so
#help-0
dawn spruce
what's your final answer?
high cypress
log2[log2(7x - 10) x logx(16)] = 3
then,
log2(7x - 10) x logx(16) = 8
dawn spruce
lmao
high cypress
😭
dawn spruce
just remember that $\frac{log_b(x)}{log_b(y)} \neq log_b(\frac{x}{y})$
ocean sealBOT
AwesomeRat
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mystic gorge
Does anybody know what f*(x) is trying to say?
lone heartBOT
Does anybody know what f*(x) is trying to say?
quasi vector
well it's just saying that the two polynomials f(x) and f*(x) have the exact same set of solutions
it's just that f*(x) can have them in different multiplicity
fallen verge
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✅
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hollow frigate
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hollow frigate
i need some help with this or atleast just need someone to confirm my answer
im getting mixed answers from 2 different sources 😭
hollow frigate
im getting mixed answers from 2 different sources 😭
my one was 29
hollow frigate
U know how to do
ye
hollow frigate
thats right just to cross check
wanton wigeon
,w area bounded by y=2x^2-8x-10,y=x^2/2 -2x-1,x=1 and x=3
ocean sealBOT
hollow frigate
oh wow
wanton wigeon
Enjoy
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granite light
The perpendicular height is root244
From Pythagoras
Formula of triangle
Half bh
1/2 x 10 x root244
wanton wigeon
What
granite light
5 root 244
granite light
5 root 244
x4=20 root 244
granite light
Thanks
rancid marten
Find all solutions to: $x^2-5y^2=\pm4$ for $x,y \in \mathbb{Z^{+}}$
ocean sealBOT
The Prophet Of The Damned
rancid marten
idk how to solve pell when one isn't on the rhs
wraith edge
you could rewrite it then
put the 5y^2 on the rhs or the x^2
but this wouldn't work if you were to square root a side
or no nvm
just remember that the range of sqrt(x) is [0, +inf [
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@rancid marten Has your question been resolved?
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@rancid marten Has your question been resolved?
keen plinth
Find all solutions to: $x^2-5y^2=\pm4$ for $x,y \in \mathbb{Z^{+}}$
this is the same as finding the units of $\Z[\f {1 + \sqrt 5} 2]$
ocean sealBOT
keen plinth
you can use dirichlet's unit theorem to help you classify what the solutions look like
theres a fundamental unit which is the smallest unit $u > 1$
ocean sealBOT
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eternal token
help
eternal token
i reached 2ab <= a^2 + b^2
but idk what happened in the part where it says "Any numbers a^2 and b^2 have have geometric mean |ab| etc... etc..."
oblique karma
That means (a-b)^2 >= 0
oblique karma
but idk what happened in the part where it says "Any numbers a^2 and b^2 have ha...
Well
It means that the arithmetic mean of any 2 numbers is always greater than or equal to the geomatic mean
This method is used to find the range of functions like x^2 +1/x^2
eternal token
and that's supposed to show that 2ab <= a^2 + b^2?
eternal token
v = (a,b) w = (b,a)
dotProduct(v, w) = v[0] * w[0] + v[1] * w[1] = a * b + b * a
alpine sable
but idk what happened in the part where it says "Any numbers a^2 and b^2 have ha...
the geometric mean of two numbers (x and y) is sqrt(xy)
in your case you have that x = a^2 and y = b^2
so the geometric mean is sqrt(a^2 * b^2) which is trivially |ab|
eternal token
it's a known fact that geometric mean <= arithmetic mean?
alpine sable
i reached 2ab <= a^2 + b^2
make a substitution x = a^2 and y = b^2 here and then you’ve shown AM>= GM
eternal token
i see i just searched it i think it's called GM-AM inequality
alpine sable
eternal token
.close
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raw jetty
2 numbers less than 100 are picked, how many ways are there to pick two numbers such that the sum of the two numbers and the product of the two numbers are perfect squares
raw jetty
the two numbers can be the same
so 1
4
9
16
25
36
49
64
81
100
121
144
169
196
1 eont work
half of the even squares work so (2,2) (8,8) (18,18) (32,32) (50,50) (72,72) (98,98)
pythag triples work also, so (9,16) (36,64)
thats 9, these are the ones i found, im not sure about ones like where for example (2,2^3) [the product are perfect squares, sum isnt, just an example if such nunber exists]
or like 6 and 24
raw jetty
thats 9, these are the ones i found, im not sure about ones like where for examp...
im not sure how to check cases like this, cause there are a shit ton
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@raw jetty Has your question been resolved?
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@raw jetty Has your question been resolved?
meager karma
have you gotten:
20 5
80 20
90 10
98 2
raw jetty
no
raw jetty
have you gotten:
20 5
80 20
90 10
98 2
how did you get these?
cloud kayak
meager karma
how did you get these?
i wish i had a method
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empty dune
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cerulean grove
harsh swallow
i think i have something that might help
maybe
Could you argue that the partial derivatives of f are +-2|x-y| and so integrating that gives the upper and lower bounds of f itself
it gives the right answer but i'm not sure it's completely justified
cerulean grove
hmmm i thought for a while but i'm too dumb to do it, how can i show that 
harsh swallow
$\int \pm 2 |x - y| dx$ and $\int \pm 2 |x - y| dy$
ocean sealBOT
Katharine
cerulean grove
i know but how can we argue that the partial derivatives of f are +-2|x-y|
tight pier
plot twist
harsh swallow
lol
f(0, 0) = 0
and the slope of f
has to lie between some values
and using that fact you want to show that at f(5,4) the value of f cannot be above some value or below some - of that value
because if it were then at some point in between f(0, 0) and f(5, 4) the slope would have to be bigger than allowed
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unborn arch
what does it mean if two quantities are directly RELATED ?
lone heartBOT
what does it mean if two quantities are directly RELATED ?
reef pier
x and y are directly related if y = kx where k is some positive constant
in other words, as x increases y increases
unborn arch
is it the same as being directly proportional ?
reef pier
and as x decreases y decreases
directly proportional is y = kx but k can be any nonzero constant
for example, as x increases, y decreases by a fixed amount
x and y would be directly proportional but not directly related
reef pier
you can't say they're either directly or indirectly related
because it doesn't satisfy either y = kx or xy = k
they're nonlinearly related
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subtle pewter
Would it be $\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ((\cos(t))(\cos^2(t) - \sin^2(t)) - (\sin(t)\cos(t))(-\sin(t))) dt$
ocean sealBOT
Calc III Victim
subtle pewter
rah whats with those gaps
$\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ((\cos(t))(\cos^2(t) - \sin^2(t)) - (\sin(t)\cos(t))(-\sin(t))) dt$
ocean sealBOT
Calc III Victim
subtle pewter
basically using this formula
normal oar
where'd you get that formula from lmao
subtle pewter
subtle pewter
mhm alr one sec
normal oar
do you know how green's theorem works
subtle pewter
we covered it this week in class. Im watching a vid rq coz I didnt attend lec
brb
subtle pewter
bet
normal oar
dont look at the formal proof just look at the geometric intuition
it's really intuitive
normal oar
Would it be $\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ((\cos(t))(\cos^2(...
oh nvm lol
you literally just do a double integral
normal oar
$$ \iint_{-\frac{-\pi}{2}}^{\frac{\pi}{2}} F dxdy $$
subtle pewter
why though
normal oar
where F = (cos(t),sin(t)(cos(t)))
ocean sealBOT
dindu
normal oar
yeah but this is probably way easier
normal oar
i used the wrong notation
normal oar
are -pi/2 and pi/2 aswell
errrr
normal oar
i fucked up
normal oar
should be F cdot [dx,dy]
subtle pewter
since all of them are equal anyways
normal oar
fuck i thought i learned this well
seems like i forgor
okay so the first formula should work but it is shit
you already have the curve parametrization
so it should just be an easy area integral
subtle pewter
so it should just be an easy area integral
but im still confused
about like how imma deal with the outer integral
wait wtf am I saying
it doesnt need to have bounds
omg i forgot about indefinite vs definite integrals
alr ty
.close
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short lichen
how to do part ii)
short lichen
<@&286206848099549185>
karmic jetty
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short lichen
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alpine sable
how to do part ii)
stop being friends with Ridho
alpine sable
how to do part ii)
Blackmail Ridho into solving his own problems
alpine sable
can someone explain me how he got 2/3)^n in left side line below the boxed thing
alpine sable
please ping when youre replying
ebon sparrow
they factored it
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alpine sable
they factored it
how ? i didnt understand that part can you elaborate some more
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mortal trellis
repeat the same argument for a_n and a_(n-1) instead of a_(n+1) and a_n
it will give you $a_n-1=\frac23(a_{n-1}-1)$
ocean sealBOT
Denascite
mortal trellis
and then you get the same formula again for a_(n-1) and a_(n-2)
and so on
all the way down to a_1
at each step you get an extra factor of 2/3
so in total you get (2/3)^n
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alpine sable
so in total you get (2/3)^n
is it like this way ?
im so sorry for responding late
thanks thanks i got it
idk how to close thread so once you available you can close the thread
cedar kelp
.close
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alpine sable
Can someone help me solve this?
lone heartBOT
Can someone help me solve this?
alpine sable
Note THIS IS NOT HW THIS IS EXAM REVIEW
I have tried alot of different ways to solve this tried using YT and it just doesn't work
I don't get the right answer no matter what I try
twilit dove
Ok do you understand that trig functions have a periodic nature? I.e they repeat after some time?
twilit dove
Kind of
If we look at a sine graph, we can see that it repeats every 2pi
Tangent also has a period like this
alpine sable
Ok yeah i understand that
twilit dove
And we can use it to solve this problem
Ok so do you know what the period is for tangent?
twilit dove
Ok try to see if you can figure it out
alpine sable
?*
twilit dove
Hm no
alpine sable
Alright then no i dont know
twilit dove
It’s just on the input
If we graph tangent, where does it repeat
Or if we look at the unit circle
twilit dove
Haha it’s ok it’s gonna help us here
alpine sable
He showed us this are we supposeed to like remember all the values?
Like i dont get it
twilit dove
If you look, you can see that after pi, the values(orange numbers) start repeating
You do need to remember values for this problem
And the period for tangent
twilit dove
So since the numbers start repeating after pi, that is the period
No after pi
It goes back to 0
alpine sable
So the Period would be 7pi/6?
twilit dove
No it would be pi
twilit dove
So for tan(9pi/4)
You can simplify that to tan(2pi+pi/4)
And since the 2 pi is when it repeats twice, it just goes back to zero
So your reference angle is pi/4
twilit dove
9pi/4 = 8pi/4+pi/4=2pi+pi/4
twin nimbus
(the reason why you'd want to do this is tan is a periodic function, with period pi, so you know that any value tan(pi k + c) = tan(c) for integer k)
alpine sable
Okay i see
twilit dove
And then we can “ignore” the 2pi since that is when the function equals 0
Yep thanks @twin nimbus
alpine sable
So could i just simplify the Tan9pi/4 and just get my answer?
twin nimbus
(also note that a function doesn't need to equal 0 at any point in order to be periodic, consider sec(x) as an example)
twilit dove
No, your answer will be equal to tan(pi/4)
alpine sable
Theyre saying pi/4 isntt correct
twilit dove
Which is on the unit circle
alpine sable
Oh ok
twilit dove
You have to find the outputted value
twin nimbus
Theyre saying pi/4 isntt correct
pi/4 is not correct, tan of pi/4 is correct, and it has a specific value
alpine sable
So its 1
twin nimbus
exactly
alpine sable
JFC im so failing this shit lmao
twilit dove
Yes exactly
alpine sable
I cant remember the whole god damn circle
twilit dove
There are a lot of resources and shortcuts to help you remember
I’d suggest trying to because a lot of math courses will require it of you
twin nimbus
you essentially only need to remember pi/4 (45 degrees), pi/6 (30 degrees), and pi/3 (60 degrees). Other values are either obvious (pi/2 is 90 degrees, and thus the point (0, 1)), or something you'd normally need a calculator for anyway, (such as 1 radian)
There's some gray area where you can, with effort, figure out an exact, closed form solution, such as sin(pi/12) using the half angle formula on sin(pi/6), but these are reasonably rare
and normally just involve repeated applications of the angle sum formulae
alpine sable
So the next question
is Sin (-240)
i assume i have to convert to Radians
and do the circle thing again'
twin nimbus
yes, but you'll notice quickly that the -240 degrees is the same as positive 120 degrees.
alpine sable
So if i have a Negative answer just divide by -2?
twin nimbus
no
sin(-240 deg) = sin(120 deg)
because the two angles are the exact same angle.
just one is going 120 counterclockwise to get to the point, while the other is going 240 clockwise.
alpine sable
Oh yeah i drew out the circle
I just didnt see it
i was being dumb for a second
Boy oh boy i SURE DO LOVE CALC
twin nimbus
this is just trig
alpine sable
Yeah im in the trig section
i hate math sm
But im gonna have to learn to like it
Ok so after simplifying i get 2pi/3
twin nimbus
yup, this is also what I got
twin nimbus
a trick I use to figure out trig values outside of the 1st quadrant is to look at reflections.
if you reflect the angle across the y axis, what angle do you get?
twin nimbus
just what angle
twin nimbus
from the positive x axis?
alpine sable
Wait ok
so Sin is Q2
across the Y
would put us in Q1
So it would be a Positive Angle right?
alpine sable
2pi/3
alpine sable
We just drop the 2?
alpine sable
Thank god my teacher taught me that very useful info im definitely not learning for the 1st time right now
twin nimbus
no worries.
well, I do hate to jet, but my wife came home, and I need to run errands 😦
wanton wigeon
No prob sir
twin nimbus
just remember, sin is the y coord
and cos is the x coord
so a reflection about the y axis doesn't change the y coord
but a reflection about the x axis does, it negates it
so sin(pi/3) = sin(2pi/3) = sin(-4pi/3)
alpine sable
Alright imma do my best
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gentle lily
hello!
gentle lily
I would like to ask for a bit of help 😅 I am quite confused on how to turn this general equation of a circle into standard form:
2x² + y² -2x + 6y - 12 = 0
it's different from the ones I've encountered (in which are equations where the x² and y² values are 1 and where the x² and y² values are the same number, and are divisible to the x and y value.)
so.. I'm in a bit of a pinch 😅
verbal blaze
I'm not sure what you mean. By "general equation of a circle", do you mean the with radius r and center (a,b) has the form?
(x-a)^2 + (y-b)^2 = r^2
Ok so start with that equation, and expand it out, and then match up terms.
Though, it's not an equation of a circle because the coefficient of x^2 is 2.
It's probably an ellipse.
you can prove this by trying to expand and match terms, and you'll end up with a contradiction.
Have a play...
https://www.desmos.com/calculator/3acksquflh
gentle lily
also! this is the first solving that I did
I already know the answer (forgive me for this, but I used chatgpt) and apparently it's this:
.flip
verbal blaze
why didn't you divide y by 2 like you did with all the other terms in the first step?
gentle lily
in a tutorial I've seen, they divided it by the coeff to isolate the variables
but since y² is already isolated maybe it didn't need any division 😅
verbal blaze
x = 2y => x = 2y/2 => x = y?
you need to apply multiplication and division to everything on both sides of the equality.
So, let's assume it's an ellipse of the following form
(x-a)^2 / A + (y-b)^2 / B = 1
Now expand everything out until you get something of the following form
(?)x^2 + (?)y^2 + (?)x + (?)y + (?) = 0
Then you can match each (?) term with your equation, and solve them indivually.
gentle lily
wait, I'm sorry, in the first equation that you sent there
that's what I'm trying to get
verbal blaze
Yes, so work it backwards
Because your equation is expanded. There's only one way to expand an equation, but multiple ways to factor it back down.
gentle lily
mmm okay okay I will try thank you
verbal blaze
Feel free to message me again if you want more help 🙂
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empty spade
Solve $8x^3-6x-\sqrt{3}=0$ by letting $x=\cos\theta$
lone heartBOT
Solve $8x^3-6x-\sqrt{3}=0$ by letting $x=\cos\theta$
ocean sealBOT
BOHO
empty spade
I got $\theta=\frac{\pi}{18},\frac{5\pi}{18},\frac{7\pi}{18},\frac{11\pi}{18},\frac{13\pi}{18},...$
ocean sealBOT
BOHO
empty spade
but im not sure how to know what three solutions are the solutions for the cubic
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empty spade
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wanton wigeon
Hlo
@empty spade
Look at the question
There will be some range given
For theta
empty spade
no range
thats the question
and im stuck on (b)
from the other parts u can assume that the sols are pi,11pi,13pi/18 but idk why they just reject 5pi/18 and 7pi/18
and well, all the other infinite possible solutions
warped marlin
you are getting all of these θ values by substituting n (or k) by integers, right?
empty spade
im just thinking quadrants in which cos3theta=root3/2 (the equation i got) is positive and then adding multiples of 2pi for the solutions of theta up to 2pi
warped marlin
for θ, they are all correct
livid sage
your derivation of cos(3theta) is correct
so, you have cos(3theta) = sqrt(3) / 2, yes?
warped marlin
But for x, only the thetas corresponding to k=0, -1, 1
are correct
if that makes some sense
livid sage
so, you have cos(3theta) = sqrt(3) / 2, yes?
essentially, the idea is you solved this equation incorrectly
5pi/18 and 7pi/18 do not satisfy cos(3theta) = sqrt(3)/2
empty spade
ohh yeah
livid sage
alright that's the first thing cleared
the second thing is that while it looks like there are infinitely many solutions, there are really only 3
when you extend the solution set to pi/18, 11pi/18, 13pi/18, 23pi/18, 25pi/18, etc...
you will notice that cos() of these values will repeat, and only the first 3 are unique
for example, cos(23pi/18) equals the previous solution cos(13pi/18), cos(25pi/18) equals the previous solution cos(11pi/18), etc
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snow walrus
how do you solve this? i thought it was gonna be super easy but apparently I got the wrong answer so now I just think im stupid
snow walrus
I laid out all the variables and kept substituting until I had a single variable
snow walrus
some dude just guess and checked and found it out in like 10 seconds
and I took like 5 minutes just to get a wrong answer lol
reef pier
how do you solve this? i thought it was gonna be super easy but apparently I got...
okay so assuming you're using variables, what system did you get?
heavy prairie
yeah if you wanted to solve this mathematically you'd want to use a system
snow walrus
im posting my work rn
x = heart
y = carrot
z = water
might be a little hard to track since I wrote everywhere
reef pier
the system at the top left looks good to me
tiny wyvern
how do you solve this? i thought it was gonna be super easy but apparently I got...
you need to discard and then compare with the last one
reef pier
if you know linear algebra and matrix stuff this is quick, otherwise you could always eliminate
snow walrus
you need to discard and then compare with the last one
Discard?
reef pier
for example, subtract the third equation from the first, and then compare that equation to the second equation. you'll have a system in just x and z which you can solve, and then go back and find y
snow walrus
for example, subtract the third equation from the first, and then compare that e...
Ohhhhh
Thats smart
heavy prairie
also is the answer 25?
snow walrus
Yes
tiny wyvern
also is the answer 25?
yes
snow walrus
I got 30.4 first time around 😭😭
heavy prairie
yeah you can just kind of look at it and figure it out
snow walrus
I have no clue where my error was
snow walrus
yeah you can just kind of look at it and figure it out
Yeah i figure im a tad bit braindead
heavy prairie
you arent
tiny wyvern
Discard?
when i say this i mean : discard the last one, my english is bad
i join just some minutes ago, i am curious about this discord
snow walrus
Im not sure where i webt wrong in the original
tiny wyvern
looks fun, i think i am bad on math so i think i can learn something from here
heavy prairie
lots of help to be had here
heavy prairie
Im not sure where i webt wrong in the original
but ok sry i will have a look at ur first one
aww man i should have payed more attention in school
im with you on the z=15-y bit
then i dont understand what happens
oh wait
now i am following again okay okay
heavy prairie
sub it in
snow walrus
I forgot to WRITE A ZERO
I multiplied 15 by 2 and then proceeded to write 3 💀
This stuff always happens
heavy prairie
yep
you are right that is correct you have forgotten a 0
you have brought great shame upon your ancestors
oh my god
snow walrus
No clue how im gonna survive in math loooool
heavy prairie
the 17 is wild
i think you will do just fine
think of it like this
who is going to carry the boats?
i mean the 1
who is going to carry the 1?
snow walrus
Huh
heavy prairie
Anyway yes sry did you get all the help you needed
You found the solution to your problem by yourself
snow walrus
Oh i see
snow walrus
Yeah it happens, im gonna grind some khan academy precalc to make myself tired so i can sleep
heavy prairie
anyway if you got all the help you needed then you can just type .close
and good idea
lone heartBOT
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gentle escarp
need hlp with part c
lone heartBOT
need hlp with part c
gentle escarp
$Let ( P(n) = \sin x + \sin 3x + \dots + \sin(2n-1)x = \frac{1 - \cos 2nx}{2 \sin x} ).
\textbf{Base Case:}
For ( n = 1 ),
[
\text{LHS} = \sin x
]
[
\text{RHS} = \frac{1 - \cos 2x}{2 \sin x} = \sin x
]
Hence, ( P(n=1) ) holds, so the base case is true.
\textbf{Inductive Step:}
Assume ( P(n=k) ) is true, where ( k ) belongs to the natural numbers:
[
\sin x + \sin 3x + \dots + \sin(2k-1)x = \frac{1 - \cos 2kx}{2 \sin x}
]
Now, prove for ( P(n = k+1) ):
[
\sin x + \sin 3x + \dots + \sin(2k-1)x + \sin(2k+1)x = \frac{1 - \cos(2k+2)x}{2 \sin x}
]
Starting with the inductive hypothesis and adding ( \sin(2k+1)x ) to both sides:
[
\frac{1 - \cos 2kx}{2 \sin x} + \sin(2k+1)x = \frac{1 - \cos(2k+2)x}{2 \sin x}
]$
ocean sealBOT
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gentle escarp
ive done this much
can someone see if this is fine and how do i move forward
do i take common denom on LHS?
livid sage
seems fairly legitimate so far
gentle escarp
can 1- cos2kx / 2 sinx be written as sinkx?
livid sage
no
quasi vector
do i take common denom on LHS?
Isolate sin(2k+1)x on one side
gentle escarp
Isolate sin(2k+1)x on one side
the method we learnt in class is to start with LHS and end up with RHS
quasi vector
What?
alpine sable
the method we learnt in class is to start with LHS and end up with RHS
doesnt matter
gentle escarp
is it also correct if we take both LHS and RHS at the same time
quasi vector
I mean, yeah, technically that's right
livid sage
is it also correct if we take both LHS and RHS at the same time
it's technically "correct" though in a proof you may opt to avoid it
quasi vector
But each step you're doing also works in the backward direction
livid sage
i've also been taught to leave one side unchanged and usually it's not too hard to fit your proofs this way
gentle escarp
it's technically "correct" though in a proof you may opt to avoid it
and this is a proof though so shouldnt we avoid it
quasi vector
Change your P to be LHS - RHS = 0
Then you can prove this for n=k+1 assuming P(n=k) is true without any problems
Although the other method is also correct, there's no logical fallacies there
gentle escarp
, rotate
ocean sealBOT
gentle escarp
This?
gentle escarp
Change your P to be LHS - RHS = 0
ive done this
quasi vector
Now you can combine the first and last terms
gentle escarp
gentle escarp
Now you can combine the first and last terms
I’m getting this then
alpine sable
wait what
quasi vector
How did you combine those two terms??
gentle escarp
How did you combine those two terms??
common denom of 2sinx
quasi vector
$\cos\left((2k+2)x\right)-\cos(2kx)\ne\cos\left(((2k+2)-2k)x\right)$
ocean sealBOT
kheerii
quasi vector
yeah they also changed the 2k to a k
gentle escarp
**kheerii**
oh okay
alpine sable
use cosc - cosd ?
quasi vector
what you actually need to do is use a sum-to-product formula
if you don't remember them you can derive them pretty easily
write $$\cos((2k+2)x)=\cos((2k+1)x + x)$$ and $$\cos(2kx)=\cos((2k+1)x-x)$$
gentle escarp
what you actually need to do is use a sum-to-product formula
what is this
ocean sealBOT
kheerii
livid sage
what is this
quasi vector
**kheerii**
again, if you haven't encountered these before you can just do this
it's the same thing
gentle escarp
, rotate
ocean sealBOT
quasi vector
yes
now do you know the angle sum and difference formulas for cos?
$\cos(A+B)$ and $\cos(A-B)$
ocean sealBOT
kheerii
gentle escarp
yep
gentle escarp
ok will use and update
I think the sum to product formulas aren’t in my syllabus though so I’ll have to find another way in the future
quasi vector
notice how the first term can be simplified
quasi vector
I think the sum to product formulas aren’t in my syllabus though so I’ll have to...
well you didn't use the sum to product formulas
gentle escarp
**kheerii**
what is this
alpine sable
I think the sum to product formulas aren’t in my syllabus though so I’ll have to...
it must be there since you know cos(a+b) and cos(a-b), its just the same formula
gentle escarp
then
quasi vector
that's just the angle sum formulas
gentle escarp
it must be there since you know cos(a+b) and cos(a-b), its just the same formula
only this is there
gentle escarp
notice how the first term can be simplified
ait how
quasi vector
only this is there
yes, the compound angle identitites
quasi vector
ait how
the sin(x) cancels from the numerator and denominator
and the 2 also cancels
lone heartBOT
gentle escarp
the sin(x) cancels from the numerator and denominator
ms does it like this
.reopen
lone heartBOT
✅
gentle escarp
is this not a simpler way
quasi vector
it's the same
quasi vector
all methods to solve this are equivalent, since they're using the same identity
lone heartBOT
whole flame
ms does it like this
sweet IB student?
gentle escarp
sweet IB student?
yeah
whole flame
👍
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Available help channel!
raw jetty
2x^3-x^2-6x+3=0, factors are sqrt(3) -sqrt(3) 1/2
lone heartBOT
2x^3-x^2-6x+3=0, factors are sqrt(3) -sqrt(3) 1/2
quasi vector
$x^2-3=(x+\sqrt{3})(x-\sqrt{3})$
raw jetty
wont rational roots theorem give us ±(3/2, 3/1, 1/2, 1/1) as the possibilities
raw jetty
$x^2-3=(x+\sqrt{3})(x-\sqrt{3})$
ysah i know
ocean sealBOT
kheerii
quasi vector
wont rational roots theorem give us ±(3/2, 3/1, 1/2, 1/1) as the possibilities
the rational roots theorem only give you the possibilities for the rational roots of the polynomial\
it doesn't say anything about the irrational roots
lapis timber
Pls help part b
lone heartBOT
Pls help part b
raw jetty
note that your b answer is only for a single triangle, and its asking for all (4) of the triangles
lapis timber
how do I find the triangle that has the base of 36
cedar kelp
Pls help part b
,rotate acc
ocean sealBOT
raw jetty
,rotate acc
whats acc i only know cw ccw
cedar kelp
whats acc i only know cw ccw
anti-clockwise
raw jetty
anti-clockwise
but thats acw?
cedar kelp
but thats acw?
i mis-typed
raw jetty
err k
lapis timber
what’s Heron’s formula
empty cedar
what’s Heron’s formula
raw jetty
what’s Heron’s formula
for sides of a triangle a,b,c, $s=\frac{a+b+c}{2}$, the formula for the area is $\sqrt{s(s-a)(s-b)(s-c)}$
ocean sealBOT
Skill_Issue
lone heartBOT
@lapis timber Has your question been resolved?
sly cedar
how do you solve question d? i don't know where to begin
lone heartBOT
how do you solve question d? i don't know where to begin
raw jetty
not sure if theres a better way, but you could let x be (0,0) and draw y=2x and see where it intersects AC
the y would be the side of the cube ans the x would be half of the side of the cube
quasi vector
There's a lot of similar and congruent triangles here
Just the relations from those should be enough to solve the problmes
lone heartBOT
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Available help channel!
lapis timber
How to solve qn6 d
lone heartBOT
How to solve qn6 d
drifting seal
Hint: quadratic equation
raw jetty
Hint: quadratic equation
you can factor tan x to make it even easier
cedar kelp
I hope that calculator is allowed
drifting seal
you can factor tan x to make it even easier
Oh ye
lone heartBOT
@lapis timber Has your question been resolved?
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rocky musk
!help
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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
rocky musk
In answer key I see the it's is k > 289/12 but solving for discriminant for the two equations I got k > -289/12 or k < 289/12
I'm confused what's wrong so any help would be appreciated 🙏
lone heartBOT
@rocky musk Has your question been resolved?
rocky musk
<@&286206848099549185>
wanton wigeon
!15min
lone heartBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
Please stick to your channel.
real gazelle
chill with the factoid spam
wanton wigeon
Sry was saying him to not ping
real gazelle
In answer key I see the it's is `k > 289/12` but solving for discriminant for th...
lemme take a look
wanton wigeon
I found a factoid list somehow
real gazelle
please stop
real gazelle
In answer key I see the it's is `k > 289/12` but solving for discriminant for th...
can you show me the two equations that you got
rocky musk
|3x^2 + 13x - 10| = k
3x^2 + 13x - 10 - k = 0 and 3x^2 + 13x - 10 + k = 0
wanton wigeon
Good
real gazelle
alright cool so then you want the values of k where those two combined have two solutions in total right? so there are a couple possibilities here
wanton wigeon
Yes
real gazelle
like maybe the first equation has two solutions and the second equation has zero solutions
wanton wigeon
Yes
real gazelle
oppa I'm not talking to you..
wanton wigeon
Or the second has zero and the first has two
rocky musk
I see
wanton wigeon
@rocky musk u can try using the condition of discrimants of the equations
Can u show your work so we can identify the mistake 🙂
rocky musk
In answer key I see the it's is `k > 289/12` but solving for discriminant for th...
^ i did that but actually that gave me values of k such that there's four distinct roots right
wanton wigeon
We should also consider the case of common roots?
When both equations have same roots
rocky musk
oh
wanton wigeon
Slender have they given anything considering the nature of the roots like real or complex
Or they have simply said that this system should have only 2 Distinct roots?
rocky musk
yes only 2 distinct roots
real gazelle
I think they probably mean real roots
wanton wigeon
Yes it seems so
rocky musk
if it was complex they'd mention it
wanton wigeon
So you can apply the conditions D>0 and D<0
Yes only that is possible
Because Both the equations will have common roots only when K=0
wanton wigeon
Yes
Now for the k>289/12 part
You can use the conditions of the discrimants
!show
lone heartBOT
Show your work, and if possible, explain where you are stuck.
wanton wigeon
Show us your work so that we can help and find the error
rocky musk
well for first equation 3x^2 + 13X - 10 - k = 0
wanton wigeon
Ok
rocky musk
discriminant:
b^2 - 4ac => 169 - (12)(-10 -k)
=> 289 + 12k > 0
=> 12k > -289
=> k > -289/12
wanton wigeon
Ok
rocky musk
for second equation i just did 169 - (-10 + k) => 289 - 12k > 0 and then k < 289/12
wanton wigeon
Now when u apply D<0 for the 2nd equation
wanton wigeon
Slender imma explain simply
See u want only two roots distinct for the equations
So only one equation can have real roots among the two
And for that we have D>0
And for the other equation which will have complex roots we will apply D<0
rocky musk
Alright thanks
rocky musk
discriminant:
`b^2 - 4ac => 169 - (12)(-10 -k) `
`=> 289 + 12k > 0 `
`=> 12k >...
but why do we not take this ?
wanton wigeon
Are u sure you got it clear right?
wanton wigeon
but why do we not take this ?
We can have D>0 only for one equation among the two
If u apply for this
Then u have to apply D<0 for the other equation
1.D>0, a quadratic equations has two distinct real roots
2.D<0, a quadratic equations has complex roots
3.D=0 a quadratic equations has equal real roots
I hope u will be able to understand now
rocky musk
discriminant:
`b^2 - 4ac => 169 - (12)(-10 -k) `
`=> 289 + 12k > 0 `
`=> 12k >...
But why are we discarding this one?
wanton wigeon
See
We are not
k>289/12 and k>-289/12 right?
But we need to satisfy both the conditions
That's why we take the larger one
k>289/12
wanton wigeon
Is your doubt resolved now?
lone heartBOT
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Available help channel!
warped topaz
lone heartBOT
warped topaz
You were like 1 seconds slower than me 😅
warped topaz
npnp haha
If I had know I wouldve let you have the channel but I didnt see you typing
kindred sundial
its fine ill go to another one
warped topaz
its fine ill go to another one
Best of luck!! 😄
kindred sundial
Best of luck!! 😄
you too mate
warped topaz
So I have some ideas for this question
I am thinking that I need to get it in the form a < p < b
And a is the first vertex and b is the second one
So I think I need to get the second derivatives and set them equal to zero to find them right?
crude totem
both parts ?
crude totem
find y value of highest intersection point where it will intersect twice and y value of lowest intersection point where it will intersect twice
sour mica
You get 3 intersections with a horizontal line if only the curve has 3 points with same y
This doesnt happen, once you cross the local minima and maxima of the function
fickle heath
So I think I need to get the second derivatives and set them equal to zero to fi...
Just the first derivative
warped topaz
Ah oops
What was second derivative gonna give me? I feel like im mixing it up with something else
crude totem
concavity
fickle heath
f''(x) = 0? The inflection point
warped topaz
Ahh oke
Well 1 moment I think I can solve it but I wont close this yet incase I run into new issues
crude totem
f''(x) = 0? The inflection point
^^
stiff copper
can anyone help with trignometry
Q- if cosA = 7/15 and A + B = 90 then find cosecB
runic glacier
can anyone help with trignometry
Q- if cosA = 7/15 and A + B = 90 then find cos...
Do you know how isolate A first?
crude totem
can anyone help with trignometry
Q- if cosA = 7/15 and A + B = 90 then find cos...
draw a triangle
with cos A = 7/15
if A + B = 90, then the other angle in the triangle would be ?
stiff copper
Do you know how isolate A first?
are you saying about
if
tanA=1
than
tanA=tan45
A=45
crude totem
find the other side
runic glacier
are you saying about
if
tanA=1
than
tanA=tan45
A=45
Well I just mean apply the inverse of Cos to both sides
stiff copper
draw a triangle
ok then?
stiff copper
if A + B = 90, then what would the other angle in the triangle be?
i dont understand
crude totem
i dont understand
how many degrees are the angles of a triangle
stiff copper
180
stiff copper
90
lone heartBOT
lone heartBOT
Available help channel!
alpine sable
lone heartBOT
sour mica
why have you cancelled the equal roots thing?
sour mica
oh okay
alpine sable
i still am confused though
sour mica
well, since the roots are real and distinct, there is only one condition that deals with this
How would you go about calculating the roots of a quadratic equation?
sour mica
Yes, so use the quadratic formula
sour mica
Well, hating something doesnt mean its not going to be helpful ¯_(ツ)_/¯
alpine sable
:(
sour mica
you gotta use it when you need it
sour mica
npnp
safe tartan
It’s clear you use discriminant
alpine sable
Hi Dear @alpine sable
The answer to this question is quite simple & easy
All you have to do is, use "Discriminant" here, which is b²-4ac. For real roots, the discriminant is greater than 0. So, just substitute the values in the formula & you will find the solution to this.
- Regards
lone heartBOT
@alpine sable Has your question been resolved?
strange heath
so the equation is
(k-3)^2 - (4 x 1 x (3-2k))
and put this greater than 0
solve for k
u can do that rught ?
lone heartBOT
lone heartBOT
✅
alpine sable
Hi Dear <@456226577798135808>
**The answer to this question is quite simple & ...
ah ok
alpine sable
u can do that rught ?
its just not natural for me to use the discriminant
we havent been taught it well
so its not somehting that's intuitive
but yea it makes sense
thanks though
.close
cerulean grove
lone heartBOT
cerulean grove
but how can i know which point is max which point is min
sharp panther
but how can i know which point is max which point is min
Um if I'm correct
There is a double derivative method for finding if it's a maxima aur minima
But I don't remember it
You can check MIT's website for multivariable calculus courses
It will be there for sure
lone heartBOT
@cerulean grove Has your question been resolved?
modest robin
but how can i know which point is max which point is min
when you sub the parametrisation z(x(t),y(t)) in the surface you get a funtion from R to R so like in single variable calculus you do the second order derivative test and you can distinguish if it’s a max or a min by checking if f’’(x*) < 0 or f’’(x*) > 0.
cerulean grove
second order derivative test of what? z?
lone heartBOT
@cerulean grove Has your question been resolved?
verbal geyser
verbal geyser
Sorry the x should be t in (product rule line)😅
lone heartBOT
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midnight wadi
can someone explain how did he get 1 - tanAtanB
lone heartBOT
can someone explain how did he get 1 - tanAtanB
hearty quartz
sin(theta)/cos(theta) = tan(theta)
This comes directly from the unit circle definition of trig functions
lean rose
Sigma
small lance
.close
lone heartBOT