you dont have to draw the triangle just where the compass is intersepting with itself
#help-36
dire glacier
shadow smelt
bro
dire glacier
but again for 60 degrees you should always use the same distance
so the triangle is equilateral
shadow smelt
so what i drew was wrong then
dire glacier
what you drew is an equilateral triangle
so its good
since you put your compas in "a" got the distance to "b" and viceversa
if you did it aleatory not the distance between points
and if you intercepted your compass
it would just bisectrate the line
what i mean is that
dire glacier
dire glacier
we usually put our protractor
yeah
dire glacier
instead of that you can just use an equilateral triangle
shadow smelt
and put it at A
dire glacier
for 60 degrees
dire glacier
no there are ways of making diferrent degrees
i can find you a spreadsheet if you want
which my teacher gave us
shadow smelt
just show me how to measure it in degress without using a protractor because i have tobuy one
dire glacier
likje depending on the special degrees there are different ways of doing so
if they allow you to use protractor
i'd recc using that
thats in spanish but you can use english ai
the one youtube implement
final saddleBOT
@shadow smelt Has your question been resolved?
barren hound
do you have a straightedge as well?
heady dagger
How to firstly calculate the length?
final saddleBOT
How to firstly calculate the length?
soft zealotBOT
south dirge
Here im gonna get a negative power??
?? whered you get the 2 and 3 from
drowsy epoch
Here im gonna get a negative power??
south dirge
Oh nvm I’m retorded
hey! everyone makes mistakes
heady dagger
The answer is sqrt37?
shadow smelt
do you have a straightedge as well?
no
south dirge
oh
here when the ask the length without mentioning the endpoints, we assume they are asking from origin
so x2,y2 = 0,0
south dirge
Now how to calculate the slope
well whats the normal formula?
,w formula for slope||
nvm its just simply put, (change in y)/(change in x)
so y1 - y2 / x1 - x2
heady dagger
(y2 - y1) / (x2-x1)?
south dirge
yup!!
ebon latch
Now how to calculate the slope
Vector is just a line
heady dagger
They calculated that the slope is 4/5?
Can you point where the slope is on the image?
south dirge
do you know whats slope for normal lines?
ebon latch
They calculated that the slope is 4/5?
Slope is 4/5 if your vector is (5,4)
south dirge
heady dagger
That’s just the length
south dirge
That’s just the length
i am not talking about the length, the tangent of the angle between the line (vector) and the x axis ( i direction)
is whats also known as slope or gradient
south dirge
oh yeah very much
south dirge
supposing your vector was (7 8)
the angle would have been different wouldnt it?
and you can see by simple trigo also that tan(angle) = change in yaxis(j)/ change in xaxis(i)
heady dagger
Anyways
That isn’t even important
How to find perpendicular vector of the following vectors.
south dirge
Oh but i think its better if you know what is slope
heady dagger
Oh but i think its better if you know what is slope
Yea it’s in what degrees the vector is
south dirge
How to find `perpendicular vector of the following vectors.`
well are you familiar with how we find a line perpendicular to a given line
heady dagger
No
south dirge
oh i see
well for now just note that for any two line to be perpendicular, the product of their slopes should be -1
heady dagger
“A perpendicular vector is obtained when a vector is rotated 90° counterclockwise in the plane.”
Oh you just switch and add minus to the first number
south dirge
“A perpendicular vector is obtained when a vector is rotated 90° counterclockwis...
true!
it could also be rotated clockwise
south dirge
Oh you just switch and add minus to the first number
uh sure
south dirge
Oh you just switch and add minus to the first number
not really sure what you meant by this but do lmk what you get your answer as
south dirge
ah thats right!!
do note that:
There can be infinite such vectors which are perpendicular to the original vector
south dirge
okay how about a vector like 5,-3
well?
there could also be a vector like 10,-6 or 5000,-3000
please ask in #discussion #serious-discussion
final saddleBOT
eternal knoll
Could anyone help me through section 4 of this assignment? I’ve got some done but I’m stuck on the other ones.
final saddleBOT
Could anyone help me through section 4 of this assignment? I’ve got some done bu...
timber leaf
which one in 4?
eternal knoll
which one in 4?
A f and I are the ones I’m stuck on
crimson pike
What have you tried on each, might I ask?
eternal knoll
For a I tried flipping the exponent, for f I’ve also flipped it, and for I I’m not sure what to try.
I don’t there’s anywhere else to go with f
eternal knoll
I don’t there’s anywhere else to go with f
Unless I raise it to the second power right?
crimson pike
Umm… 1/2 exponent does something else
eternal knoll
What’s it do?
eternal knoll
Umm… 1/2 exponent does something else
Yo?
crimson pike
So I’d square both 4 and 25?
Not square 
eternal knoll
Does it give it a radical?
crimson pike
Does it give it a radical?
To the top and bottom of the fraction yea
eternal knoll
Alr, could you help me with a?
crimson pike
A has a similar idea
crimson pike
The negative is separate; I’d leave it til the end
eternal knoll
Ok so how’s 6/5 effect -64?
crimson pike
6/5? I thought it was 5/6?
crimson pike
Take g for example; what did you do with the 7^(2/5)?
(I know g was done already)
eternal knoll
I subtracted 1/3 and 2/5
crimson pike
So with negative exponents, they get sent to the opposite side of the fraction
crimson pike
By sent to the opposite side
Like if there’s a negative on the top, it gets sent to the bottom with opposite sign
eternal knoll
Yeah
crimson pike
So 7^-(2/5) becomes 1/(7^(2/5))
eternal knoll
Ok
crimson pike
Similar can be done with the -5/6 or whatever exponent a has
crimson pike
The exponent just switches signs, no flipping is needed
crimson pike
Wait where is 6/5 coming from?
crimson pike
It still stays a 5/6, but it’s on the bottom and positive
crimson pike
So far yea
eternal knoll
Ok so what would the next step be?
crimson pike
Remember when i said an exponent of 1/2 was basically a square root?
eternal knoll
Yeah
crimson pike
Something like that can be said for any fraction exponent
The bottom is just whatever root
So if it’s #/2 it’s a square root, #/3 is a cube root, etc.
As for the top of the fraction; remind me how you can split up 5/6?
eternal knoll
By division?
crimson pike
5/6 can’t exactly reduce
eternal knoll
You multiply by 6/5?
crimson pike
Not quite
eternal knoll
5/6?
crimson pike
What kind of root do we need to take? Let’s start with that
crimson pike
Close…
crimson pike
It would be a sixth root if that’s what you’re asking
eternal knoll
Yeah
crimson pike
However, we do still have a 5 in the exponent
eternal knoll
What do we do with it?
crimson pike
We already delt with the /6 part; remember how you can have stuff like (a^b)^c?
eternal knoll
Yeah
crimson pike
How would that simplify?
eternal knoll
(-64^6)^5
crimson pike
Um… no
crimson pike
We already delt with the /6 part; remember how you can have stuff like (a^b)^c?
Let’s try this with letters;
What happens to the exponents b and c after simplifying?
eternal knoll
They add into each other
crimson pike
That’s a different exponent rule
B and C multiply
So (a^b)^c becomes a^(bc)… Now let’s set one of b or c to be 1/6… what do you multiply by 1/6 to get 5/6?
This might say where the 5 resides… kind of
eternal knoll
In the top half on the fraction
crimson pike
Well yea… the 5 would stay in the exponent
Hence why I tried asking how you’d separate 5/6… 5 and 1/6, to form it into an (a^b)^c thing
eternal knoll
So 1^5/-64^6
crimson pike
Hence why I tried asking how you’d separate 5/6… 5 and 1/6, to form it into an (...
64^(5/6) can be written as 64^(5*1/6)… looks like the power^power thing huh
eternal knoll
Yeah
crimson pike
I wouldn’t split the 64 tbh… 5/6 is the thing we’re splitting into a product
Although…
That also could work
And now 2 and 5/6 could multiply
Actually this is interesting 
Notice anything about 8?
eternal knoll
So the answer is 32
crimson pike
And said 2^5 is in the bottom
eternal knoll
Or -1/32
crimson pike
… we still have a negative outside
crimson pike
So this would be the answer?
Indeed
eternal knoll
Alr thanks for the help
crimson pike
Happy to help 
eternal knoll
.close
final saddleBOT
soft zealotBOT
final saddleBOT
still ruin
yeah nvm i already got it
stone flint
^
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Available help channel!
untold robin
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untold robin
How would I start this
lone void
If three vectors span r3, then that means none of them point in the same or opposite direction
opaque ember
How would I start this
ur notes should give easy computational criteria for n vectors to span R^n
untold robin
In my class my prof just reads off the textbook. I am not sure how to interpret this kind of problem
opaque ember
In my class my prof just reads off the textbook. I am not sure how to interpret ...
ideally ur lecture notes or textbook should explain what it means for some vectors to span some space and list some theorems that help u easily verify that
final saddleBOT
@untold robin Has your question been resolved?
loud sundial
Since $\dim \mathbb{R}^3=3$, we can say that the three vectors span $\mathbb{R}^3$ if and only if the three vectors are a basis of $\mathbb{R}^3$. So, ${v_1, v_2, v_3}$ must be a linearly independent set.
soft zealotBOT
Civil Service Pigeon
loud sundial
atp it's a direct application of the defn of linear independence
untold robin
Ohhh okay
How would I write the proof for linear independence as I can see none of the vectors are redundant and they are unique
loud sundial
atp it's a direct application of the defn of linear independence
What is the definition of linear independence?
untold robin
Linear independence if no vector can be a linear combination
loud sundial
give me the definition with an equation
untold robin
Umm how would I do thjs
untold robin
So the vectors multiple by a constant = vector with b1 b2 b3?
loud sundial
So the vectors multiple by a constant = vector with b1 b2 b3?
If you work with the strict definition of span, then yes - the system
$$c_1 v_1+c_2 v_2+c_3 v_3=\begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix}$$
must always have a solution $(c_1, c_2, c_3)$ for any $(b_1, b_2, b_3)$. You could definitely find $c_1$, $c_2$, $c_3$ in terms of $b_1$, $b_2$, $b_3$ and that would be sufficient, but using the linear independence route is easier in my eyes since you only have to work with the case where $b_1=b_2=b_3=0$ (also homogeneous systems are usually easier to work with by nature).
soft zealotBOT
Civil Service Pigeon
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opaque ember
.reopen
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✅ Original question: #help-36 message
opaque ember
@untold robin do u have a theorem relating row reduction to linear independence and spanning sets?
untold robin
Would it have something to do with gaussian elimination and then looking at the pivots for the image?
opaque ember
im not sure wym by image bc u didnt mention any function but yes im talking about pivots
im gonna give u theorems that use pivots to determine independence and spanning sets. theyre the computational criteria that i mentioned before
scarlet nimbus
U can take an arbitrary vector and see if you can write a linear combination using those vectores
Which means (x,y,z)=av1+bv2+c*v3 and find a b and c in function of x y and z
Maybe they already told u this but its an easy method to always find a family is a span
opaque ember
given $m$ vectors $v_1,\dots,v_m$ in $\bR^n$ (ie the vectors have $n$ components), let $A$ be the matrix whose columns are these vectors and compute its row echelon form $B=\mathrm{ref}A$. then:
\begin{enumerate}
\item the vectors are linearly independent if and only if $B$ has a pivot in every column
\item the vectors span $\bR^n$ if and only if $B$ has a pivot in every row
\end{enumerate}
soft zealotBOT
ロケット・ジャンプ
untold robin
.close
final saddleBOT
final saddleBOT
Available help channel!
sharp yew
Is that possible I got asked that 🥺
X^(X^X) = (2^32)^(1/5)
final saddleBOT
Is that possible I got asked that 🥺
X^(X^X) = (2^32)^(1/5)
proud igloo
lofty willow
$x^{x^x}= (2^{32})^\frac{1}{5}$
soft zealotBOT
Maddie
lofty willow
interesting
final saddleBOT
@sharp yew Has your question been resolved?
proud igloo
lwky have to use lambert W
i can't find a way that just bs's this
But im too silly for that fr so we are gonna try newton raphson
trail mango
proud igloo
the answer I get is ||2.196979425074||
proud igloo
<:AA_Sus:1451985218306048026>
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Available help channel!
sturdy kindle
@grim raft
final saddleBOT
<@981969441875169360>
ivory vessel
<@981969441875169360>
do you have a question?
sturdy kindle
do you have a question?
none
ivory vessel
then please close this channel
inland kettle
none
don’t chat in the help channels and such. we have #discussion and #chill for those purposes
unless you have a specific question, close this channel using .close
final saddleBOT
@sturdy kindle Has your question been resolved?
final tangle
rustic wedge
$\lim_{n \to \infty} (1- \cos x)^{-n} (1+ \sin x)^{-n}$, where $x \in \left(0, \frac{\pi}{2} \right)$
final saddleBOT
$\lim_{n \to \infty} (1- \cos x)^{-n} (1+ \sin x)^{-n}$, where $x \in \left(0, \...
rustic wedge
apparently this is one?
soft zealotBOT
rak³en
timber leaf
you mean the limits?
timber leaf
just try while n approach infinity and simplify the equation
rustic wedge
you mean the limits?
the limit evaluates to 1 for every x in that given range, apparently
i was looking at som integrals from iitb icup and one of them involved this limit
timber leaf
have you tried evalute the equation when n approach inf
woven ledge
for x=pi/3 (1+ sin(pi/3))(1-cos(pi/3)) is less than one so it should diverge right? did i make a mistake
@rustic wedge
final saddleBOT
@rustic wedge Has your question been resolved?
onyx peak
apparently this is one?
It depends on what x is
For a fixed x, its just const^(-n)
So u really only need to solve (1-cosx)(1+sinx) < 1
At pi/2 its 2 already, so surely it doesnt go to 1 everywhere
rustic wedge
have you tried evalute the equation when n approach inf
??? isnt that evaluating the limit?
rustic wedge
for x=pi/3 (1+ sin(pi/3))(1-cos(pi/3)) is less than one so it should diverge rig...
hmm i see
ty
.close
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rustic wedge
can someone explain how these 2 are equivalent, c=cosx
final saddleBOT
can someone explain how these 2 are equivalent, c=cosx
lyric obsidian
can someone explain how these 2 are equivalent, c=cosx
whats the lower bound on the right integral?
cant see it
severe canyon
can someone explain how these 2 are equivalent, c=cosx
Because you can split the first integral into two parts. The first one has an even function on a symmetric interval, hence it is equivalent to double of the integral from 0 to π. The remaining part, instead, is an odd (and continuous) function over a symmetric interval, hence it is 0
After noting this, you can use the double angle formula, and write cosx = 2cos²(x/2) - 1 and then doing a u-sub, such as u = x/2
daring lion
I was making the even part 0 and odd part 2 times 
final saddleBOT
@rustic wedge Has your question been resolved?
rustic wedge
whats the lower bound on the right integral?
0
rustic wedge
Because you can split the first integral into two parts. The first one has an ev...
oh u split it i see
ty
.close
final saddleBOT
severe canyon
oh u split it i see
Yeah I also had to think a quite long while before noticing it lol
At first they seemed two completely unrelated integrals to me, especially for the numerator without sinx that disappeared
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Available help channel!
jagged flare
$$a^2+b+c+d=142$$
$$a+b^2+c+d=184$$
$$a+b+c^2+d=234$$
$$a+b+c+d^2=292$$
Find $a+b+c+d$
final saddleBOT
$$a^2+b+c+d=142$$
$$a+b^2+c+d=184$$
$$a+b+c^2+d=234$$
$$a+b+c+d^2=292$$
Find $a+...
soft zealotBOT
ihave<skissue>
jagged flare
how would you do this
inland kettle
i think
i haven't tried it yet
just an idea
subtract each eq from another, so for example for 1 subtracting 2 we get
a(a-1) - b(b-1) = -42
for say, 1 subtracting 4, we get a(a-1) - d(d-1) = -150
and we substitute like. w = a(a-1), etc
then you get a system of linear eqs
then just apply the quadratic formula once you figure out w, x, y, and z
should work @jagged flare
inland kettle
happens to the best of us
jagged flare
i immediately skipped it and tried factoring it like (a-b)(a+b+1)
mmm i get the idea tho
okay okay ty!!
.solved
final saddleBOT
inland kettle
okay okay ty!!
ofc
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gentle zephyr
how do i prove this
final saddleBOT
how do i prove this
gentle zephyr
spark swan
have you made any progress on any of these?
spark swan
well how would you typically solve a problem like this?
gentle zephyr
dunno is the first exercise involving index unions
spark swan
ok lets do the first one first
we need to show both sides are subsets of each other, so first assume some $x \in B \backslash \bigcup_{i \in I} A_i$
soft zealotBOT
Green
gentle zephyr
ok, any idea how x is an element in the right hand set
spark swan
by definition, x belongs to B, but not any A_i
spark swan
in the intersection on the right hand side, you have B in all the sets, so you can say at least x belongs to B
versed crater
(let them do some of the work!!)
gentle zephyr
in the intersection on the right hand side, you have B in all the sets, so you c...
intuitively yes, formally idk
spark swan
(let them do some of the work!!)
mb i was about to ask them to the fill the rest
versed crater
in the intersection on the right hand side, you have B in all the sets, so you c...
what he means here is that if $x\in \bigcap_{i\in I}(B\setminus A_i)$ then $x$ is at least $\in B$ because $\bigcap_{i\in I}(B\setminus A_i) \subseteq B$
soft zealotBOT
frosst
spark swan
can you see why x is not in the right side if it is in any A_i?
gentle zephyr
can you see why x is not in the right side if it is in any A_i?
ya
the assumption is that x is not in Ai
spark swan
okay now you need to show if x is in the right side, it is also in the left, are you able to do that?
gentle zephyr
right side guarantees x is in B and is not in any Ai
what i mean is, right side is a subset of B
now i need to show that x is in none of the Ai
spark swan
right side guarantees x is in B and is not in any Ai
you need to show that the left side also implies that so x is in the left side
versed crater
to give you an example, one might say something like this:
Let $x\in B\setminus \bigcup_{i\in I}A_i$, then $x\in B$ and $\nexists i\in I$ such that $x\in A_i$. Since $\nexists i\in I$ such that $x\in A_i$, then $x\in B\setminus A_i$ for any $i\in I$, since it lies in every $B\setminus A_i$, it must also lie in their intersection by definition. Hence $x\in \bigcap_{i\in I}(B\setminus A_i)$ and $B\setminus \bigcup_{i\in I}A_i \subseteq \bigcap_{i\in I}(B\setminus A_i)$
soft zealotBOT
frosst
gentle zephyr
to give you an example, one might say something like this:
Let $x\in B\setminus...
i see, the intersection part is the only doubt i have with this
versed crater
that it must also lie in the intersection?
gentle zephyr
yeah
versed crater
okay what's the definition of the intersection
gentle zephyr
is in both
soft zealotBOT
frosst
gentle zephyr
if $x\in A_i$ for every $i\in I$, then $x\in \bigcap_{i\in I}A_i$
can the same thing be said about unión
?
versed crater
well, union is a bit different
while it is true for the union, it doesn't characterise everything in the union
versed crater
**frosst**
this is actually an iff
gentle zephyr
this is actually an iff
i see
versed crater
$\forall i\in I,,x\in A_i \iff x\in \bigcap_{i\in I} A_i$
soft zealotBOT
frosst
gentle zephyr
while it is true for the union, it doesn't characterise everything in the union
yeah because they dont need to be in every Ai to be in the union
versed crater
yeah for the union you need to be in at least one
that's where the existence comes in
$\exists i\in I$ such that $x\in A_i \iff x \in \bigcup_{i\in I}A_i$
soft zealotBOT
frosst
gentle zephyr
i shee
versed crater
i mean the whole goal of these exercises here is to make you familiar with these ideas and notation
so if you weren't able to write these down then that's why you're doing the exercises
gentle zephyr
so if you weren't able to write these down then that's why you're doing the exer...
yeah
i shee
do you mind I show you my proof and you help me with this stuff?
stiff dagger
i see i see
gentle zephyr
versed crater
hmm lemme think on it
dont troll in help channels
stiff dagger
okay
versed crater
do you mind I show you my proof and you help me with this stuff?
yeah go ahead although i might sleep soon
gentle zephyr
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \text{Take an arbitrary } x \in \left(B \setminus \bigcup_{i \in I} A_i \right) \ \text{ then it follows that } x \in B \ \text{ and that } \forall i \in I \text{ it is satisfied that } x \not\in A_i
versed crater
\setminus
that dont sound right
if x is in every A_i, then x is clearly in U A_i
that also deosn't sound right
if x is in some A_i, then x is clearly in U A_i
okay just because x doesn't live in one of the A_i's doesn't mean it can't live in any of the other A_i's
gentle zephyr
ok
versed crater
if A_i's were disjoint then this would automatically be true that there exists some i \in I such that x \notin A_i
soft zealotBOT
Renato
versed crater
yes
or in other words, there does not exist an i \in I such that x \in A_i
but that's the same statement
gentle zephyr
ok
now what
well if its in B its in bigcap B
thats fo shure
what about bigcap B \ Ai
versed crater
well if its in B its in bigcap B
what is this
what is the indexing on thie bigcap
there also is no bigcap B anywhere on the problem
so whether it's true or not isn't actually relevant here
gentle zephyr
I am just trying to get to the right side from the left side
gentle zephyr
yes but the right side is a subset of B
versed crater
how do you know that
gentle zephyr
what I need to say to finish this is:
since x is in B and x is not in Ai forall i in I
it follows that x is in (B \ Ai) forall i in I
versed crater
**Renato**
that's what needs to come after this
gentle zephyr
and from here just wrap up saying that if its in all (B \ Ai) then its in the intersection by definition
versed crater
also whenever you use words like "it follows" you're sweeping the reasoning under the carpet
so make sure you know why it's true
at least have it make sense in your head
gentle zephyr
it does, I never use it in a handwavy manner, Is there any different wording that would be preferred?
versed crater
nah it's fine
gentle zephyr
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \text{Take an arbitrary } x \in \left(B \setminus \bigcup_{i \in I} A_i \right) \ \text{ then it follows that } x \in B \ \text{ and that } \forall i \in I \text{ it is satisfied that } x \not\in A_i \ \text{from here it follows that } x \in (B \setminus A_i) \quad \forall i \in I \ \text{we conclude that } x \in \bigcap_{i \in I} \left(B \setminus A_i\right)
soft zealotBOT
Renato
versed crater
sure
but i'll say, you use a lot of meaningless connecting words
it's like you're writing an essay and dont want to repeat yourself
gentle zephyr
what do you recommend?
versed crater
i just think words like "from here it follows" can be replaced with "hence"
but that doesn't really describe much reasoning. I believe that if you use english in maths you want to give an insight that isn't already written down, or to express the notation in words
using a lot of connecting words just make you sound verbose and can become confusing
gentle zephyr
yeah I also want to use less words and more symbols
hence seems like a much a better replaccement to from here it follows
versed crater
look at where i use english and what purpose it serves
"since it lies in every ..., it must also lie in their intersection by definition"
this is just english for the symbols bit before and after this sentence
gentle zephyr
yeah
versed crater
Since $\nexists i\in I$ such that $x\in A_i$,
then $x\in B\setminus A_i$ for any $i\in I \implies x\in \bigcap_{i\in I}(B\setminus A_i)$
soft zealotBOT
frosst
versed crater
here i've just replaced that sentence with \implies
gentle zephyr
sure, that works
versed crater
but it might not be clear why this is the case, which is why i used words to describe what is happening
(especially because i was writing this for the purposes of explaining it)
gentle zephyr
ok so
dusty siren
hi
gentle zephyr
do you want me to rewrite it or can we continue with b)
versed crater
**Renato**
we didn't even say the left side is a subset of the right side
dusty siren
can someone help me with my math question i dont get it so good
versed crater
we haven't even done that the right side is a subset of the left side
we don't have equality yet
we're not even half way through part a D:
gentle zephyr
however, we just noticed that by grabbing an arbitrary element in the LHS we can get to the RHS, isnt that enough to show that the entire LHS is a subset of the RHS
final saddleBOT
can someone help me with my math question i dont get it so good
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
versed crater
like i mean the statement of part a is true
you could just say "look at it it's true, im done here"
gentle zephyr
ok, let me add that last sentence to the proof I made and we can continue with b), I will promise to try to use less innecessary words
versed crater
no man
we havne't done the other direction
for A = B we need A \subseteq B and B \subseteq A
we've only done A \subseteq B
gentle zephyr
alright
versed crater
these sorts of proofs are the kind of "do it once and then never again" type problems
gentle zephyr
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \text{Take an arbitrary } x \in \left(B \setminus \bigcup_{i \in I} A_i \right) \ \text{ then it follows that } x \in B \ \text{ and that } \forall i \in I \text{ it is satisfied that } x \not\in A_i \ \text{from here it follows that } x \in (B \setminus A_i) \quad \forall i \in I \ \text{we conclude that } x \in \bigcap_{i \in I} \left(B \setminus A_i\right) \ \text{We conclude } B \setminus \bigcup_{i \in I} A_i \subseteq \bigcap_{i \in I} (B \setminus A_i)
soft zealotBOT
Renato
versed crater
good
gentle zephyr
ok let me do right side inclusion
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \text{Take an arbitrary } x \in \left(\bigcap_{i \in I} (B \setminus A_i) \right) \text{, then } \ \text{Since } x \in \bigcap_{i \in I} (B \setminus A_i) \text{ then, } x \in (B \setminus A_i) \text{ , } \forall i \in I \
\text{ then, } x \in (B \setminus A_i) \text{ , } \forall i \in I \implies \begin{cases} x \in B } \ x \not \in A_i \text{ , } \forall i \in I \end{cases}
versed crater
that's nice and all
and gets you to the definition you have on the left side
but it would be nice to get some more words and explanation why
but if you think you understand why, that's okay as well
gentle zephyr
care to give an example?
versed crater
im just not sure if it's immediately obvious that $\bigcap_{i\in I}(B\setminus A_i) \subseteq B$
soft zealotBOT
frosst
gentle zephyr
i shee
let me re do this second to last statement
maybe like this?
no, I dont think that clarifies anything really
and then since x is not in Ai for every i in I
versed crater
now you've lost the part where you say it's in B
gentle zephyr
let me fix that
versed crater
it might feel to you like you're just typing the same shit backwards and forwards
and the truth is, you are
that's the whole exercise
shuffle the definitions around until you get to the other side
gentle zephyr
it might feel to you like you're just typing the same shit backwards and forward...
am I doing something wrong?
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \text{Take an arbitrary } x \in \left(\bigcap_{i \in I} (B \setminus A_i) \right) \text{, then } \
\text{Since } x \in \bigcap_{i \in I} (B \setminus A_i) \text{ then, } x \in (B \setminus A_i) \subseteq B \text{ , } \forall i \in I \
\text{Furthemore, if } x \in \bigcap_{i \in I} (B \setminus A_i) \text{ then, } x \in (B \setminus A_i) \text{ , } \forall i \in I \
\text{ Lastly, } x \in (B \setminus A_i) \text{ , } \forall i \in I \implies x \not \in A_i \text{ , } \forall i \in I
versed crater
am I doing something wrong?
no no
im just noting that if you feel like you're just shuffling the definitions around, don't worry because that's exactly what you're aiming to do with these problems
no
im saying your 2nd last line
doesn't say "hence x \in B" anymore
after you added the new thing
now the "x \in B" just shows up as your conclusion
it just appears, without notice or anything
gentle zephyr
ok let me try to fix it yeah I preferred the previous way of doing it but it was very handwavy
let me try to use the best parts of both approaches
versed crater
you're just making it up
gentle zephyr
ok ok
versed crater
$x\in B\setminus A_i \subseteq B$
gentle zephyr
let me fix this
soft zealotBOT
frosst
gentle zephyr
$x\in B\setminus A_i \subseteq B$
this
versed crater
so clearly x is in B
soft zealotBOT
Renato
versed crater
have you read your 2nd last line
read it and then read your third one
bro is writing an essay with these furthermore and lastly's
versed crater
we dont actually need the 2nd last row
we already said the same thing in the 3rd row
lastly can just be replaced with "and"
there's no need to get so fancy with the writing
once you get to the $x\in B\setminus A_i\subseteq B,,\forall i\in I$, here you can immediately say hence $x\in B$ and $x\in A_i,,\forall i\in I$
soft zealotBOT
frosst
gentle zephyr
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \text{Take an arbitrary } x \in \left(\bigcap_{i \in I} (B \setminus A_i) \right) \text{, then } \
x \in \bigcap_{i \in I} (B \setminus A_i) \implies x \in (B \setminus A_i) \subseteq B \text{ , } \forall i \in I \implies x \in B \
x \in \bigcap_{i \in I} (B \setminus A_i) \implies x \in (B \setminus A_i) \text{ , } \forall i \in I \implies x \not \in A_i \text{ , } \forall i \in I \
\text{Since } x \in B \text{ and } x \notin A_i \text{ , } \forall i \in I \text{ it follows } x \in \left(B \setminus \bigcup_{i \in I} A_i \right) \
\text{We conclude } \bigcap_{i \in I} (B \setminus A_i) \subseteq B \setminus \bigcup_{i \in I} A_i
versed crater
why in the 2nd last line you use "since ... then" and in the last row you use an implication sign
also the "then" should be after the comma not before
"since ..., then..."
gentle zephyr
why in the 2nd last line you use "since ... then" and in the last row you use an...
should I remove the sincce .. then?
versed crater
you should be consistent
that's fine
that's fine
and then afterwards you gather them together and say aha it must also lie in the left side
and then...?
soft zealotBOT
Renato
gentle zephyr
we good?
versed crater
well, we've yet to combine it all and then say the sets are equal
gentle zephyr
,, B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i) \ \noindent\rule{\textwidth}{0.4pt}
\ \text{Take an arbitrary } x \in \left(B \setminus \bigcup_{i \in I} A_i \right) \ \text{ then it follows that } x \in B \ \text{ and that } \forall i \in I \text{ it is satisfied that } x \not\in A_i \ \text{from here it follows that } x \in (B \setminus A_i) \quad \forall i \in I \ \text{we conclude that } x \in \bigcap_{i \in I} \left(B \setminus A_i\right) \ \text{We conclude } B \setminus \bigcup_{i \in I} A_i \subseteq \bigcap_{i \in I} (B \setminus A_i) \ \noindent\rule{\textwidth}{0.4pt}
\ \text{Take an arbitrary } x \in \left(\bigcap_{i \in I} (B \setminus A_i) \right) \text{, then } \
x \in \bigcap_{i \in I} (B \setminus A_i) \implies x \in (B \setminus A_i) \subseteq B \text{ , } \forall i \in I \implies x \in B \
x \in \bigcap_{i \in I} (B \setminus A_i) \implies x \in (B \setminus A_i) \text{ , } \forall i \in I \implies x \not \in A_i \text{ , } \forall i \in I \
\text{Since } x \in B \text{ and } x \notin A_i \text{ , } \forall i \in I \text{ it follows } x \in \left(B \setminus \bigcup_{i \in I} A_i \right) \
\text{We conclude } \bigcap_{i \in I} (B \setminus A_i) \subseteq B \setminus \bigcup_{i \in I} A_i \ \noindent\rule{\textwidth}{0.4pt}
\ \text{Having } \bigcap_{i \in I} (B \setminus A_i) \subseteq B \setminus \bigcup_{i \in I} A_i \text{ and } B \setminus \bigcup_{i \in I} A_i \subseteq \bigcap_{i \in I} (B \setminus A_i) \ \text{ we conclude } B \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (B \setminus A_i)
versed crater
bit of a latex issue at the end but should be good
soft zealotBOT
Renato
gentle zephyr
bit of a latex issue at the end but should be good
we good?
versed crater
yeah looks good
gentle zephyr
help me with b) please
versed crater
it's 2:30am so im probabaly gonna go sleep unfortunately
gentle zephyr
yeah looks good
.solved
final saddleBOT
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desert rain
the vectors (x, y, z) whose components satisfy each of these conditions are not subspaces of V3, can someone tell me why?
final saddleBOT
the vectors (x, y, z) whose components satisfy each of these conditions are not ...
drowsy epoch
consider x=y=z=0
desert rain
i think the first two arent subspaces because they give each element another element, but what about the third one?
desert rain
consider x=y=z=0
what?
drowsy epoch
Is the subspace supposed to satisfy 6. 7. and 8. or are they separate tasks?
desert rain
they are separate
drowsy epoch
bruh okay
ok indeed consider x=y=z=0 for 8. tho
you are checking if the null vector is contained
desert rain
oh i see
and the second one does not follow the axiom of negative element, am i correct?
the first one too
versed crater
and the second one does not follow the axiom of negative element, am i correct?
Give an example
versed crater
the first one too
Give an example
desert rain
oh i'm wrong actually
but such an element is not unique
for the first one we would have A = (y, y, z) or A = (z, y, y), then B + A = 0 if B = (-y, -y, -z) or B = (-z, -y, -y)
versed crater
(z, y, y)?
desert rain
for the first one we would have A = (y, y, z) or A = (z, y, y), then B + A = 0 i...
for the first one we would have A = (y, y, z) or A = (z, y, z), then B + A = 0 if B = (-y, -y, -z) or B = (-z, -y, -z)
desert rain
wait, then i didnt understand what the "or" is supposed to mean
versed crater
The set contains {(x, y, z) | x = y or x = z}
So it’s the set {(y, y, z)} ∪ {(z, y, z)}
versed crater
Maybe it is a subspace
desert rain
nvm i still cant see
versed crater
You can try to prove that it’s a subspace
desert rain
oh youre right
in order to be a subspace we need to make sure the elements follow the ten axioms
but it does
and the set is a subset of R^3
it is still a subspace of V3 in my head
WAAAAAAAAIT
i think i figured it out
if i choose like A = (1, 1, 2) and B = (2, 0, 2), then A + B = (3, 2, 4), which does not follow the rules { x = y or x = z }
and if i choose A = (1, -1, 2) and B = (1, 1, 2), then A + B (2, 0, 4) which does not follow { x = y or x = -y }
and @versed crater thank you so much
.close
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keen hare
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keen hare
have to prove this
final saddleBOT
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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
keen hare
Factor out n+1 from denominator
which denominator
keen hare
Then write 2n choose n in terms of factorials
also i did
final saddleBOT
@keen hare Has your question been resolved?
vital crag
oh your work is quite confusing since you're changing both the left and right sides
i recommend changing just one side to end up at the other
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forest lodge
Can someone help me with this exercise. They ask me to find the lateral area of that figure.
final saddleBOT
Can someone help me with this exercise. They ask me to find the lateral area of ...
forest lodge
I could do it with area inbetween two curves, but I really dont understand.
final saddleBOT
@forest lodge Has your question been resolved?
drowsy epoch
I could do it with area inbetween two curves, but I really dont understand.
You already computed the front area, so what's left is the cap, the sides and from below the bottom of the object
forest lodge
You already computed the front area, so what's left is the cap, the sides and fr...
I think I get it
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lime crest
For a positive integer $n$, let $\Phi_n(x) = \prod_{(k,n) = 1, 1 \le k \le n } (x - e^{\frac{i2\pi k}{n}})$. Prove that for all distinct primes $p,q$, [ \Phi_q(x^p) = \Phi_q (x) \Phi_{pq}(x) ]
final saddleBOT
For a positive integer $n$, let $\Phi_n(x) = \prod_{(k,n) = 1, 1 \le k \le n } (...
soft zealotBOT
Copter
lime crest
is this something about expanding the x^p - e^... term?
final saddleBOT
@lime crest Has your question been resolved?
lime crest
<@&286206848099549185>
graceful spindle
Oops
vital crag
Holy gpt wall
graceful spindle
Holy gpt wall
Lol
timber leaf
!noai
final saddleBOT
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
timber leaf
I'm too lazy to calculate
your brain is shrinking 💀
vital crag
I'm too lazy to calculate
Take that a step further and be too lazy to help
loud sundial
<@&268886789983436800>
loud sundial
is this something about expanding the x^p - e^... term?
Yeah you're essentially splitting each of these into p linear factors
formal trail
@graceful spindle Please don't post copy-pasted GPT output in the help channels, people here want to hear from a human
obsidian lichen
<@&286206848099549185>
I can not help
final saddleBOT
@lime crest Has your question been resolved?
timber leaf
I can not help
then dont text 💀
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neon quail
where is the second 2^399 going?
final saddleBOT
where is the second 2^399 going?
spark sundial
just cancel
the two terms
oh
thats factoring
take the common thing
and push it aside
so by method of distribution its still the same
spark sundial
alr
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left trail
can someone provide some suggestions on how I can evaluate $\int_0^2 \frac{x^3}{\sqrt{x^2+5}}dt$ Since right now I have the definite integral equal to $\frac{19}{5} - 4\int_0^2 \frac{x^3}{\sqrt{x^2+5}}dt$
final saddleBOT
can someone provide some suggestions on how I can evaluate $\int_0^2 \frac{x^3}...
soft zealotBOT
BigBen
zinc igloo
can someone provide some suggestions on how I can evaluate $\int_0^2 \frac{x^3}...
the cleanest path here is substitution ig
gimme a sec to write in one note and help u ou if that s ok?
vital crag
can someone provide some suggestions on how I can evaluate $\int_0^2 \frac{x^3}...
you need to show your full work and not just the start and end.
keen storm
you need to show your full work and not just the start and end.
its fine for this problem
left trail
you need to show your full work and not just the start and end.
sorry let me send it
left trail
the cleanest path here is substitution ig
for the whole thing?
keen storm
for the whole thing?
no the root thing
zinc igloo
can someone provide some suggestions on how I can evaluate $\int_0^2 \frac{x^3}...
sorry for the ugly ass writing
zinc igloo
is this for the whole thing or just the integral
just the integral
keen storm
just the integral
yeah i didnt evaluate i just wrote the u to 1/2 - 5....
zinc igloo
need to plug it to the given 19/5...
left trail
This was my work
zinc igloo
iwas thinking smth like
left trail
I am confused on how you have x^3 = u-5. we have x^2 + 5 = u and then u -5 = x^2
keen storm
ok so we u sub x²+5
but before that we change the integrals nominator to x² times x
after sub we get
left trail
why is the 5^2 here
ye thats wrong. my bad
left trail
I don't understand how you do your third line
also if we are doing u = x^2+5 why not just do u -5 = x^2. sqrt (u-5) = x. so then x^3 = (u-5)*sqrt(u-5)
so (u-5)^3/2
vital crag
they're wrong. you're supposed to use ibp again
compare this with
a^2 is still 5
find x and n in your new integral
left trail
x is 2 and n is 3
zinc igloo
yeah but for that inner integral u sub is cleaner than ibp (??)
vital crag
x is 2 and n is 3
right so use a) again
proper lion
yeah but for that inner integral u sub is cleaner than ibp (??)
but the question says use part a
zinc igloo
oof sorry
left trail
right so use a) again
ok so I have $3I_3(2) = 12 - 10 I_1(2)$. Then we have $I_3 = 4 - \frac{10 I_1(2)}{3}$ We have that $I_1(2) = 3$ so then we have -6 for our final answer
soft zealotBOT
BigBen
proper lion
**BigBen**
I1(2) feels wrong
left trail
I1(2) feels wrong
Ye I realized a's condition is for n≥2
vital crag
n=1 you can finally use substitution
left trail
Ye so that's what we get
but then our final answer is 19/5 - 26 - 10sqrt 5 which is not close to what they have
proper lion
What I wrote earlier was wrong. But the final answer I have is off by 1/5
,, 5I_5(2) = 2^{5-1} \sqrt{2^2 +5} - 20I_3(2)
=> 48 - 20I_3(2)
soft zealotBOT
ghost
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sick citrus
Yo
sick citrus
Ok, that's correct.
negative infinity to 3
,close
leavve
/leave
leave
I also need help in log functions
But its okay
somber fog
But its okay
You need help with de 2x+10=16 domain or with the practice exercises?
sick citrus
You need help with de 2x+10=16 domain or with the practice exercises?
Its fin It was wrong question
somber fog
Its fin It was wrong question
Ok, np you're right your domian it's +-infinite, do you know why?
proper lion
ts is chem '
proper lion
!done
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sick citrus
Ok, np you're right your domian it's +-infinite, do you know why?
Yea any equation is negative to positive infinity
no domain restrictions
if square root
x greater or equal to 0
This has no restriction
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sick citrus
!done
final saddleBOT
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sick citrus
.close
sick citrus
You already close it, have a nice day.
u too
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bold plover
yo i had a help earlyer but the guy helping me went afk but didnt come back and it closed i wanna know if someone can help me with my geometry
final saddleBOT
yo i had a help earlyer but the guy helping me went afk but didnt come back and ...
bold plover
why are you just telling me the answer? but ok
like how am i ment to solve it on my own?
ill jsut find out my self
.close
final saddleBOT
modest sequoia
final saddleBOT
its 12
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
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vale wave
vale wave
I need help on this question
final saddleBOT
@vale wave Has your question been resolved?
radiant igloo
okay so most of what you did here is correct
we know that the distance between the two centres will be 2d or 8
but since this line segment lies on y=x we can write $\sqrt{(h-0)^2 + (h-0)^2} = 8$
soft zealotBOT
Itsuki
radiant igloo
from here we get $h=4\sqrt{2}$
soft zealotBOT
Itsuki
radiant igloo
hence the centre will have coordinates $(4\sqrt{2},4\sqrt{2})$
soft zealotBOT
Itsuki
radiant igloo
we know the equation of a circle whose coordinates are a,b and has radius r is given by $(x - a)^2 + (y - b)^2 = r^2$
soft zealotBOT
Itsuki
radiant igloo
hence we get $(x - 4\sqrt{2})^2 + (y - 4\sqrt{2})^2 = 25$
soft zealotBOT
Itsuki
radiant igloo
now the rest is just expanding both brackets and we are done
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pine sand
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pine sand
How can I find the radius of this equal circle?
Given data of sides ..prove this is a right angle triangle
desert mantle
just use pythagoras
a^2+b^2=c^2 iff its a right triangle
it works in both directions
worldly spruce
What is an equal circle
candid pulsar
i think congruent circles
worldly spruce
,rccw
soft zealotBOT
pine sand
i think congruent circles
Yes
rancid idol
draw some radii maybe?
proper lion
why do we need circles
worldly spruce
How can I find the radius of this equal circle?
To find its (their?) radius I guess
quick ruin
proper lion
oh missed that
worldly spruce
I mean I am still a bit confused by the question
rancid idol
draw some radii maybe?
.
timber leaf
What does question ask
zinc igloo
i'm so lost
timber leaf
Find radius of congruent circles bro
proper lion
looks like there were two parts
find radii
prove triangle is right
wary juniper
rancid idol
me seeing anyone but op post their working 😔
rancid idol
my solution is use tan(x) and tan(x/2)
sure but you should be guiding the user in doing this rather than doing it for them and giving the answer
final saddleBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
neat wolf
sure but you should be guiding the user in doing this rather than doing it for t...
im wondering if the op is still online or not
so just make him curious and let's he ask what he's thinking
@pine sand
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pine sand
.reopen
final saddleBOT
✅ Original question: #help-36 message
pine sand
I just want to learn the steps
leaden moon
I just want to learn the steps
that's a good mindset you should have
neat wolf
I just want to learn the steps
my method?
pine sand
What should I do next?
pine sand
my method?
Sure
proper lion
too messy, lengthy method, what is done :
divide into 4parts trapezium AEDC,
truangle ABE, triangle BED, triangle BDC
you can see ED = 2r, AC = 25 cm so area of AEDC = 1/2(r+25)r
area of ABE = 15r1/2
area of BDC = 20r1/2
in triangle BED, height = BF - r
you can find BF by equating area of triangle ABC to 25BF/2
150 = 25BF/2
=> BF = 12
so area of triangle BED = 2r(12-r)/2
adding all equaltions will give you area of triangle = 150
you get an ugly expression but the solution will satisfy r
neat wolf
Sure
you see that the angle ECO'=1/2 ECA
we can easily have tan ECA
so we can find tan ECO'
tan ECO' = r/EC
pine sand
you see that the angle ECO'=1/2 ECA
Why it is half?
neat wolf
Why it is half?
you see at the two triangle O'EC and O'KC
O'E = O'K
O'C = O'C
O'EC = O'KC
O'EC = O'KC
so the angle O'EC = O'KC
that is
neat wolf
tan ECO' = r/EC
from here we can find EC = a.r , you will need to find a. Do the same thing to the other circle. We now have BF + FE +EC =25
final saddleBOT
@pine sand Has your question been resolved?
maiden rapids
how can we prove this
final saddleBOT
how can we prove this
shell condor
what are they asking here?
autumn mountain
how can we prove this
what's the exact ques?
rancid idol
prove 9 divides thing
maiden rapids
what have u tried
dont know where to start tbh
maiden rapids
what's the exact ques?
autumn mountain
ion speak french 🙁
drowsy epoch
I recommend inductio
maiden rapids
what proof techniques
they dont have names man
rancid idol
ofc they have names
maiden rapids
I recommend inductio
whats that
rancid idol
yes induction is probably the best one
drowsy epoch
yes induction is probably the best one
then why are you clowning
rancid idol
because i was not asking you
rancid idol
what proof techniques
this was a leading question
somber fog
yes induction is probably the best one
Yeah, it-s a good option.
rancid idol
they dont have names man
do you have a set of notes; what class is this
maiden rapids
do you have a set of notes; what class is this
arithemeitic dans Z
rancid idol
university?
maiden rapids
no
drowsy epoch
this feels like an interrogation
maiden rapids
how do i share something here
rancid idol
if they haven't been taught induction we can only suggest something else?
you can share a screenshot
"Raisonnement par récurrence" <-- you have not been taught this?
or told this
rancid idol
then yes, i recommend you try proving using this
by the way --- proof techniques generally have names. you don't need to know them all, but having a rough idea of the common ones is important
the first 4 probably the ones that matter most
maiden rapids
i got itttt
maiden rapids
by the way --- proof techniques generally have names. you don't need to know the...
ik them in french 💔
anyways i got it
thanks @rancid idol @drowsy epoch @somber fog @autumn mountain @shell condor your help is appreciated