and $\frac{100}{180}$ can be simplified
#help-13
surreal cave
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
0.5555...
surreal hamlet
5/9
surreal hamlet
yea
surreal cave
ok now what is the radius of the circle?
surreal hamlet
2?
surreal cave
yes
and now
we have the components we need to plug and chug
$S=r\theta$, we know $\theta=\frac{5\pi}{9}$ and $r=2$
wraith daggerBOT
XxMrFancyu2xX
surreal cave
now just solve for $S$!
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
so the answer is just 5 pi over 9?
surreal cave
not quite
surreal hamlet
2 = 5 pi/9?
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
idk
surreal hamlet
no
surreal cave
do you know what pi is?
surreal hamlet
yes
surreal cave
ok
good
that is something to work with
id be very concerned otherwise π
so $\pi\approx3.141\ldots$, right?
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
yes
surreal cave
ok good
so pi is defined as the circle's cirumference, the outside, divided by the diameter
so $\pi=\frac{C}{d}$
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
yes
surreal cave
ok so know I will do the following rearrangement:\
$\pi=\frac{C}{d}\implies\pi\cdot d=C$
wraith daggerBOT
XxMrFancyu2xX
surreal cave
do you agree?
surreal cave
d is the diameter
it can be anything
but here
we use a variable, a placeholder
that can variate
hence the name
variable
So if I were to ask you what is the circumference of a circle with diameter of 3. What would be your answer?
surreal hamlet
9.42477 round it becomes 9.42
surreal cave
you are completely correct! (but try to keep it in terms of pi) 3pi is a totally acceptable answer and actually is preferred at higher level maths
surreal hamlet
alright
surreal cave
ok now what about the circumference of a circle with diameter 5?
surreal hamlet
5pi
surreal hamlet
alright
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
yes
surreal cave
$\frac{d}{2}\implies$\fbox{$d=2r$}
wraith daggerBOT
XxMrFancyu2xX
surreal cave
right?
surreal hamlet
yes
surreal cave
So how can we use the newly obtained knowledge: $d=2r$ and use it for $C=d\pi$?
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
JK is the radius correct?
in circle J so J is the midpt
JK is radius which is 2
so diameter has to be 4
right?
surreal cave
so diameter has to be 4
well in your case yea but im talking about this formula
surreal cave
**XxMrFancyu2xX**
^
surreal hamlet
im very confused now
surreal cave
Okay fine, so we have $C=d\pi$ right?
surreal hamlet
here is the image so you dont have to scroll all the way up again
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
yes
surreal cave
and $d=2r$ right?
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
yes
surreal cave
So now we have $C=2\pi r$
wraith daggerBOT
XxMrFancyu2xX
surreal cave
and THAT is the formula for the circumference of a circle
surreal hamlet
yes
surreal cave
So if I told you that the radius of my pizza is 4in, then what is it's Circumference? (also include units)
surreal hamlet
25.13 inches
surreal cave
yes and because its a word problem good job not using pi here! in a word problem use real world units because it is about the real world
unless its a word problem or stated otherwise, DO NOT ROUND PI
that should be drilled into your mind π
surreal hamlet
got it
so if it does not say round it, you dont round it? if it says leave in terms of pi, you dont round it right?
ok i got it,
surreal cave
if it says round it, round it
surreal hamlet
ok
surreal cave
alright good, now if were to ask you to compute the circumference of a circle with radius 2, what would your answer be?
note this not a word problem and im not asking you to round
surreal hamlet
2pi2
surreal cave
$2\cdot\pi\cdot2=4\pi$
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
yes
surreal cave
and that is your answer
now I will do a series of computations and explain them after:
$C=2\pi r\implies S=2\pi r\cdot\frac{\theta}{360}\implies S=\cancel{2\pi} r\cdot\frac{\theta}{\cancel{2\pi}}\implies S=\theta\cdot r$
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
hello?
surreal cave
yes?
surreal cave
well you kinda ghosted me π
sorry
anyways
the first step is just the formula we had
then the second is change it to the formula for arc length with respect to the angle
notice that is the same
expect for a theta/360ΒΊ factor
this just makes it so it multiplies by whatever fraction of the circumference you want, this is arc length
third step is converting into radians
then the fourth the 2pis cancel
and we are left with the final formula
ok
now do you understand where that formula came frmo?
S=arc length
surreal hamlet
yes
surreal hamlet
ok thank you
surreal hamlet
surreal cave
so earlier you had found the angle was $\theta=\frac{5\pi}{9}$
wraith daggerBOT
XxMrFancyu2xX
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
2
surreal cave
yes so knowing both of those things what is $S$ when $S=r\theta$?
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
$S=r\theta$
wraith daggerBOT
E'
surreal hamlet
what is the 0 looking thing?
surreal hamlet
2 angle wha???
surreal hamlet
yea
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
thanks, i got it correct
surreal cave
where $\theta$ is in radians
wraith daggerBOT
XxMrFancyu2xX
surreal hamlet
thank you very much
surreal cave
yw!
surreal cave
also don't forget to close the channel
surreal hamlet
and have a nice evening
surreal cave
sorry to 1 hour away from you
no worries i got time you too
surreal hamlet
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
obtuse hinge
how to do part c?
cedar kilnBOT
how to do part c?
obtuse hinge
that notation means C -> B
cedar kilnBOT
@obtuse hinge Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
crimson sedge
cosmic steppe
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
feral egret
Hi
radiant topaz
You have to find answer yourself
feral egret
I think it's 'a' option
tropic oxide
Please tell me the answer with explanation
we don't give answers here.
radiant topaz
So what is this server for
For maths
tropic oxide
bye bye andrew tate fan
radiant topaz
L rating for this server
Great
tropic oxide
note he does not yet appear to have left.
radiant topaz
His hopes needs to be crushed
dire geode
L rating for this server
Lol your rating is worthless
cedar kilnBOT
radiant topaz
Not for free answers. Go pay for a tutor
I assume Andrew tate fan should be able to atleast do that 
cedar kilnBOT
Available help channel!
undone star
"A car can hold 3 ppl in the front and 4 ppl in the back seats. How many ways can 7 people be seated if 2 people must sit in the back seat and 1 in the driver seat"
undone star
I might just be missing something obvious but i dont particularly understand the information about 2 ppl must sit in the back and 1 in the driver
its not as if there are extra seats available
all 7 seats will be occupied regardless so im kind of confused why that information is given
astral bay
...maybe they mean specific people?
as in it's already chosen who's going to drive and that person must be in the driver seat
astral bay
well i guess if there's an answer we can try to reverse-engineer it
hmm
2^5 * 3^2 apparently
cedar kilnBOT
@undone star Has your question been resolved?
sudden stratus
"A car can hold 3 ppl in the front and 4 ppl in the back seats. How many ways ca...
um
since driver is seated in front
let us fix his seat
there will be 6 people
then the 2 people who gonna sit in the back
we have 4 seats
out of that they have to choose any 2
thats 4C2
and then choosing seats among them
which is 2!
and remaining 4 ppl can choose whatever seat they want so 4!
4C2 * 2 * 24
,w 6224
wraith daggerBOT
sudden stratus
Is 288 the answer?
gentle cape
cedar kilnBOT
gentle cape
im trying to calculate u'(1)
but I get 3
im sure everything should be positive
im not sure why the textbook would say 0
upper abyss
I agree with you. Answer is a typo.
kind lodge
Hello. I have problem that I would like to know how I write this excel formula in math. "=POISSON.DIST(C5;C2;TRUE) " Poisson distribution that is cumulative. Thank you.
cedar kilnBOT
Hello. I have problem that I would like to know how I write this excel formula i...
@kind lodge Has your question been resolved?
dire geode
Hello. I have problem that I would like to know how I write this excel formula i...
there isn't a clean way to do it
cedar kilnBOT
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Available help channel!
limpid plume
Let $f : [a, b] \rightarrow \mathbb{R}$ be monotone. Given a partition on $[a, b]$, show that $\sum_{i = 1}^{n} \abs{f(x_i +) - f(x_i -)} \leq \abs{f(b) - f(a)}$. Use this to show D(f) (the set of discontinuities) is at most countable
wraith daggerBOT
rikusp2002
limpid plume
I tried to assume f is Increasing WLOG
what to do afterwords
I don't think Triangular inequality works
what it does is infact it proves the opposite
|f(b) - f(a)| is less than the sums
$\abs{f(b) - f(a)} \leq \abs{f(b) - f(x_n +)} + \sum_{i = i}^{n} \abs{f(x_i +) - f(x_i-)} + \abs{f(x_1 -) - f(a)}$
wraith daggerBOT
rikusp2002
limpid plume
With first and last terms in RHS are zero
because f(x_n+) = f(b) and f(x_1-) = f(a)
please help somebody π₯²
i have exams tomorrow
wraith daggerBOT
sacred grail
limpid plume
limpid plume
oki
sacred grail
[ \abs {f(x_i+) - f(x_i-)} = f(x_i+) - f(x_i-) ]
wraith daggerBOT
sacred grail
there is no need for the absolute values
limpid plume
.....lmao.
sacred grail
so triangle inequality isnt exactly useful here
limpid plume
I completely forgot
oh wait, so you mean the question is correct?
it didn't get the inequality wrong?
sacred grail
how could the sum of the jumps be larger than the difference between the endpoints
limpid plume
.......omg.
of course it won't attain global maxima in a point x \in (a, b)
why am I so idot most of the times
yeah so
and even if it doesn't, the sum of jumps can't be greater unless it breaks monotonicity
sacred grail
its kinda quite clear if you draw a diagram
but ofc proving it is a different matter
you can try think about it for a bit
limpid plume
Also yeah I see that equality is only achieved in case of step function
Any other monotone functions will have overall less jumps
That by drawing ofc
sacred grail
π§βπ¨
limpid plume
I mean if that's true the function is continuous 
sacred grail
if the function is continuous then $f(x_i+) - f(x_i-)$ would be 0 \thonk
wraith daggerBOT
sacred grail
there would literally be no point
limpid plume
Oh wait true
sacred grail
you know what would telescope
[2\textwidth] [
{\c bf(x_1+) - f(x_1-)} + {\c r f(x_2-) - f(x_1+)} + {\c b f(x_2+) - f(x_2-)} + {\c r f(x_3-) - f(x_2+)} + \dotsb
]
wraith daggerBOT
sacred grail
its a shame we only have the blue terms 
limpid plume
Wait a minute.
if i take the open interval (f(x_i-), f(x_i +)), then the intervals we get from the partition like this has to be pairwise disjoint
limpid plume
Oh wait, that proves the second statement but we don't use the first statement to prove it
Because the open intervals will form a countable collection
....wait isn't that the original proof
sacred grail
limpid plume
WAIT I SHOWED
a jump is less than f(b) - f(a)
I mean uh
For a < x < y <b
f(x+) - f(x -) and f(y+) - f(y-) both are less than f(b) - f(a)
limpid plume
Trivial
sacred grail
the 2 point version
limpid plume
sacred grail
of what you're meant to show
limpid plume
We have to add everything and still show its less than the f(b) - f(a)
sacred grail
i mean
these red terms
they are like
well they dont exist in your sum but
theyre all positive
so what i mean is
[2\textwidth] \begin{align*}
\MoveEqLeft
{\c bf(x_1+) - f(x_1-)} + {\c b f(x_2+) - f(x_2-)} + \dotsb \
& \le {\c bf(x_1+) - f(x_1-)} + {\c r f(x_2-) - f(x_1+)} + {\c b f(x_2+) - f(x_2-)} + {\c r f(x_3-) - f(x_2+)} + \dotsb
\end{align*}
limpid plume
wraith daggerBOT
limpid plume
So you mean rhs is just f(b) - f(a)
sacred grail
yeah lol
limpid plume
It's not equal, but less than equal to
Got it
Now we gotta use this stuff to show D(f) is at most countable
limpid plume
I just defined the sum over D(f), it's still finite
Less than f(b) - f(a)
Can we use this somehow
I think as the series sums upto something finite and positive, every jump must be positive and finite.
So, we define {x in [a, b] such that f(x+) not equal to f(x -)} = D(f)
And the first set is equivalent to the set of all intervals (f(x-), f(x+)) for x in D(f), which is a pairwise collection of open intervals in R
Hence D(f) is countable
sacred grail
do you know how to prove that the sum being finite implies its at most a countable sum
limpid plume
no
sacred grail
normalise f(b) - f(a) to 1
then there can be at most 1 jump of size >=1
2 jumps of size >=1/2
3 jumps of size >=1/3
4 jumps of size >=1/4
etc
countable union of finite sets is countable
sacred grail
define something like
[ D_n(f) = \set {x \in \cci ab \where f(x+) - f(x-) \ge \f1n} ]
wraith daggerBOT
sacred grail
then
[
D(f) = \set {x \in \cci ab \where f(x+) - f(x-) > 0} = \Union_{n = 1}^\infty D_n(f)
]
wraith daggerBOT
limpid plume
then there can be at most 1 jump of size >=1
Wait I don't understand this part
sacred grail
suppose there were two
limpid plume
2 jumps, okay
sacred grail
then f(x_1+) - f(x_1-) + f(x_2+) - f(x_2-) >= 1 + 1 = 2
but by assumption, f(b) - f(a) = 1
contradiction
limpid plume
But why are we putting greater than equal
sacred grail
oh
well
you dont know the actual size of the jumps
but you can be like
lets look at all those jumps >= 1/n for some n
sacred grail
yeah
you can WLOG assume f(b) - f(a) = 1
so you contradict the inequality from earlier
limpid plume
That just says no jumps can be greater than 1
sacred grail
it says the sum of all the jumps cannot be greater than f(b) - f(a) = 1
[
\sum_{i = 1}^n \abs {f(x_i+) - f(x_i-)} \le \abs {f(b) - f(a)}
]
wraith daggerBOT
sacred grail
thats the inequality
limpid plume
Yes
sacred grail
LHS is sum of jumps over all x_i
but like
say you just pick these x_i to be where the jumps are >= 1
then you cant have more than 1 of them
well i mean like
limpid plume
then you cant have more than 1 of them
That jump also has to be exactly equal to 1
sacred grail
yeah
limpid plume
I mean otherwise it exceeds
sacred grail
if it exists
it might not
but you certainly cant have more than 1 of them
thats all that matters
and its a similar argument for jumps >= 1/2
like
if you try to fill a gap of 1 using stairs of size 1/2
you cant use more than 2 stairs
sacred grail
yeah but we dont care about it exactly
just an upper bound
because all you need is all the D_n(f)s to be finite
so that when you take their countable union
its at most countable
cedar kilnBOT
Available help channel!
proud spindle
cedar kilnBOT
proud spindle
I have no idea how to even attempt this q
tranquil oracle
I would seriously consider plugging in values of nice angles to have a feel how the left hand side behaves
at least plug in 0 and pi/6 and pi/4
tall mica
from first glance either trig identity shenanigans
or try some x that give exact trig values
proud spindle
so i tried a graphical approach
tall mica
or try some x that give exact trig values
i think this is a better idea because you notice some stuff working nicely with each-other
proud spindle
so i tried a graphical approach
which gave me pi/8 and 7pi/24
mortal wagon
try multiplying by \sqrt{\sqrt 2 + 1} because it will cancel some things
proud spindle
yep i did that and the LHS became 2 * sqrt(4+2sqrt2)
and the rhs = cosx+(sqrt2)sinx+sinx/cosxsinx
cedar kilnBOT
@proud spindle Has your question been resolved?
cedar kilnBOT
@proud spindle Has your question been resolved?
forest portal
Explain me please, what is phase shift in trig function
cedar kilnBOT
Explain me please, what is phase shift in trig function
hollow gorge
It's the horizontal shift of a sin or cos function on a graph
forest portal
It's the horizontal shift of a sin or cos function on a graph
Okay, how would you draw y = cos(x - 2*pi) ?
.close
graceful moss
graceful moss
<@&286206848099549185>
pulsar sequoia
Yo
pulsar sequoia
<@&286206848099549185>
What do u need
graceful moss
a bit of advice
pulsar sequoia
Go ahead
graceful moss
i need to find all the question marks
pulsar sequoia
<@&286206848099549185>
cedar kilnBOT
@graceful moss Has your question been resolved?
graceful moss
<@&286206848099549185> multivariable calculus
manic scaffold
I would exploit the symmetries of the problem
for example the functions dont depend on y in this case
so the easiest way to do this
would be to work out the area between 1-x^2 and x^2
then multiply by the difference in y
oh I missed the planes, but I'm sure you can do something similar
graceful moss
thanks
cedar kilnBOT
@graceful moss Has your question been resolved?
trim vale
Hello this is what i have so far but i dont know the next step in order to solve this problem
trim vale
livid hound
simplify / cancel common factor
snow zodiac
hey flame there's a mistake in your work
livid hound
also ^
trim vale
where?
livid hound
missing ()
snow zodiac
exactly
livid hound
(a+b)(a-b) = a^2 - b^2
the - sign applies to the whole of b^2
not just whatever term appears first in the expansion
$-\sqrt{4-x} \cdot \sqrt{4-x} = -(4-x) \redneq -4 -x$
trim vale
slightly lost as to what your talking about
wraith daggerBOT
βamonov
livid hound
do you agree that
$$-\sqrt{4-x} \cdot \sqrt{4-x} = -\br{\sqrt{4-x} \cdot \sqrt{4-x}}$$
wraith daggerBOT
βamonov
trim vale
no, shouldnt they cancel out the square root?
livid hound
this is just an application of the associative property
which allows you to multiply in any order you want
doing this first makes things clearer
similar to if you had
77 * 2 * 5
for efficiency you can first multiply the 2 by 5
= 77 * (2 * 5)
same idea here
trim vale
ah ok yea i understand
livid hound
and those roots when multiplied give 4-x
but note that the - is applied to the whole expression
$ = -\br{{=underbrace{\sqrt{4-x} \cdot \sqrt{4-x}}_{4-x}} = x(4-x)$
trim vale
So would it be like this?
livid hound
yes
but the whole expansion is actually unnecessary
as the use of conjugates is the application of a difference of two squares and you can just apply the identity for that
(a-b)(a+b) = a^2 - b^2
trim vale
Like this?
trim vale
i tried to use the (a-b)(a+b)=a^2 -b^2
but i dont know how to use it so im gonna try to simplify it with the whole expansion
livid hound
that's for the numerator
trim vale
ah
livid hound
there's no such thing present on the denominator
and remains as x(2 + sqrt(4-x))
technically the same, but that - isn't needed
trim vale
i messed up the sqrt of the numerator but i fixed it so that it gets the -x
but is it meant to look like this?
livid hound
yes, better/clearner if that - i circled isn't there
and have your root cover the whole thing you're rooting
instead of what appears to be just the 4
trim vale
sorry but that property i dont know how to use or how to continue working with it from there so i went back to the whole expansion and this is what i got
wraith daggerBOT
βamonov
trim vale
that would equal to (sqrt(this))^2 = this
livid hound
yes
and to do the full expansion you would've had to know how to expand
$$\red{2^2} - \blue{(\sqrt{4-x})^2}$$
those parts individually
wraith daggerBOT
βamonov
livid hound
the difference of two squares cuts out the need to even get those two other terms that as evident above end up cancelling out anyway
trim vale
livid hound
no
trim vale
trim vale
yea so (sqrt(4-x))^2 = 4-x no?
trim vale
my b
livid hound
the identity pretty much allows you to jump from \
$(a-b)(a+b)$ to $a^2 - b^2$ directly which in your case is
$$(2 -\sqrt{4-x})(2 +\sqrt{4-x}) = \red{4} - (\blue{4-x})$$
since:
$$\red{2^2 = 4}$$
$$\blue{\sqrt{(4-x})^2 = 4-x}$$
wraith daggerBOT
βamonov
trim vale
livid hound
fix stuff like
don't have a temporarily empty numerator
and make your radicals clearer
trim vale
but i have a 1 as my numerator
livid hound
yes, and you should write that 1 as you're cancelling
as just outright crossing that x out, there's temporarily nothing on there
cedar kilnBOT
@trim vale Has your question been resolved?
trim vale
ty for the help @livid hound
regal rain
Hi! I have been trying to figure out this problem on my homework for about 15 minutes and eventually just gave up. Do I just substitute x for a negative 3 minus a 3 or what
ornate hearth
twilit flame
yes is that !
regal rain
I got -78 when I did that and got it wrong still π
twilit flame
I got -78 when I did that and got it wrong still π
it's not a problem if it's negatif
wait i remake it on my book
no @regal rain it's 76
twilit flame
f(-3) - f(3) = 52 - (-24) = 52 + 24 = 76
f(-3) = 52
f(3) = -24
i'm sure about it
regal rain
oh okay thanks
twilit flame
oh okay thanks
you're welcome !
regal rain
f(-3) - f(3) = 52 - (-24) = 52 + 24 = 76
It was 78 and I got it wrong π
twilit flame
It was 78 and I got it wrong π
76 you meen no ?
ancient talon
cedar kilnBOT
ancient talon
what's the easiest way to do this problem?
the only thing I can think of is expressing the derivative of x and y wrt theta individually and divididing
seems like a lot of work tho and i wonder if theres a quicker way to do it?
actually ig its not that bad
.close
bold vine
Think about the possibilities of what happens when each person takes and returns a random counter from the bag
What can happen? What are the odds that each of those events happens?
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
Available help channel!
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
@crimson sedge Has your question been resolved?
frigid fjord
Help: the base of a solid is a circle of radius a and every plane section perpendicular to a diameter is a square. Find the volume of the solid.
frigid fjord
Diagram
Work so far
Idk where to move on from here
The answer is (16/3)a^3 but idk why
frigid fjord
The cross-section is a square
flint hemlock
tf
frigid fjord
So I think it's just 2a
frigid fjord
Yeah okay, thank you!
nova snow
cedar kilnBOT
nova snow
Yeah that helps thanks a lot
flint hemlock
np[
cedar kilnBOT
@nova snow Has your question been resolved?
glossy escarp
For this question am I just crossing/taking out the Log?
cedar kilnBOT
For this question am I just crossing/taking out the Log?
vernal scarab
glossy escarp
Oh
surreal cave
have you considered $\log_{3}(3)=1$?
wraith daggerBOT
XxMrFancyu2xX
glossy escarp
have you considered $\log_{3}(3)=1$?
How did you figure out that 1 is log3 3
surreal cave
because one is equal to log base three of three
glossy escarp
So u did this?
surreal cave
pretty much and $\frac{\ln(3)}{\ln(3)}=\log_{3}(3)$ by the change of base formula
wraith daggerBOT
XxMrFancyu2xX
surreal cave
but my intuition was that $\log_{b}(b)=1$
wraith daggerBOT
XxMrFancyu2xX
surreal cave
and hence $\log_{3}(3)=1$
wraith daggerBOT
XxMrFancyu2xX
glossy escarp
and hence $\log_{3}(3)=1$
Was it because of the two logs in the equation?
That's why it was 1
In the quo
Question
surreal cave
?
I wanted ot convert 1 into the form of a logarithm with base 3
because all the others in the question have base 3
cedar kilnBOT
@glossy escarp Has your question been resolved?
tranquil star
S
tranquil star
Angular features
I got a question again
I have to calculate without my calculator. Sin30* +Cos60*
the star is meant as degrees
How do i calculate it
Do you mean this
obsidian coral
Does that say 1/1?
obsidian coral
What does 1/1 simplify to?
tranquil star
To 1
obsidian coral
Yes
obsidian coral
,rotate
wraith daggerBOT
tranquil star
So4x45?
obsidian coral
No
Use the table to find sin45 and cos45
Replace sin45 and cos45 with that value
Then do the math
tranquil star
?
obsidian coral
Replace sin45 with that value, in that expression you have
Then do the same for cos45
tranquil star
Yes
obsidian coral
What happened to the 4 and 2?
tranquil star
I dont know where to put em
obsidian coral
Replace sin45 with that value, in that expression you have
Then do the same for cos45
obsidian coral
As what?
obsidian coral
You wrote $3\sqrt{2}$, what is that suppose to mean?
tranquil star
I dont even know
wraith daggerBOT
dldh06
obsidian coral
Where did that number even come from?
tranquil star
I calculated it on calculator tried to figure it out
obsidian coral
4sin45 + 2cos45
That's the expression you have, let's start with 4sin45. What does sin45 equal to?
tranquil star
To this
obsidian coral
4sin45
Replace sin45 with that value, what do you have now?
tranquil star
What i dont get here is replace where do i put the 4
Oh so muliply it
Ohh thanks
So basicly
I hope this is right cuz if not um stupid
obsidian coral
I hope this is right cuz if not um stupid
You only did 4sin45
What about the 2cos45?
obsidian coral
Yes
obsidian coral
ctg = 1/tan
tranquil star
Alr good to know
tranquil star
So that means i multiply them or
obsidian coral
So that means i multiply them or
Yes
tranquil star
Oh okay
So lemme try an we will see
Do i have to put em all 3 to same number
Or no
Cuz its multiplying
obsidian coral
Yes
tranquil star
Lemme see if i have anything more i dont know
accurately calculate the length of the remaining sides in the 90degree triangle. a=20cm alpha=30*
How do i do that
Hell is this
obsidian coral
accurately calculate the length of the remaining sides in the 90degree triangle....
Draw your diagram
obsidian coral
Then use SOH CAH TOA
tranquil star
Yes but i dont know what are the lenghts of sides
obsidian coral
That's what the question wants you to do, find the missing sides using SOH CAH TOA
tranquil star
Ye but how to find them with SOH CAH TOA if i cant calculate without hypotenuse
obsidian coral
What does your diagram look like?
obsidian coral
Use SOH CAH TOA, with the angle and side you have to form the equations you need for the other sides
tranquil star
What do you mean form the equations how do i form them
jagged viper
cedar kilnBOT
jagged viper
Is anything Iβve done correct
cedar kilnBOT
@jagged viper Has your question been resolved?
jagged viper
<@647885227364188161> Has your question been resolved?
β
final ore
If I have a tournament that has 10 distinct tables (numbered 1-10) and 20 players and 3 rounds. Is it possible for each player to always play at a different table?
cobalt pier
π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§βπ¦±
π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦
π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§βπ¦±
Just move each person to the table to their right for each round and if they're at the end they move in fronbt of them
oh nvm to always play at a different table, mb
gentle flower
very interesting visuals
dull oxide
π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§β𦱠π§βπ¦±
π¦ π¦ π¦ π¦ π¦ π¦ οΏ½...
I thought this was spam at a glance. Nice move
cobalt pier
maybe they can just move 2 in a rotational motion instead
final ore
Oh sorry. Also a player cannot play the same player as we use strength of schedule
cobalt pier
if they just move 2 spots it would work right?
dull oxide
Just move each person to the table to their right for each round and if they're ...
This is pretty close. A half-answer almost
cobalt pier
yeah i realize it was wrong bc he said for each round to be at a different table
final ore
(This is a real life application. We play Star Wars legion in 20 man tournaments with 3 rounds)
dull oxide
This is pretty close. A half-answer almost
Wait actually, why is the wrong? If they move, are they not always at a different table?
cobalt pier
if you're at the end, you move to the one in front and are at the same table
dull oxide
if you're at the end, you move to the one in front and are at the same table
same table of what?
cobalt pier
the same table you were on the other side of originally
dull oxide
How were you there originally? There's 3 rounds max
final ore
If I have a tournament that has 10 distinct tables (numbered 1-10) and 20 players and 3 rounds. Is it possible for each player to always play at a different table?
Also each round you play a different opponent
dull oxide
cobalt pier
π’ β
π¦ π¦
π΄ π€
the red is at the same table
the question mark is bc i didnt want to show the entire thing so it's whoever was to the left of that spot
final ore
the question mark is bc i didnt want to show the entire thing so it's whoever wa...
Does this work if you must play different players?
cobalt pier
the example where you move once, no
but i thought of something else
if you move the sides individually, with one side moving to the left once per round and the other moving to their left one per round, it should work
final ore
I guess you can have red player move up a table and blue player down a table
final ore
At one of our 2 stores yes
cobalt pier
1οΈβ£ 2οΈβ£ 3οΈβ£ 4οΈβ£ 5οΈβ£ 6οΈβ£ 7οΈβ£ 8οΈβ£ 9οΈβ£ π
π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦
π€ π’ π΄ β« π΅ π π£ βͺ π‘ βοΈ
and the next round it'd be:
π 1οΈβ£ 2οΈβ£ 3οΈβ£ 4οΈβ£ 5οΈβ£ 6οΈβ£ 7οΈβ£ 8οΈβ£ 9οΈβ£
π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦ π¦
π’ π΄ β« π΅ π π£ βͺ π‘ βοΈ π€
dull oxide
Players inside the circle move once clockwise, players outside the circle move once counterclockwise
dull oxide
Nah this is the best way
cobalt pier
like if you dont actually care about the table but just the opponent, you can move one side over once per round and it's fine
dull oxide
You'd get a repeat after 5 rounds
final ore
Ok. I am really sorry I am bad at asking questions!!!!
cobalt pier
it's fine
do you care about them being at the same table or just playing against a different person
final ore
If I have a tournament that has 10 distinct tables (numbered 1-10) and 20 players and 3 rounds. Is it possible for each player to always play at a different table?
Also each round you play a different opponent
Winners play winners and losers play losers
I am not asking for you to solve how to do it
I am just asking is there a way I can find out if itβs possible
Like a formula or something
cobalt pier
not sure about a formula
but i think it's impossible for winners to always play winners and losers to always play losers
or if the first round doesnt affect the results of the second, you'll have 10 losers and 10 winners for rounds 1 and 2 and you can match them up
final ore
A competition looks like three rounds and 20 players will try to pair up skilled (folks who win) players with each other and avoid players playing the same table
cobalt pier
ok, let's say that round 1 you have 10 winners and 10 losers
you pair those 10 losers against each other, 5 to 5
if one of the round 1 losers wins round 2, do they go against someone who won both round 1 and round 2?
cobalt pier
ok then it's fine
you can just match them up 5 to 5 each round
and they can easily have different opponents each time
you dont even have to worry about different opponents until round 3 because for round 2 it'd be impossible to go against your round 1 opponent
final ore
Basically, with my constraint of performance game matching and 3 rounds can I guarantee no one will play the same table?
cobalt pier
i believe so, yes
obsidian coral
Google them
You can still google them
Something like "trig identities"
Yes that's fine
Then you can combine the numerator into one expression using common denominator
Not quite
If you have 1/2 - 1/4, could you find the common denominator?
Multiply what by 2?
Close, you have 1/cos(t) - cos(t)
What's the common denominator?
No
This is what you have in the numerator, at the moment, you want to combine them
1/cos(t) - cos(t)
Also it's $\frac{1}{cos(t)} - cos(t)$
wraith daggerBOT
dldh06
obsidian coral
No
Just focus on 1/cos(t) - cos(t)
What is the common denominator?
Good, can you combine those two terms into one expression?
Don't forget the denominator
Not that denominator
obsidian coral
**dldh06**
The one when you combined this
So what would that expression be?
If you typing the expression in text form, you need to add parentheses to separate the numerator and denominator
Can you see that you can simplify that numerator?
Use that, and look closely at the pythagorean identities
Yes, exactly
Yes
Now, do you know that division saying for fractions, "keep, change, flip"?
So you keep the numerator, you change the division to multiplication, and you flip the denominator
Almost there
What's sin/cos equal to?
Yes
So that simplifies to?
That referring to this
Yes
I was saying you can simplify the sin/cos here, and what it is equivalent to
And that's it, because the question ask to simplify to a single trig function, and tan(t) is a single trig function, right?
Are there any other trig functions that you see with tan(t)?
Looks good
You're primarily using these
And these I believe
If you know tan = sin/cos, cot is just 1/tan so then 1/tan = 1/(sin/cos) which is cot = cos/sin
sin^2(x) + cos^2(x) = 1, if you just know that one, you can get the other ones, by either dividing by sin^2 or cos^2
You should know the reciprocals, those aren't too hard to remember
Almost, if you divided by 8 on both sides, you forgot about the 3
Make $cos(2\beta)$ to be in terms of sin
wraith daggerBOT
dldh06
obsidian coral
If you divided by 8 on both sides, you divide all the terms by 8, including the 3
Yes
dire geode
,tex .double angle
wraith daggerBOT
riemann
obsidian coral
You have an expression with both cos and sin existing. It's easier when you just have one trig function involved
So you need to use those identities above, to make $cos(2\beta)$ be in terms of sin
wraith daggerBOT
dldh06
obsidian coral
Yes
Now you can group all the terms with the beta to one side and everything to the other
Not exactly
You don't divided by sin^2
It's getting added, so you do the opposite of that
If you had 1 - 2x^2 = x^2 + 3/8, how would you group all the variable terms to one side and everything else on the other?
I meant on the right side it's +sin^2, it's added on that side
Not that you add it to bring it to the other
Look at this
1 - 2x^2 = x^2 + 3/8
How would you bring the x^2, on the right, to the left side?
Yes
Same process with
How would you bring sin^2, on the right, to the left side?
dire geode
Look at this
> 1 - 2x^2 = x^2 + 3/8
@inland blade solve this first
dire geode
How would you bring the x^2, on the right, to the left side?
And answer this
obsidian coral
What's -2x^2 - x^2?
No
-2x^2 - x^2 = -2x^2 + -x^2
Does that help?
Yes
Continue to isolate for sin^2
Then how would you solve for beta?
Not quite, a hint would be inverse trig functions
Yes
From there you should be able to figure out the rest of the answer
You're suppose to
Notice how it says accurate to at least 2 decimal points, that means a calculator should be used
crimson sedge
cedar kilnBOT
crimson sedge
How can i start this
rain vine
what's the question about?
stray epoch
Solving that π
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!