In that case it is 2
obsidian monolithBOT
21 Sausage:
winged shell
i need to subtract the two
uncut mulch
consider factorising the denominator of the fraction on the left
winged shell
$\frac{2x-1}{(x-2)(x+1)}-\frac{1}{x-2}$
LCD: (x-2)(x+1)
obsidian monolithBOT
21 Sausage:
winged shell
what do i do next
uncut mulch
express it such that you have two fractions with the same denominator
i.e pretty much the same process as combining 1/2 + 1/4
uncut mulch
numerator looks off
winged shell
what is wrong with it
uncut mulch
how'd you get it?
winged shell
so the 2x-1 part already had the LCD in its original denominator so i did not multiply anything with the 2x-1
oh wait
obsidian monolithBOT
21 Sausage:
winged shell
correct?
uncut mulch
stray parentheses but other than that, looks ok
obsidian monolithBOT
21 Sausage:
obsidian monolithBOT
ramonov:
uncut mulch
which does simplify to 1/(x+1) and you should make a note that its for x \neq 2
winged shell
yes yes but my teacher was not looking for excluded value
values
and i was too lazy to put the parentheses
uncut mulch
i usually recommend that you state restrictions regardless if they are no longer implied in your final expression
winged shell
will do
i cant seem to find what im doing wrong here
ill send pictures of my work
give me like 5 minutes i need to try to finish the problem and upload the pictures
wait nvm
switched up the signs of the factored form of x^2-x-12
verbal pebble
final agate
sick steppe
with what exactly?
weary wing
Isn’t this basically the same? Should I get my instructor to view it and see if he can change my grade(s)
viscid thistle
can you show more context
weary wing
It was just the answer, I put in as “1” but it marked it as wrong and the correct one was supposed to be “x=1”
Isn’t that basically the same?
peak path
@weary wing not really, asymptotes can be parallel to the x or y axis (i.e. horizontal or vertical). if you do not specify it’s x=1, it could mean y=1, which is wrong. plus just the number 1 doesn’t really mean anything, is it an x-intercept? the gradient at one point? hope it helps!
weary wing
Ah okay, thanks for clarification! I’ll make sure not to mess that up.
peak path
no worries :)
quaint mason
Im lost on how to solve this
Trying different things and getting different solutions, none seems to be working.
peak path
@quaint mason you’re not doing this by hand right
peak path
like, without a calculator
quaint mason
i have a calculator
quaint mason
a
peak path
for a, you plug in 3 into n(t) then plug in the answer into s(t)
lemme use my calculator and see what i get
quaint mason
thats what i did
peak path
okay
quaint mason
wait so
u plugged 3 in for t right? for time
and then plug into other equation as n
peak path
yes
plugged it into n(t) first then plug it into s(n), don’t worry about the variables, just plug the number in
peak path
nope you don’t have to
quaint mason
ah repeat process
quaint mason
?
peak path
you gotta multiply what u get by n(2)
quaint mason
multiply what
peak path
s(n(2)) x n(2)
actually
the question said average, even though they didn’t say “average annual earnings per person”, i’m guessing they mean the same thing?
if so, then it’s just s(n(2))
if its asking for total earning, then it’d be s(n(2)) x n(2)
peak path
which one did you try?
peak path
ahh ok
peak path
@quaint mason oh no i forgot about this question im sorry oops
quaint mason
Nah its all gud haha
peak path
@quaint mason wait so how did u get it ?
quaint mason
Its the same step, my calculation was off thats all
Btw are u gud with graphs
f(g(x))
peak path
composite functions
quaint mason
Cuz im lost and tryna ask for help but apparently other peeps r lost too
quaint mason
nope
that is the entire problem
unfortunately, its estimating and ive tried every number so far that seems logical
someone recommended me findin the function itself but idk how i would be able to
peak path
ah i would recommend finding the fucntion itself
i found it
ok so
f(x) would be a cubic right ?
peak path
y=ae^x
peak path
have u learnt euler's number?
quaint mason
that one
peak path
yup the little e
quaint mason
i have
peak path
ok lets look at g(x) first
do u know how the basic exponential graph looks like?
y=e^x
quaint mason
yes
quaint mason
yes and its also raised up by 4
peak path
yup
hmm
i know what you're trying to say
but its called a dilation not translation, do u agree?
quaint mason
yeah
peak path
you know the y intercept is (0,5), u plug the values in and solve for a
yup!
ok g(x) = 5e^-x
quaint mason
now find f(x) right?
peak path
yup
quaint mason
well quadratic functions?
quaint mason
yes
peak path
do u know how to factorise a quadratic function?
to make the function look like this basically y=(x-a)(x-b)
quaint mason
yea
peak path
a and b represent the root of the function, i.e. the x-intercepts or where the graph cuts the x-axis
quaint mason
so whenever it crosses the x intercept?
peak path
because if you let y=0 to solve for the x-intercept, (x-a) = 0 or (x-b) = 0 because Null Factor Law
whenever it crosses the x-axis
quaint mason
Null Factor Law?
peak path
ahh ok, ill explain that later, but for now, just know that a and b is where the graph cuts the x axis
quaint mason
ok
quaint mason
c is if it crosses the 3rd time?
peak path
either one of a or b or c will equal to where f(x) cuts the x-axis (you'll have to approximate from the graph given)
mm
yes
but that wouldnt be possible for a quadratic right? because the maximum number of times a quadratic function can cut the x-axis is 2 times (since x can only equal to a maximum of two values in a quadratic)
therefore it has to be a cubic or a higher degree polynomial
(pls tell me if its confusing cuz idk how much u know hahah)
peak path
thats good
so lets say f(x) cuts the x-axis at (3.3,0)
do you know what f(x) will look like? (its not gonna be the final answer so theres still gonna be some variables in it)
peak path
either one of a or b or c will equal to where f(x) cuts the x-axis (you'll have to approximate from the graph given)
@peak path
peak path
thats a good question, i was bout to get to that ahah
at x=0, its a turning point, do u agree ?
have u learnt about turning points?
quaint mason
turning point?
quaint mason
ah yes
peak path
when there's a turning point at the x-axis, it basically means the graph "cuts the x-axis twice at the point"
quaint mason
twice? so ur saying a and b are both 0 and c would be 3.3
peak path
in simpler words, when the graph cuts the x- axis at point c, it will looks like (x-c)
when there is a turning point at the x-axis at point c, it will look like (x-c)^2
in this case, since c=0, its just x^2
sorry my wifi is slow rn
@quaint mason all good :)
quaint mason
x times x
peak path
x times x
@quaint mason ?
quaint mason
to get x^2
peak path
it should be -(x^2)(x-3.3)
peak path
do u know why there's a negative sign?
quaint mason
sounds good
fleet yew
i don't really know where this goes
but i was never taught combinatorics or anything like that
how do you solve this problem
my initial intuition was 5/6 * 4/5 so i got the right answer by being numerically close
but wth am i actually supposed to do here
<@&286206848099549185>
tfw self ping
round fossil
5!
peak path
yes!
peak path
@quaint mason yup
@fleet yew if there are 3 people sitting in a row, how many different ways can they sit? do u know this? :)
fleet yew
@peak path sorry for late reply
i do know about factorials
and to answer your question, that would be 3 starting people multiplied by 2 people following
so 3!
that problem is easy intuitively but i just don't know enough to solve that kind of problem analytically
fleet yew
yeah that's what i meant
peak path
there are two type of questions, arrangement of objects in a row and arrangement of objects in a circle
for row questions, its n! , n=number of objects
for circle questions, its (n-1)!
fleet yew
unique?
peak path
for the same n value, circle would have less possible combinations because its a circle and the two ends are joined up
unique?
@fleet yew ?
fleet yew
the unique number of arrangements
like let's say you're in a circle
is ABCDEF different from BCDEFA
@peak path
?
<@&286206848099549185> sorry for ping but anyone else willing to explain?
peak path
let Kenji = K and Mary = M and the remaining 4 people b A B C D
K and M cant be together
so the probability is (5! - number of ways that K and M are sitting together) / 5!
peak path
i would write it out
K and M is considered a group now
6 people now technically becomes 5 people
peak path
sorry if i keep going on and off im in a class lol
fleet yew
it's alright lol, better than nothing
peak path
(5-1)! 2! is how many possible arrangements there are if K and M will sit together
2! is because there are 2 people in the bracket, so 2!
fleet yew
right because it could be (K,M) or (M,K)
peak path
yes
4! 2! = 48
(5! - 48) / 5! will get u the answer
so the probability is (5! - number of ways that K and M are sitting together) / 5!
@peak path
fleet yew
dude thanks so much
i was clueless going into this lol
they don't teach this shit at schools but expect you to know it on the ACT
I was never taught about binomial coefficients or factorials or any of that stuff
fleet yew
@peak path are you in HS or college?
when are you supposed to learn this
i should've taken stats lol
peak path
last year of high school
i leant it last year
which is yr 11
which is sophomore ?? idek
ahh its junior
fleet yew
what was the class called
fleet yew
yeah that's not really a specific course here
like do you know what the unit was called
peak path
probability
fleet yew
ok yeah that makes sense
i never really learned about probability asides from the basic multiplying
like P(a+b) = P(a) * P(b)
here in america our math classes are calculus focused
inland igloo
does anybody know how to solve pythag identities
peak path
c^2 = a^2 + b^2?
fleet yew
for this one you need a very basic proof
and then the other identities like cot^2 + 1 = sec^2 etc follow naturally
inland igloo
k im currently on this type of question
fleet yew
draw a right triangle with some arbitrary angle theta
in quadrant 1 of the unit circle
peak path
@inland igloo have u learnt the unit circle
inland igloo
yes
peak path
so u know cos is x and sin is y ?
inland igloo
yes
but that has nothing to do with the unit circle as im
i would prove the identities
peak path
to prove it ?
peak path
do u know what i did in the photo ?
peak path
OHH
inland igloo
not the one im having trouble with