what's the exact wording of the quesiton
#help-13
vocal raptor
Bro it's on Georgian and Im translating
They ask for CD
Answer is 4π/3
And the area of full ABC will be 4π since its quarter of full circle
livid hound
ok, so this isn't about area at all
they want the length of this circular arc marked in red
livid hound
in which case, you'd want to first determine <DAC using trig
followed by applying the formula for arc length
vocal raptor
How to determine <DAC
with a cos?
Actually it's not necessary
sin<DAC is enough
Which makes it 60
crimson sedge
Answer is 4π/3
wait am i tripping or does this not look right
livid hound
diagrams aren't to scale
vocal raptor
<DAC = 60, what now Ramonov
vocal raptor
wait am i tripping or does this not look right
Idk, that's answer
crimson sedge
4pi/3 would be the answer if <DAC = pi/6 no
livid hound
no
sin<DAC is enough
Which makes it 60
assuming you meant that <DAC is 60°
that's fine
vocal raptor
Yh cause sin<DAC = 2√3/4 which is √3/2 which is sin60
livid hound
ideally you'd want the angle in radians
vocal raptor
well, how do u get that answer
crimson sedge
omg mbmb i thought it was sector area
livid hound
ideally you'd want the angle in radians
crimson sedge
sleep deprivation 
livid hound
can you convert 60° to radians
vocal raptor
its π/3
cedar kilnBOT
cedar kilnBOT
Available help channel!
pulsar igloo
help
pulsar igloo
i know III is false but i forgot how you would explain that
small dust
bc the max slope it has is 3 so (5,6) to (7,y)
y can be at most 6 more than what it was and 6+6<13
pulsar igloo
why it can be at most 6 more?
small dust
Bc the derivative is 3 so for every +1 in x u get +3 in y
so do that twice
3+3 is 6
small dust
yeah derivative is slope
pulsar igloo
ah yea
small dust
Np
pulsar igloo
.close
sudden willow
How do you do question 7?
cedar kilnBOT
How do you do question 7?
crimson sedge
no but like, that equality only holds when theta = pi/2 seemingly
livid shell
How do you do question 7?
Try to use some trig identities
hasty fulcrum
no but like, that equality only holds when theta = pi/2 seemingly
but
we cant even have tan(pi/2)
crimson sedge
Oh I see okay my graphing software didn't graph this correctly, I just did it on desmos
hasty fulcrum
How do you do question 7?
are ya still stuck
hasty fulcrum
okay
so sec - tan
write that in terms of sin and cos
thats usually a good place to start
what do ya get
sudden willow
I started from the left side and got 1-sin/root cos
But it’s meant to not the have root
hasty fulcrum
right
you shouldve had a cos^2 at some point
square root of cos^2 is cos
hence you get what you wrote
vast cobalt
The area of a sector of a circle with a central angle of 140° is
67 m2.
Find the radius of the circle.
hasty fulcrum
Something about circular sectgor?
u know the formula for sector?
or circular sector
vast cobalt
nay but I can google
hasty fulcrum
$\frac12\theta r^2$
wraith daggerBOT
blanket
hasty fulcrum
vast cobalt
ooo this one
vast cobalt
r= the square root of theta?
south tundra
**blanket**
Before using this make sure theta is in radians
hasty fulcrum
Before using this make sure theta is in radians
oh shit right
convert 140 deg to radians
vast cobalt
im confused 😦
hasty fulcrum
not 1/2
vast cobalt
I could have sworn Ive done it before with 1/2
quiet plover
hope this helps (:
quiet plover
so theta is 7pie/;9 radians
yes
vast cobalt
67 = 7pie/9 rad over 2 pie times n squared?
quiet plover
wait lemme type it and show
vast cobalt
I remember this being so easy yet so hard to explain
quiet plover
$67 = {\frac{\frac{7\pi}{9}}{2\pi}}\pi r^2\$
wraith daggerBOT
quiet plover
where $\theta$ is $\frac{7\pi}{9}$
wraith daggerBOT
quiet plover
you just need to know that the area of the circle is proportional to the angle
for eg, if theta is pi radians, the sector is half the area of the circle, or (pi)r^2/2 and for pi/2 radians, it is (pi)r^2/4
subtle peak
Hello
dusk finch
,rotate
wraith daggerBOT
prisma gull
Apply ln both sides?
eternal crag
Have you used natural log?
subtle peak
Is given that the equation is lambert
But it given like this
I used ln but i didnt get anywhere
dusk finch
I mean the original problem is definition of product log, aka lambert W function
If you use log, and apply few properties you should get to $\ln\left(y\right)+y=\ln\left(x\right)$
wraith daggerBOT
MathIsAlwaysRight
dusk finch
where y=f(x)
subtle peak
I want to prove that the range is from [0,+infinity)
eternal crag
I want to prove that the range is from [0,+infinity)
Sorry for awful writing but these are all you need
subtle peak
Thanku, now do u know from this how do i prove that the range is from [0,+infinity)?
eternal crag
Thanku, now do u know from this how do i prove that the range is from [0,+infini...
You need to calculate f(x) first. It's easy to prove it's range after that.
subtle peak
Idk how to do this
eternal crag
I think no. Where you got f(x) + f(x) = Ln x
eternal crag
I said f(x)=ln(x/y)
That's not true.
f(x) = ln(x/f(x)) not that
Remember e^f(x) = x/f(x) meaning natural log of x/f(x) is f(x)
cedar kilnBOT
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autumn notch
Hey, I would like some help getting to the solution to this problem: A spam filter is designed by looking at commonly occurring phrases in spam. Suppose
that 80% of email is spam. In 10% of the spam emails, the phrase "free money" is used,
whereas this phrase is only used in 1% of non-spam emails. A new email has just arrived,
which does mention \free money". What is the probability that it is spam?
raw elm
6 + 9
autumn notch
<@&268886789983436800> is this not trolling?
autumn notch
yeah idk
raw elm
sry for that
autumn notch
i get that i need to use bayes rule and rule of total probability
dusk finch
Hey, I would like some help getting to the solution to this problem: A spam filt...
Suppose B means that "free money" is in email, and A means that the email is spam.
Now you know P(A)=0.8, P(B | A)=0.1, P(B | A')=0.01 and you need to know P(A | B)
A' means not A
autumn notch
yeah
dusk finch
at this point you can just plug it in the second equation
autumn notch
ah i see
wintry oyster
I was reading https://math.stackexchange.com/questions/37398/what-is-the-geometric-interpretation-of-the-transpose
and came across this
Is this true generally? I tried to see but for some reason it only works out as the transpose in orthonormal basis
$$
\langle Au, v \rangle = A^i_j u^j v^k G_{ik}
$$
Where A is a linear operator and G is the metric tensor
In an orthonormal basis $G_{ik} = \delta_{ik}$
So
$$
\langle Au, v \rangle = A^i_j u^j v^i = u^j (A^T)^j_i v^i = \langle u, A^Tv \rangle
$$
Does this mean transpose is only meaningful in an orthonormal basis? But the determinant identity $det(A^T) = det(A)$ holds regardless of the basis so it's gotta have some more meaning to it right
wraith daggerBOT
Frisk17
wintry oyster
Another thing I noticed was for a linear operator to preserve the inner product
$$
\langle A(u), A(v) \rangle = (Au)^T G Av = u^T G v
$$
Which also becomes in an orthonormal basis
$$
u^T A^T A v = u^T v \forall u, v \in V
\implies A^T A = A A^T= I
$$
wraith daggerBOT
Frisk17
wintry oyster
this again raises the question if the transpose is meaningful in a non-orthonormal basis
cedar kilnBOT
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rancid jacinth
I need help with part C
rancid jacinth
<@&286206848099549185>
cedar kilnBOT
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wispy cave
Car A and Car B starts at the same position, car A goes first, after some time, car B starts to catch up with car A, car A is n kilometres away from car B when car B left. We know the time it takes for car B to go and catch up with Car A and Car A's speed, Calculate car B's speed
(All speed is in km/hr and time is in hrs)
cedar kilnBOT
Car A and Car B starts at the same position, car A goes first, after some time, ...
@wispy cave Has your question been resolved?
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next glen
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birdy
crimson sedge
no
why would it be
x = 0
x isnt an element of the first set but is an element of the 2nd
-5 < -1
11+6 = 5
where did all the infinities go?
x^2 + 11 < -6
x^2 + 5 < 0
velvet mortar
Isn't intersection the overlap and union includes all?
crystal raptor
at one point you timesed both sides of the equation by x-2
you need to know the sign of x-2
in this case you assumed the fraction was negative
the numerator is always positive
so the denom must've been negative
so all your inequalities flip when you times by it
?
no?
what do i mean by what
because x^2+3x+11 is positive for all x
so if you assumed the fraction was negative it mustve been caused by the denom
?
you assumed a/b < 0
but a > 0
so it must be that b < 0
thats pretty much a separate thing
(-a/b) = -a/b = a/-b
but it doesnt matter which you didnt it wouldnt have made a difference
crystal raptor
you assumed a/b < 0
this is what matters
velvet mortar
Inequalities and absolute values are crazy
crystal raptor
you assumed a/b < 0 to be in the "negative" case on the right
i dont know what youre asking
and im telling you the direction of the inequality depends on the sign of x-2, because you timesed both sides by it, timesing both sides of an inequality by a negative flips it
look everything you did after timesing by x-2 would be perfectly fine if x-2 > 0
but it's not
for reasons i have repeated
no worries
cedar kilnBOT
@leaden shore Has your question been resolved?
crystal raptor
?
oh are you working it out again with the fix
yes
but now -x+2 is positive because its -(x-2)
well they cant both be correct
because one is the negative of the other
this line was fine
i was just telling you that by assumption -x+2 is positive because as we discussed earlier x-2 is negative
hence at this point when you times both sides by -x+2, the inequality does not flip
i am practically telling you what to do
crystal raptor
carry on from here
times both sides by -x+2, which is positive for reasons discussed many a time
yes
yes
how are you getting the same result?
what have you done?
you know that cant be right. -10 <= 0 and -2 <= 0 but (-10)*(-2) = 20 > 0
the only way to get something negative is it only one of the brackets is negative
crystal raptor
how are you getting the negative on the left hand region
when x <= 5
Within which range? The middle one?
Your inequality is <=
Less than OR equal to
cedar kilnBOT
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karmic epoch
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grave axle
i need help with a tangent line equation question
cedar kilnBOT
i need help with a tangent line equation question
grave axle
i have the solution on this practice but after plugging in the x im not sure whats happening
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grave axle
help
lethal jackal
you mean you're not sure what's happening after this step?
lethal jackal
okay
so do you have an idea of what it means for a line to be tangent to a graph at a given point x?
grave axle
yes
lethal jackal
what two properties does it have?
grave axle
just the slope of that specific point right?
grave axle
what two properties does it have?
oh ig im not sure
lethal jackal
but you could draw a tangent line to some curve I gave you right?
just like sketch it?
grave axle
i think so yeah
lethal jackal
does it make sense that the line tangent to a graph at the point (x, f(x)) would have the same slope as the "slope" of the f at the point x and it would also pass through the point?
basically it has the same slope and passes through the point?
lethal jackal
yeah it's okay, the whole point is to get you to visualize it
so naturally, you might expect that mathematically, we might say something like:
A line is tangent to the graph of f(x) at x_0 if:
- the line passes through (x_0, f(x_0))
- the line's slope = f'(x_0)
right?
grave axle
right so it has to pass through both
lethal jackal
no, it has to have the given slope
and has to pass through a given point
but it has to satisfy both conditions yes
grave axle
ohhh okok
lethal jackal
now consider the line given by $y = f'(x_0)(x-x_0) + f(x_0)$
wraith daggerBOT
Saccharine
lethal jackal
what's the slope of that line?
grave axle
to be honest im not really sure what x_0 means entirely
so thats probably one reason im confused
lethal jackal
x_0 is basically just a letter
let's say I replace it with k
$y = f'(k)(x-k) + f(k)$
wraith daggerBOT
Saccharine
lethal jackal
not quite
grave axle
oh wait k would cancel k
lethal jackal
$y = f'(k)(k-k) + f(k)$ right?
wraith daggerBOT
Saccharine
wraith daggerBOT
Saccharine
grave axle
yeah but why do we try to make that middle part 0
now i realize since in the original problem x was 2 we did x-2 and that remindes me of this now
lethal jackal
well, I'm getting there
grave axle
okok
lethal jackal
so it would be fair to say that the point (k, f(k)) is on that line right?
grave axle
because we have it to equal f(k) ?
lethal jackal
no, because if you plug in x=k to the equation of the line, you get y=f(k)
like let's say I have a line y = 3x+2
the point (2, 8) is on that line, because if I plug in 2 for x, I get y = 3(2) + 2 = 8
grave axle
oh so the plugged in value and also the solved value together make the point on a line
lethal jackal
if I have $y = f'(k)(x-k) + f(k)$, the point (k, f(k)) is on it for basically the same reason
wraith daggerBOT
Saccharine
lethal jackal
oh so the plugged in value and also the solved value together make the point on ...
basically
grave axle
ok i see that now thank you
lethal jackal
okay so remember when I said this:
so naturally, you might expect that mathematically, we might say something like:
A line is tangent to the graph of f(x) at k if:
- the line passes through (k f(k))
- the line's slope = f'(k)
would you agree that the line $y = f'(k)(x-k) + f(k)$ is tangent to the graph of f(x) at $x = k$?
wraith daggerBOT
Saccharine
grave axle
yes
lethal jackal
and your problem gives you an f and asks you to find the equation of a tangent line at x = 2, right?
grave axle
yes
lethal jackal
so naturally, if we plug in k = 2, the statement becomes that the line $y = f'(2)(x-2) + f(2)$ is tangent to the graph of f(x) at $x=2$, right?
wraith daggerBOT
Saccharine
grave axle
yes
lethal jackal
would you agree that the steps shown here
and here
are simply calculating f(2) and f'(2)?
grave axle
i understand the first half but the 8(-1/2)(3x-2)^2 . 3 confuses me
i dont really know how he got -1/2
also the 4 i got by plugging x does that mean there is a point (4,2) somewhere?
wraith daggerBOT
lethal jackal
if I told you that f'(2) = -3/2 and f(2) = 4
could you find the tangent line of f at x=2?
lethal jackal
**Saccharine**
you agreed with this statement right?
grave axle
yeah i think i know what that means
lethal jackal
so if I plug in the values for f'(2) and f(2)
the statement becomes the line $y = \frac{-3}{2} (x-2) + 4$ is tangent to the graph of f at x = 2
wraith daggerBOT
Saccharine
grave axle
ooooo so then it would be -3/2(x-2)+4
okok
i was so slow at typing that
but yeah it acc makes sense now i think i get f' confused with f
lethal jackal
and you can simplify that to y = -3x/2 + 7
so really, the only questions that remain are how did I calculate f'(2) and f(2), right?
grave axle
yeah
lethal jackal
so you can calculate f(2) by just plugging 2 in for x
$f(2) = \frac{8}{\sqrt{3(2) - 2}} = 4$
wraith daggerBOT
Saccharine
grave axle
which is the very first step when you get 4
grave axle
yeah this is honestly my main confusion
lethal jackal
it actually makes sense to figure out f'(x) for all x and then just plug in 2
so we need to take the derivative of f
i.e. we need to take the derivative of 8/sqrt(3x -2) with respect to x, right?
grave axle
yeah but how
lethal jackal
we'll get there
but all of this framing information is important
because it gives you a path for solving the problem as given
as for actually finding the derivative, that's the easy part if you remember a few rules
we can write $\frac{8}{\sqrt{3x-2}} = 8(3x-2)^{-1/2}$ right?
wraith daggerBOT
Saccharine
grave axle
yeah
lethal jackal
have you heard about the power rule and the chain rule?
lethal jackal
should be in your notes somewhere, but I'll give you a very abbreviated explanation:
the power rule states that if you have something like x^n, the derivative of that is nx^(n-1)
the chain rule says that if you need to differentiate f(g(x)), you get f'(g(x))g'(x)
you should review this later, and you should ponder why this means the following statement
but I'm going to skip over that
grave axle
ill do the proofs for those later yes
lethal jackal
in this context, it means that if you have (something)^n, the derivative of that is n(something)^(n-1) x (derivative of the something)
in this case, the something is 3x-2
so we write the derivative as $8 \times \frac{-1}{2} \times (3x-2)^{-3/2} \times 3$
wraith daggerBOT
Saccharine
lethal jackal
where the 8 just gets carried through (this is because you can factor constants out of derivatives)
n = -1/2
the something is (3x-2)
n-1 = -3/2
and the 3 is the derivative of the something
grave axle
ahhhh so chain rule is basically since its in a bracket all then it works as a whole thing kinda i get it
that was a horrible explanation of my thoughts but yeah this makes a lot of sense now
lethal jackal
the chain rule is basically what makes the extra x 3
and then if you plug in x = 2, stuff the thing through a calculator
you get -3/2 I hope
grave axle
i see so first we got the normal solution then we got the derrivative solution which is the slope of the tangent line
the normal solution helped us get the y intercept?
lethal jackal
so f(2) tells you that (2, f(2)) is a point on the line
f'(2) tells you the slope of the line
the equation of the line that I wrote out before is just the point-slope form of the line, given these two things
grave axle
i seeee
lethal jackal
in general, if you know a point on a line and the slope of the line, you can find the equation of the line
but I sorta tried to explain it a little backwards, where I gave you an equation of a line and guided you through verifying that this line is indeed tangent to the curve
grave axle
yeah thank you so much, probably gonna study chainrule and read over everything a little
i want to do computer science too someday
lethal jackal
I regret it lol
grave axle
oop
lethal jackal
okay but if you apply the formula to that situation
it does indeed come up as 0.553
so why don't you think that it works for the given problem?
,w calculate (3.5 - 1.2 * 0.495 - 2.5) / (3.5+1.2+2.5) * 9.8
wraith daggerBOT
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glossy grail
i dont understand how to get from here
cedar kilnBOT
i dont understand how to get from here
glossy grail
to here
short blade
looks like they multiplied by the conjugate but im not 100% sure
that is, they multiplied by $\frac{2\sqrt{x^8+x^6}+2x^4+x^2}{2\sqrt{x^8+x^6}+2x^4+x^2}$
wraith daggerBOT
maximo
glossy grail
oh ok makes sense
but what is exactly the conjugate? its not the same thing but all positive right?
short blade
if you have a+sqrtb, the conjugate is a-sqrtb
(or -a + sqrtb)
the idea is that
$(a+\sqrt{b})(a-\sqrt{b})=a^2 - (\sqrt{b})^2 = a^2 - b$
wraith daggerBOT
maximo
short blade
it "gets rid" of the square roots
civic coral
Can someone put those x+2y+3z=0 and x-z=0 and -2x+4y-2z=0 into calculator and find the (x,y,z)
civic coral
What I got is (0,0,0)
latent bloom
Yep indeed
Doesn't need a calculator
Solve using x-z = 0 then plug in and keep solving
Or you can go for matrix simplifying
cedar kilnBOT
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flint plinth
clearly (0,0,0) is a solution, but the more interesting fact is that it's the only solution.. to see that, you need to show that the three equations are linearly independent
scarlet ginkgo
cedar kilnBOT
scarlet ginkgo
I don't understand why the Manila - Honolulu edge is present in the Pseudograph, but not in the Simple Graph
The instructions are the same, aren't they?
If so, the edge must be present (or absent) in both diagrams.
cedar kilnBOT
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scarlet ginkgo
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scarlet ginkgo
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lethal jackal
probably a typo I think
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charred horizon
Guys I have a question : On a sphere , if four random points are assumed and joined then a three dimensional figure will be formed . What is the probability of the center of the sphere coming in the figure. (I know the answer but I want a mathematical proof and how should i write it)
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crimson sedge
How do i find number of solutions of the equation a+b+c=11 when 0<a,b,c <7
crimson sedge
And a,b,c are natural nos
crimson sedge
How do i find number of solutions of the equation a+b+c=11 when 0<a,b,c <7
Hint: Use binomial theorem for rational index
crimson sedge
Hint: Use binomial theorem for rational index
We have not studied binomial theorm yet
Any alternate methods beside making the cases?
crimson sedge
We have not studied binomial theorm yet
*edit
These are the total natural solutions that i found
But i have one more condition too
A,b,c should be less than 7
If i somehow find the number of cases where a,b,c are more than 6, i will be able to subtract it from these total cases.
crimson sedge
If i somehow find the number of cases where a,b,c are more than 6, i will be abl...
But i am not sure how should i do that
I thought to start with a,b,c being any whole numbers.
We want their sum as 11.
Consider this case equivalent to 11 identical balls laid down in a straight line.
Introduce two identical flags in between the balls, to partition them into 3 groups.
Total entities are now 13.
Arranging all the entities now will be equivalent to the number of solutions a,b,c belonging to the whole number set.
Solutions are 13!/(11!2!) (Use arrangement of like objects), which is basically 13C2 solutions.
Now impose the condition of 0<a,b,c<7, and exclude cases.
7+3+1 (3! arrangements)
8+2+1 (3! arrangements)
9+1+1 (3!/2! = 3 arrangements)
These three will be excluded as one of the numbers is >7
Now fix one number as 0.
We now need x+y=11.
Again think of 11 identical balls laid down in a line. Now we have to only introduce 1 flag to partition the balls into 2 groups. Total entities are now 12. Arrangements are 12!/11! = 12. Now we have to multiply by 3 too as we have in total 3 such cases where one of the numbers is 0.
A case with two zeroes is not possible.
Hence the answer is 13C2 - 3! - 3! - 3 - 12×3 = 13C2 - 51 = 27
crimson sedge
I thought to start with a,b,c being any whole numbers.
We want their sum as 11.
...
But the given answer is 27
All the numbers are greater than 0 here
So i dont think that we can fix one number 0
We can also think it of like 3 dice
A,b,c
Whose sum is 11
crimson sedge
So i dont think that we can fix one number 0
I did that to exclude such cases
crimson sedge
Did you try this out
Try how?
Hmm but thinking of them as dice would not really help out, cuz cases would still arise
crimson sedge
If i somehow find the number of cases where a,b,c are more than 6, i will be abl...
What about this tho
crimson sedge
I thought to start with a,b,c being any whole numbers.
We want their sum as 11.
...
Only methods I could think of is by binomial or this
Checking
crimson sedge
Only methods I could think of is by binomial or this
Oh hmm
crimson sedge
These are the total natural solutions that i found
Total is now 10C2. Now find cases where a', b', c'≥6 (because a = a' + 1, same for b and c, which are less than 7).
6 + 1 + 1 (3!/2! = 3 arrangements)
7 + 1 + 0 (3! arrangements)
8 + 0 + 0 (3!/2! = 3 arrangements)
6 + 2 + 0 (3! arrangements)
Answer: 10C2 - 3! - 3! - 3 - 3 = 45 - 18 = 27
crimson sedge
Total is now 10C2. Now find cases where a', b', c'≥6 (because a = a' + 1, same f...
Ooooohhh got ittttt
So we just had to remove 6,7,8 and their position shifts
worn prism
can I propose using the ball analogy again
dividing 11 balls into 3 piles, with at least 1 ball in each pile
the first 3 balls have to go in separate piles, so they each have 1 way of placement (3 ways)
(oh maybe it's 3+2+1)
crimson sedge
can I propose using the ball analogy again
Sure
worn prism
the remaining 8 can go into any of the 3, and since all these events cannot occur simulatneously, just need to add 8x3 to the 3 ways before
wait there's an upper restriction
crimson sedge
Mhm
worn prism
crimson sedge
crimson sedge
Total is now 10C2. Now find cases where a', b', c'≥6 (because a = a' + 1, same f...
But we can do this to deal with that
We call that method the "beggar method" here lol
First distributing 3 balls (like giving them to the beggar, satisfying the condition first) then doing the later distribution. The upper bound problem comes. If the upper bound wasn't there, for this question only, that method would've surely worked
crimson sedge
These are the total natural solutions that i found
What is did here was basically the ball analogy that you are talking about but reduced into a formula
crimson sedge
We call that method the "beggar method" here lol
First distributing 3 balls (lik...
We also call that beggar method here which was proposed by the father of coaching industries here
I see
crimson sedge
What is did here was basically the ball analogy that you are talking about but r...
Yes
crimson sedge
What is did here was basically the ball analogy that you are talking about but r...
Basically this if anyone is interested
This question is actually good for revising distribution of like objects with empty groups allowed
crimson temple
How do i find number of solutions of the equation a+b+c=11 when 0<a,b,c <7
Assuming (a,b,c) are unordered pairs here
for a,b,c < 7 you can just easily brute force.
Theres actually a way to find the number of unordered (a,b,c) to a+b+c=11 for a,b,c in the positive integers efficiently, (actually it works for all a+b+c=n)
Imagine spitting 11, with 2 dividers, then you have three sections. There are in total 10x9 ways to spilt 11 into 3 sections of length at least 1.
Then you separate them into classes of sets
S_3 where all a,b,c are the same
S_2 where at least 2 are of the same value
S_1 where a,b,c are distinct
You notice here 2( |S_3|+3|S_2|+6|S_1| )= 9x10
Then from here you notice |S_3|=0,
|S_2| = 5 because (1,1,9),(2,2,7),…,(5,5,1)
Then from here just sub and solve for |S_1|
With this, just add |S_1|+|S_2|+|S_3| are ur done
crimson sedge
Assuming (a,b,c) are unordered pairs here
for a,b,c < 7 you can just easily brut...
Very creative
crimson temple
You can use this to solve for a+b+c=10000 in like one minute lol, its very nice
crimson sedge
crimson sedge
Assuming (a,b,c) are unordered pairs here
for a,b,c < 7 you can just easily brut...
Damn that's neat
Ok
Ok so we found 4 methods today
- Beggar method
- knock off beggar method
3.Big brain method - Poor man's method
crimson sedge
In 4ABC, AB ∼= AC. Let D ∈
−→BA such that B − A − D. Let
P ∈ int CAD. If −→AP||BC then prove that −→AP bisect ∠CAD.
lusty grotto
do u have a picture
crimson sedge
No sorry
lusty grotto
this is quite unclear
lusty grotto
what is 4ABC
lusty grotto
i dont know what B - A - D means
lusty grotto
well have u drawn a picture with all this information
crimson sedge
No
lusty grotto
can u
lusty grotto
send it
cedar kilnBOT
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chrome vigil
Can someone explain how these expressions are equivalent, I’m having trouble figuring it out?
grand forge
Whats the function you were orginally trying to integrate / differentiate
cos^2x = 1 - sin^2x
chrome vigil
cos^3(x) was the function I integrated
chrome vigil
alr thx
cedar kilnBOT
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flint folio
cedar kilnBOT
flint folio
am I not expressing this properly? Is it okay to bring the 1/3 outside of the limit ? Is there a better way to have written this ? Thanks a lot.
mighty drift
you don't need to write it as a limit since you've most likely already defined improper integrals. Also then writing it the way you did, with a limit, an equal sign that shouldn't bet there, and then the result, is ugly at best, wrong at worst
also the integral for u doesn't start at 1
flint folio
also the integral for u doesn't start at 1
tHanks for you're response ! This last part rings a bell but I'm not sure how I would write it in the u world. What should 1 Be then and how can I know? Tysm
mighty drift
1 becomes u(1) = 4
flint folio
ohhh i see.
of course
So I could have solved without resubbing using this, right ?
mighty drift
so then you can just write [-1/3u]_4^inf = 1/3*4 = 1/12
flint folio
@flint folio
mighty drift
did you just ping yourself ?
flint folio
i cant see the channel on my other device !! Haha I thought the ping would help
Thank you, though.
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Available help channel!
devout steeple
Algebra question: Can someone please check this for me?
cedar kilnBOT
Algebra question: Can someone please check this for me?
mighty drift
yes
30Z = {3k, k in 10Z} and k in 10Z => 3k in 10Z
may I ask what class introduces the concept of ideal and doesn't write the answers in a concise manner for Q2 ?
devout steeple
Higher Algebra
mighty drift
what level is that ? First year of undergrad ?
mighty drift
never seen a group ring before
mighty drift
yes
you'll probably see pretty soon that aZ + bZ = gcd(a, b) Z
btw
which can also be used to define the gcd
it's usually how the gcd of polynomials is defined btw
crystal finch
I’m really confused on this
cedar kilnBOT
I’m really confused on this
crystal finch
I’m putting in the value in the calculator but it’s not getting the same value as my teacher
crystal raptor
what are you getting and what is your teacher getting
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crimson sedge
,rccw
wraith daggerBOT
cedar kilnBOT
@crystal finch Has your question been resolved?
plucky holly
cedar kilnBOT
plucky holly
does this question mean 10choose2?
inland fossil
thats the number of ways of permuting the ppl
not the number of ways of partitioning them
plucky holly
so how would i tackle this problem?
inland fossil
so how many ways are there to choose our first group consisting of 2 ppl
picked from 10
(uses choose)
plucky holly
so is it 10choose2 now?
inland fossil
yep
but ur not done yet
because the ordering of the pairs dont matter
let the groups be ABCDE
each with 2 ppl inside
it doesnt matter whether the groups are ordered ABCDE or DECAB or whatever
so we divide by the number of ways to permute the groups to account for overcounting
cedar kilnBOT
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torpid mountain
hello, I need some help figuring out how to do this, idk if im missing a formula here but im not sure how to solve
torpid mountain
its a homework quiz but like ok, just needed an explanation 😔
tired inlet
How would I expand this? 1/x (x+2)?
cedar kilnBOT
How would I expand this? 1/x (x+2)?
wary grotto
1/(x^2+2x)
red pumice
brackets 🥹
tired inlet
1/(x^2+2x)
?
wary grotto
assuming you meant 1/(x(x+2))
tired inlet
assuming you meant 1/(x(x+2))
Yes
tired inlet
But why is x in the bracket now?
wary grotto
because you distributed it
crimson sedge
$\frac{1}{x(x+2)}$
tired inlet
But isn’t it x^-1
wraith daggerBOT
hibyehibye
wary grotto
**hibyehibye**
is it not this?
wraith daggerBOT
hibyehibye
tired inlet
**hibyehibye**
Yes that is it
wary grotto
okay so 1 + 2/x
tired inlet
**hibyehibye**
Can I also write it like this?
red pumice
1/(x(x+2)) means the first
red pumice
Can I also write it like this?
not the same
tired inlet
Oh
crimson sedge
$\frac{1}{x} * (x+2) = \frac{x+2}{x} = \frac{x}{x} + \frac{2}{x} = 1 + \frac{2}{x}$
wraith daggerBOT
hibyehibye
tired inlet
But how would I do (1/x) * x?
crimson sedge
2/x
tired inlet
Oh so do I basically multiply the numerator by whatever it is being multiplied by?
wraith daggerBOT
hibyehibye
tired inlet
Alright thanks so the answer is 2/x + 1
And what about (2/x)(x^2+3x)?
Is it equal to 2x + 6
crimson sedge
$\frac{2}{x} * \frac{x^2+3x}{1}$
wraith daggerBOT
hibyehibye
crimson sedge
$\frac{2}{x} * \frac{x^2+3x}{1} = \frac{2x^2+6x}{x} = 2x+6$ where $x \not = 0$
wraith daggerBOT
hibyehibye
tired inlet
Ok thanks
And last question
(2+1/x)(x+3)
How am I supposed to do this
When I expand I get: 2x+6 + 1 + (3/x)
@crimson sedge
cedar kilnBOT
@tired inlet Has your question been resolved?
cedar kilnBOT
@tired inlet Has your question been resolved?
royal loom
@tired inlet if you still want help, let me know what your exact question is and I'll see what I can do
short agate
cedar kilnBOT
short agate
how do I prove this using just the definition of matrix multiplication
All I have is that AB = [AB_1 AB_2 ... AB_k], where each B_i is the ith column of B
I don't know how to manipulate the sum to get to the formula the textbook has
@lusty birch
<@&286206848099549185>
I just need to prove that a matrix times a column vector is a linear combination of its columns where the weights are given by the rows of the column vector
But I dont know how
please please please please please im going insane
cedar kilnBOT
@short agate Has your question been resolved?
nimble veldt
isnt it obvious from the definition of the maxtrix multiplication? the elements of gamma_j, the jth column of C, are the c_ij for a fixed j.
short agate
Where do I go with that fact? I agree with it, but I am trying to figure out how to get from the definition of matrix multiplication into the equation the textbook (in the image)
I would imagine that this is almost purely algebraic manipulation, something to do with indices in the sum using the fact that it is a fixed j
And another thing that is confusing me is the fact that matrix multiplication is defined by multiplication of the entries, whereas the equation that they have in the image includes a matrix in the sum
wraith daggerBOT
ThM
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
nimble veldt
look at this with a j_1 and a j_2
short agate
i can see that with a fixed j, when ranging L from 1 to n, we have that the j-th column of C will consist of linear combinations of the columns of A with weights coming from the j-th column of B
makes sense to me, it just doesn't feel right to immediately jump from that to the equation that the image has
i feel like there needs to be more algebraic manipulation, or maybe i'm being too precise?
nimble veldt
you just have to switch between vector notation (gamma_j, alpha_r) and element-wise notation.
$\gamma_i = <c_{ij}>=<\sum_{r=1}^na_{ir}b_{rj}>\rightarrow \sum_{r=1}^nb_{rj}<a_{ir}>=\sum_{r=1}^nb_{rj}\alpha_r$
short agate
i guess that's truly my key area of confusion, when is it okay to switch between vector notation and element-wise notation
wraith daggerBOT
ThM
nimble veldt
try to express each element of C in both notations (matrix multiplication and linear combination of A-Coloumns) and see if there are differences or not.
(if you need it to be more comfortable)
cedar kilnBOT
@short agate Has your question been resolved?
cedar kilnBOT
cosmic steppe
Wait why
Why did you delete it
I meant the other guy
Not you
Go get a new channel lol
crimson sedge
I remember how now
cosmic steppe
Ah
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Available help channel!
formal lynx
hi so with SSA triangles it is an ambiguous case meaning that there can be 0 1 or 2 triangles but if a triangle is SAS for example or any other triangle then is it only possible for it to have 1 triangle
cedar kilnBOT
@formal lynx Has your question been resolved?
slow thicket
maybe nobody knows lol
small dust
IMO is that it's kinda redudant/useless to care about those features bc the only parts that matter for similarity are SSA side-side-angle and side angle side, not the height other stuff bc those just follow when u look at the trigangles
formal lynx
this is not about congruency and similarity
small dust
SAS and SSA? yes it is
small dust
I see
slow thicket
its precalc
pre calc whaaa? I'm in calc 3 and we havent used triangles at all
formal lynx
its laws of sin / laws of cosine ambiguous cases
slow thicket
I've forgotten completely what those SSA are
small dust
pre calc whaaa? I'm in calc 3 and we havent used triangles at all
I've not touched triangles and two column proofs since middle school and have almost never needed that so, felt lol
slow thicket
exactly lol
formal lynx
yeah but it isnt proofs and all that
tbf idk why we are doing this topic
u've prob done laws of sin and laws of cosine right
slow thicket
yeah
formal lynx
its like a continuation of that unit
formal lynx
yeah but this isnt about congruency
woven aspen
mhmm
formal lynx
we just use SSA bc in some cases it can form 2 triangles
and u have to check the height in comparison with the a and b value to check
slow thicket
hi so with SSA triangles it is an ambiguous case meaning that there can be 0 1 o...
okay i think i get your question
if you know SAS
formal lynx
like if it forms two trignales then the problem can i have 2 solutions
slow thicket
that means you can use law of cosines to get the third side
formal lynx
im pretty sure SAS can only have 1 triangle
slow thicket
so there is only one sas triangle
formal lynx
ok
cedar kilnBOT
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cedar kilnBOT
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onyx barn
May I ask how I can even begin to try to find X (length of AC) and Y (Length of BC)?
onyx barn
There's surely something but I feel clueless.
mental mural
There's surely something but I feel clueless.
Any given information?
cedar kilnBOT
@onyx barn Has your question been resolved?
cedar kilnBOT
@onyx barn Has your question been resolved?
cedar kilnBOT
@onyx barn Has your question been resolved?
small dust
There are no numbers or anything so it's kinda impossible
small dust
Only the image
so not even a question? u just have an image and no context?
cedar kilnBOT
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jade stag
lol its not touching though?
cedar kilnBOT
lol its not touching though?
small dust
lol its not touching though?
it's continuous since it doesn't have any discontinuties such as holes, jumps, or asymptotes from the graph
jade stag
true but they asked at -6,6
and if u look at graph it doesnt go there
not at -6 atleast, maybe -5.5, 6
slender fog
(x)^2=x+x+x+x__ x times. Taking derivative of both sides. 2x=1+1+1__x times. Hence, 2x=x. Hence, 2=1. Where is the problem in this?
small dust
(x)^2=x+x+x+x__ x times. Taking derivative of both sides. 2x=1+1+1__x time...
yo u have ur own channel
yo Doug, u been doing math for a while, u studying or doing hw
also how late is it for u rn
jade stag
.close
crimson sedge
I assume here you mean grad$(f) = O$ where $O$ is the zero vector. Use the definition of the gradient vector.
wraith daggerBOT
45
crimson sedge
grad$(f(x, y)) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$
wraith daggerBOT
45
crimson sedge
Okay
You have a simultaneous equation here, no?
Yes
$6x - 6x^2 + 6y = 0$ and $6x - 6y = 0$
wraith daggerBOT
45
crimson sedge
$x - x^2 + y = 0$ and $x - y = 0$
wraith daggerBOT
45
crimson sedge
In fact, you can solve this by adding the two equations together
Yes there is the trivial solution x = y = 0.
Yes.
No
(0,0) and (2, 2) are the only solutions.
waxen crater
cedar kilnBOT
waxen crater
so how do i continue from what ive done
small dust
choose the most restrictive interval
here u have (-2, 5) and (3, 7)
what interval satisfies both?
waxen crater
here u have (-2, 5) and (3, 7)
but why do we put (-2,5) together aren't they both x(s)?
small dust
yeah those are the roots and since ur a value is positive the quadratic is upward facing meaning that the quadratic is <0 in between the roots
aka (-2, 5)
waxen crater
what interval satisfies both?
satisfy what?
small dust
here u have (-2, 5) and (3, 7)
these two intervals
what values are in both of these?
i gtg but it's ||(3, 5)||
waxen crater
ok bye and thx
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latent bloom
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covert island
cedar kilnBOT
covert island
for this one
i did u sub
my u = 1-p^2
and my du would then be 3pdu = 6dp
is this the right step
?
cedar kilnBOT
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cedar kilnBOT
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crimson sedge
Number 3
crimson sedge
So when solving with A when a 3 is under it, how does it change from when your solving just with A with a 1 below it
,rotate
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tacit phoenix
Why is it that when I repeat the Lloyd's algorithm, in this case 1024 times, the cells of the Voronoi diagram move to the upper left corner?
tacit phoenix
hmm, maybe it's because of integer rounding
nvm, ill try to increase the precision of the calculations and squeeze the result to the size of the screen (now I count at coordinates <= window size)
thx!
.close
hollow kraken
how to solve this kind of problem ?
cedar kilnBOT
how to solve this kind of problem ?
vagrant elbow
Multiply both sides with x and then use the discriminant of a quadratic
dusk finch
yes, now multiply both sides by x