@lethal raft
#help-41
uneven seal
arctic fulcrum
whats the question
lethal raft
I dont know I guessed on it
lethal raft
Yes
lethal raft
1 is not congruent right?
arctic fulcrum
yes
9
and i guess maybe the last one could work but id use HL
sorry i havent done these in a llong time so i may be wrong
lethal raft
Its alright im just trying to pass geometry 😭
For 2 is it not congruent and 3 is AAS?
And 9 is HL?
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knotty dirge
amber waspBOT
knotty dirge
i did 0 to 1 pi (outer^2-inner^2) dy in calculator and its wrong how tf is that wrong
elder harbor
can we see the full work
knotty dirge
knotty dirge
thats what i did
amber waspBOT
what kind of prank is this by my teacher
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
knotty dirge
what
i get it the boundaries are -1 to 1
i didnt even bother graphing it
thanks for help
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amber waspBOT
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floral terrace
amber waspBOT
floral terrace
I have no clue on what to do
do i find y from first equation and then repla?
replace
amber waspBOT
@floral terrace Has your question been resolved?
amber waspBOT
@floral terrace Has your question been resolved?
jade marsh
I’m having an issue with proofs of overlapping triangles. I don’t know where to go after the given.
amber waspBOT
I’m having an issue with proofs of overlapping triangles. I don’t know where to ...
@jade marsh Has your question been resolved?
jade marsh
<@&286206848099549185>
violet raven
what do u need to prove ?
sterile nymph
^ but also, can you see any other equal angles?
jade marsh
oh mb
jade marsh
^ but also, can you see any other equal angles?
no
full question (top hidden point is S)
sterile nymph
There's one pair you can get immediately
jade marsh
There's one pair you can get immediately
3 and 4 vertical angles?
violet raven
bro
u can directly prove them congruent
if 3 = 4
think abt the congruence criteation
sterile nymph
3 and 4 aren't directly angles of SVR and SUT
jade marsh
Which implies what two triangles are similar?
ur and vt?
sterile nymph
Yeah and the middle point they share
jade marsh
yeah
sterile nymph
Oh true, but I was thinking SRV
jade marsh
true also
and does reflexive property work with angles too or just segments
like angle s
sterile nymph
So let's see
You have 1 = 2, and 3 = 4, but we need 5 = 6, which we can get simply from supplementary angles
Reflexive property works for angles
i.e angle TSR = angle RST
jade marsh
You have 1 = 2, and 3 = 4, but we need 5 = 6, which we can get simply from suppl...
can we just assume they're supplements?
sterile nymph
I guess there's no need to show 3=4 or the similar triangles
sterile nymph
can we just assume they're supplements?
We have implicitly that SUR and SVT are straight lines because we are told that SVR and SUT are triangles
So therefore the angles are supplementary. They are on a straight line.
jade marsh
alr
jade marsh
Oh true, but I was thinking SRV
what property would be used for this
sterile nymph
Angles of similar triangles have equal measure
jade marsh
that's not an option
we've learned it but it's not an option in aleks
here's all the rules
actually i figured it out lol
thanks
10/10 service would try again
have a good halloween
.close
amber waspBOT
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Available help channel!
keen pawn
keen pawn
We first attempt to prove that ${0}$ and $\mathcal{L}$ are two sided ideals of $\mathcal{L}$ after which we will prove that there are the only two-sided ideals.
\
We first prove that $\mathcal{L}(V)$ is a two -sided ideal.
\
let $U(v)=T(E(v): V \to V$. Thus by the definition of $\mathcal{L}(V), T(E(v)) \in \mathcal{\epsilon}$
\
Similarly $H(v)= E(T(v)) : V \to V$ . Thus again, by definition of $\mathcal{L}(V), E(T(v)) \in \mathcal{\epsilon}$
\
\
therefore $\mathcal{L}(V)$ is a two -sided ideal of itself.
grizzled pagodaBOT
A dense set
dire hemlock
Wat class is this?
keen pawn
We first attempt to prove that ${0}$ and $\mathcal{L}$ are two sided ideals of $\mathcal{L}$ after which we will prove that there are the only two-sided ideals.
\
We first prove that $\mathcal{L}(V)$ is a two -sided ideal.
\
let $U(v)=T(E(v): V \to V$. Thus by the definition of $\mathcal{L}(V), T(E(v)) \in \mathcal{\epsilon}$
\
Similarly $H(v)= E(T(v)) : V \to V$ . Thus again, by definition of $\mathcal{L}(V), E(T(v)) \in \mathcal{\epsilon}$
\
\
therefore $\mathcal{L}(V)$ is a two -sided ideal of itself.
\
\
\
We now prove that ${0}$ is a two sided ideal
\
To do so,we must first establish what ${0}$ really is .
\
${0} \in \mathcal{L}(v)$ is a linear transformation, such that $ V+{0} =V$. It's evident that one such transformation is the transformation $Q(v) = 0_v$.
\
As established earlier, the additive identity is unique, thus ${0} =Q(v)$ and it's the only such transformation.
\
We thus have established that ${0}$ is a singleton set. Thus $T=E=Q$. So $Q(Q(v)) = Q(0_v)=0_v$.
\
\
therefore ${0}$ is a two-sided ideal too
keen pawn
Wat class is this?
Undergraduare maths
dire hemlock
That doesn’t narrow things down
grizzled pagodaBOT
A dense set
keen pawn
That doesn’t narrow things down
This is from Axler, so technically for a second course in LA
dire hemlock
Ahh ok linear
Dont remember much from the transformation area of linear algebra but i wish soneone else can help u
Gl soldier o7
keen pawn
I mean if no one does, it's fine, I won't be tested on most of this anyway, doing ti just for fun
keen pawn
Gl soldier o7
Thanks!
vast spade
hows the exam prep going?
acoustic bane
amber waspBOT
acoustic bane
I'm not really sure how to go about this
also the notation is a bit different
f[B] is the image of a set B under f
amber waspBOT
@acoustic bane Has your question been resolved?
acoustic bane
Is anyone able to help?
I think I have gotten a start
$f^{-1}[B] = { x \in \mathbb{Z}^2 : f(x) \in B }$
grizzled pagodaBOT
Curry
acoustic bane
so I want to find (a,b) such that gcd(a,b) = 12 and lcm(a,b) = 50 or gcd(a,b) = 5, lcm(a,b) = 175
I know that 12 does not divide 50
so theres a contradiction
so we cant find (a,b) for that case however I am unsure for the second case
lethal steppe
What is the exercise?
acoustic bane
This is the context
acoustic bane
this is the exercise
Well its one part of the exercise but I have done everything else I am just stuck on this bit
I can assure you other parts do not help
lethal steppe
this is the exercise
This doesn’t give me any information on what you’re supposed to do
Are you supposed to denote the preimage of B under f?
acoustic bane
yes
lethal steppe
How is 3 in the greatest common divisor of a and b but not in their least common multiple 
acoustic bane
How is 3 in the greatest common divisor of a and b but not in their least common...
yeah 12 does not divide 50
so I ruled out that we cant find any values of (a,b) such that gcd(a,b) = 12 and lcm(a,b) = 50
acoustic bane
**Curry**
from here
I'm not on how to start from trying to find values such that gcd(a,b) = 5 and lcm(a,b) - 175
lethal steppe
I would stick to the prime factorization
acoustic bane
sorry if I seem dumb I've had a lot of coursework due in so I'm very tired
acoustic bane
oh
so would 25 and 35 work?
as one example
could we generalise this to a whole set?
or would it just be the set containing (25,35)
lethal steppe
Well we could work out the set
lethal steppe
or would it just be the set containing (25,35)
It would at the very least contain its mirror
acoustic bane
yeah (35,25)
lethal steppe
There’s another possibility (and it’s mirror again)
acoustic bane
I was thinking just 5 and 175
lethal steppe
How do we guarantee these are the only 4 options?
acoustic bane
I'm not sure
lethal steppe
Invoke gcd(a,b) = 5 to write a and b as multiples of 5
lethal steppe
You have a*b = 5^3 * 7
Then use this
acoustic bane
ohh
I see
thank you for taking your time out to help me
I should be able to that
lethal steppe
Yw
acoustic bane
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amber waspBOT
amber waspBOT
Available help channel!
acoustic fox
(\textbf{R}) has an order relation (<) with the following properties: \newline
R1. ((Trichotomy)) Given any (x,y \in \textbf{R}) exactly one of the following holds: (x=y), (x<y), (y<x). \newline
R2. ((Transitivity)) If (x < y) and (y < z) then (x < z). \newline
R3. If (x < y) then (x+z < y+z). \newline
R4. If (x < y) and (z > 0) then (xz<yz). \newline
R5. If (x < y) then (-y < -x). \newline
\textbf{Definition.} (x > y) means (y < x). \newline
(\Cooley) \newline
The problem is to prove (x < y \implies x^3 < y^3). \newline
(\Cooley) \newline
I have tried a shit ton of case based approaches (direct and contrapositive), I can't figure out how to proceed without taking cases. Taking cases itself is not helping me 💀. Any hints?
grizzled pagodaBOT
usopper.
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acoustic fox
Also I have proved various corollaries resulting from the 5 properties of the order relation < in R so assume (x \neq 0 \implies x^2 > 0), (0 < x < y \implies 0 < \frac{1}{y} < \frac{1}{x}), and other basic stuff
grizzled pagodaBOT
usopper.
acoustic fox
"Basic". Even what I am trying to prove is basic but for some reason I can't figure out how to approach it
I am gonna try case bashing again because my ego is hurt
keen pawn
Okay, so just to be clear
using the field axioms, you want to prove $x<y \implies x^3<y^3$
grizzled pagodaBOT
A dense set
keen pawn
okay, cool
acoustic fox
using the field axioms, you want to prove $x<y \implies x^3<y^3$
Do field axioms talk about ordered relations?
keen pawn
wait
no
teh field axioms + ordered relations
It's easy to prove that $x^2<y^2$
yes?
grizzled pagodaBOT
A dense set
acoustic fox
Yes
keen pawn
so it also follows that $x^3<xy^2$ and that $x^2y<y^3$ assuming x,y>0
grizzled pagodaBOT
A dense set
keen pawn
adding them
acoustic fox
**A dense set**
The issue after this is that
grizzled pagodaBOT
usopper.
acoustic fox
So it doesn't help us through transitivity
acoustic fox
give me a minute
Bet
grizzled pagodaBOT
A dense set
keen pawn
now we know that $x^3-y^3< 0 \implies (x-y)(x^2+xy+y^2)<0$
grizzled pagodaBOT
A dense set
keen pawn
It's easy to prove that $(x^2+xy+y^2)>0$
grizzled pagodaBOT
A dense set
keen pawn
@acoustic fox
can you take it from here
not the most elegant method, I know
but it works
acoustic fox
**A dense set**
I agree
acoustic fox
**A dense set**
How does (x - y < 0 \implies x^3 - y^3 < 0)? Isn't that what we are trying to prove?
keen pawn
I'm not saying it does
grizzled pagodaBOT
usopper.
keen pawn
I'm saying that we know That $x^3-y^3 = (x-y)(x^2+y^2+xy)$
grizzled pagodaBOT
A dense set
keen pawn
do we know that or no?
keen pawn
we can prove that $(x^2+y^2+xy)>0$
grizzled pagodaBOT
A dense set
acoustic fox
So you are saying ((x-y)(x^2 + xy + y^2) > 0 \implies x - y > 0)?
grizzled pagodaBOT
usopper.
keen pawn
no
I mean yeah
The contrapositive
That's not what I'm exactly saying though
but this works too, sure
acoustic fox
I mean yeah
But how do we know that ((x-y)(x^2 + xy + y^2) > 0) to begin with?
grizzled pagodaBOT
usopper.
acoustic fox
but this works too, sure
How?
keen pawn
OK, let me sketch out a proof
lethal steppe
**usopper.**
I think the arrow has to go in the other direction
acoustic fox
Ohk
I mean that's what we are trying to prove
That's the original statement
Equivalent to it I mean
lethal steppe
We are assuming x-y<0
acoustic fox
I have proven a corollary that (x < y \iff x - y < 0). So to prove (x < y \implies x^3 < y^3) is same as proving (x - y < 0 \implies (x-y)(x^2 + xy + y^2) > 0), I think.
grizzled pagodaBOT
usopper.
acoustic fox
We are assuming x-y<0
Yes
lethal steppe
**usopper.**
Well yeah okay with the signs flipped that way I suppose this is correct
keen pawn
We already know $(x-y)<0$
\
Now as the reader can verify $(x-y)(x^2+y^2+xy)= x^3-y^3$
\
We also ask the reader to verify that $x^2+y^2+xy \geq 0$
\
From this it follows that $(x-y)(x^2+y^2+xy)<0$
grizzled pagodaBOT
A dense set
lethal steppe
Well yeah okay with the signs flipped that way I suppose this is correct
But the weaker form. So >=
acoustic fox
Should I consider the contrapositive? (x^3 \geq y^3 \implies x \geq y)?
grizzled pagodaBOT
usopper.
acoustic fox
I tried but couldn't make it work using cases earlier
keen pawn
I feel a direct proof would work better
lethal steppe
I don’t see the advantage to the contrapositive either
acoustic fox
Yeah I will try what dense set said
amber waspBOT
@acoustic fox Has your question been resolved?
amber waspBOT
acoustic fox
.reopen
amber waspBOT
✅
acoustic fox
Ok let's prove this shi and get it over with 💀
I need to prove (x^2 + y^2 + xy > 0)
grizzled pagodaBOT
usopper.
acoustic fox
We know (x^2 > 0) and (y^2 > 0). So (x^2 + y^2 > y^2 > 0)
grizzled pagodaBOT
usopper.
acoustic fox
(\frac{1}{2} > 0)
grizzled pagodaBOT
usopper.
grizzled pagodaBOT
usopper.
acoustic fox
Now (1 \cdot (x^2 + y^2) > \frac{1}{2} \cdot (x^2 + y^2))
grizzled pagodaBOT
usopper.
acoustic fox
Ok nice. I think I'm done
The equivalent statement to prove is (x - y < 0 \implies x^3 - y^3 < 0 ; (\because P1)). Now, (x^3 - y^3 = (x-y)(x^2 + xy + y^2)). We know ((x + y)^2 \geq 0 ; (\because P7) \implies x^2 + y^2 + 2xy \geq 0 \implies x^2 + y^2 \geq -2xy ; (\because R3) \implies x^2 + y^2 \cdot \frac{1}{2} \geq -2xy \cdot \frac{1}{2} ; (\because R4) \implies \frac{x^2 + y^2}{2} \geq -xy).
grizzled pagodaBOT
usopper.
acoustic fox
Works? @keen pawn
Wait
I need to remove the or equal to sign
Because at least one of x,y is non-zero
Because x<y
keen pawn
I'm eating, give me 3 minutes
proven vapor
The equivalent statement to prove is \(x - y < 0 \implies x^3 - y^3 < 0 \; (\bec...
you should separate your lines with \\
indigo cloud
no. either double enter or stay as a paragraph. \\ shouldnt be used
proven vapor
no. either double enter or stay as a paragraph. \\\\ shouldnt be used
why is that?
indigo cloud
just some style advice I picked up early. and frankly I dont like \\
acoustic fox
Wait I feel like making it greater than equal to is making my proof messy
I am trying this approach
The equivalent statement to prove is (x - y < 0 \implies x^3 - y^3 < 0 ; (\because P1)). Now, (x^3 - y^3 = (x-y)(x^2 + xy + y^2)). Now, (x < y \implies x \neq y ; (\because R1) \implies x - y \neq 0 ; (\because R3) \implies (x-y)^2 > 0 ; (\because P7)).
grizzled pagodaBOT
usopper.
acoustic fox
no. either double enter or stay as a paragraph. \\\\ shouldnt be used
What if I am on overleaf?
indigo cloud
the proof isnt unreadable because its missing \\, its unreadable because of the symbol overload
acoustic fox
Yeah I don't like symbols usually as well but I want to not overload the solutions the problems in this worksheet with words so
Usually I avoid symbols
indigo cloud
What if I am on overleaf?
same
acoustic fox
I mean what if you need to have a paragraph break
indigo cloud
line 1
line 2
acoustic fox
So double enter works?
On overleaf
Tf
Noted
The equivalent statement to prove is (x - y < 0 \implies x^3 - y^3 < 0.) Now, (x^3 - y^3 = (x-y)(x^2 + xy + y^2)). Now, (x < y \implies x \neq y \implies x - y \neq 0 \implies (x-y)^2 > 0). Therefore (x^2 + y^2 - 2xy > 0).
grizzled pagodaBOT
usopper.
acoustic fox
Now I have to work with this to prove that (x^2 + y^2 + xy > 0)
grizzled pagodaBOT
usopper.
indigo cloud
complete the square
grizzled pagodaBOT
usopper.
acoustic fox
Hmm
indigo cloud
how would you complete the square for x^2+3x+9
keen pawn
sorry denascite, didn't realise you were already here
indigo cloud
I mean we could also do cases depending on whether x,y are positive or negative
the only interesting case is that they are both positive
acoustic fox
how would you complete the square for x^2+3x+9
((x+3)^2 - 3x)
grizzled pagodaBOT
usopper.
indigo cloud
thats not how completing the square works
indigo cloud
you want (x+something)^2 + number
acoustic fox
(2 \cdot x \cdot something = 3x) so we get (something = \frac{3}{2}). Hence, (x^2 + 3x + 9 = (x+\frac{3}{2})^2 + \frac{27}{4})
grizzled pagodaBOT
usopper.
grizzled pagodaBOT
usopper.
indigo cloud
and is that >0 ?
acoustic fox
Yes 💀
Bruh
Ok thanks
Let me just complete the original problem then I'll close it in a bit
The equivalent statement to prove is (x - y < 0 \implies x^3 - y^3 < 0.) Now, (x^3 - y^3 = (x-y)(x^2 + xy + y^2)). Notice that (x^2 + xy + y^2 = (x + \frac{y}{2})^2 + \frac{3y^2}{4}).
Case 1: y = 0
Then (x \neq 0), so ((x + \frac{y}{2})^2 + \frac{3y^2}{4} = x^2 > 0).
Case 2: (y \neq 0)
Then (y^2 > 0 \implies y^2 \cdot \frac{3}{4} > 0 \cdot \frac{3}{4} \implies \frac{3y^2}{4} > 0 \implies \frac{3y^2}{4} + (x + \frac{y}{2})^2 > 0 + (x + \frac{y}{2})^2 = (x + \frac{y}{2})^2 \geq 0). Hence we have ((x + \frac{y}{2})^2 + \frac{3y^2}{4} > 0).
Therefore ((x-y)(x^2 + xy + y^2) < 0 \implies x^3 - y^3 < 0).
grizzled pagodaBOT
usopper.
grizzled pagodaBOT
usopper.
acoustic fox
Also, is there a simpler proof?
This was much harder than I expected. Even after completing the squares there are a lot of steps compared to the other corollaries
indigo cloud
all things considered the steps after completing the square are just a bit of cleanup
the hard part was before
acoustic fox
I mean true
It was just write-up after you notice completing the square
Okay thanks
.close
amber waspBOT
amber waspBOT
Available help channel!
split sail
i’m stuck
molten shale
where r u stuck on
slender blaze
you should find the equation of line M
remember product of perpendicular slopes is -1
then you should have the slope of line M, and a point it goes through
amber waspBOT
@split sail Has your question been resolved?
solemn escarp
might sound like a dumb question, but is there any other way we can know that a triangle is right angled other than a^2 + b^2 = c^2 without drawing?
tough cradle
what information is given to you?
pliant copper
yes there are lots of ways depending on other additional info, see pranjals answer for your specific case
amber waspBOT
might sound like a dumb question, but is there any other way we can know that a ...
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
warm burrow
law of cosines probably
amber waspBOT
@solemn escarp Has your question been resolved?
split sail
amber waspBOT
split sail
Not really a physics question but I would required help with this neverthelesss
here's what I have done uptill now:
$$N_y = N_x_\text{initial} - N_x_\text{left}$$
$$N_y = N_x_\text{initial} - \frac{N_x_\text{initial}}{2^\text{H}}$$
grizzled pagodaBOT
Edmund Cloudsley
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primal holly
split sail
here H is supposed to be the number of half lives
primal holly
would you just not use the formula for time directly here
primal holly
the one with logarithms?
primal holly
alright
split sail
could you tell me what it is?
primal holly
Nuclear decay follows first order kinetics, that is the decay speed is proportional to the first power of the number of nuclei left
grizzled pagodaBOT
neon
primal holly
Well you can find it yourself
But first I must ask are you fine with this procedure
split sail
I would require half life for it tho
primal holly
Considering you haven't covered it yet
split sail
Ofcourse! I would love to learn something new
primal holly
I would require half life for it tho
We also have the differential equation
quick ridge
am i bugging or is this all terribly unnecessary
primal holly
Ofcourse! I would love to learn something new
Alright that's great
primal holly
am i bugging or is this all terribly unnecessary
Well go ahead then
This is the first thing that came to mind
quick ridge
Amount of X remaining after x half lives is (1/2)^x, Amount of Y after x half lives is 1 - (1/2)^x
so just two half lives
3/4 : 1/4
12 minutes = 2 half lives
1 half life = 6 minutes
primal holly
smart
split sail
Amount of X remaining after x half lives is (1/2)^x, Amount of Y after x half li...
I got this bit
I just didn't quite understand how you arrived at 3/4:1/4
quick ridge
yea well i just went through the half life proportions in my head
and the second one worked
split sail
is there anyway to do this analytically?
quick ridge
Amount of X remaining after x half lives is (1/2)^x, Amount of Y after x half li...
i divided the Y by X
split sail
yeah I was doing it in my notebook and I got this equation too
thanks so much guys
@primal holly and @quick ridge
have a wonderful rest of your day
.close
amber waspBOT
amber waspBOT
Available help channel!
keen pawn
I would like to prove this
amber waspBOT
I would like to prove this
keen pawn
We start by assuming that dim(V)=n. We then assume that $dim(null(T))=r$
grizzled pagodaBOT
A dense set
proud ruin
don't know the proof of the top of my head, but if u get no where, this is a rlly good resource for this proof https://www.youtube.com/watch?v=yZ52iqxaiCc
keen pawn
I'll look at it if I get no where
thganks
We start by assuming that dim(V)=n. We then assume that $dim(null(T))=r$.We now attempt to prove that dim(range (T)) = n-r
grizzled pagodaBOT
A dense set
shrewd elm
so far so good
keen pawn
To do so we consider $\beta_1 = {e_1,e_2,\dots,e_r,\dots, e_n\}$ to be the basis of $V$. Let $e_1,e_2,\dots, e_r$ form a basis of null (T)
grizzled pagodaBOT
A dense set
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shrewd elm
^ this is ok but i might suggest wording it slightly differently, start with the basis of null(T) and use the theorem that says you can extend it to a basis of V
shrewd elm
not necessarily no
for example, (1,0,0), (0,1,0), (0,0,1) is a basis for R^3, but you can't reduce it to a basis for span(1,1,1)
grizzled pagodaBOT
A dense set
keen pawn
We now extend this to a basis of $V$, let that basis be $e_1,e_2,\dots, e_r,f_1,f_2,\dots f_{n-r}$
grizzled pagodaBOT
A dense set
keen pawn
The range of $T$ would be given by $\alpha_1T(e_1) + \dots + \alpha_n T(e_n)+ \beta_1T(f_1)+\beta_2T(f_2)+ \dots +\beta_{n-r} T(f_{n-r})
grizzled pagodaBOT
A dense set
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keen pawn
we now attempt to prove that $T(f_1),T(f_2), \dots T(f_n)$ is a linearly independent set
grizzled pagodaBOT
A dense set
shrewd elm
yep this is the right strategy
keen pawn
I was thinking contradiction
I assume it's not LI and show that we then arrive at a contradiction
shrewd elm
(small correction, that should be T(f_(n-r)) at the end, not T(f_n)
keen pawn
that's pretty easy
If $T(f_1),T(f_2), \dots T(f_{n-r})$ was linearly depndent, then by linearity, that would imply that $f_i$ is linarly depndent too, which is obviously wrong , thus the OG set must be LI
grizzled pagodaBOT
A dense set
shrewd elm
you need to show the details
why would that imply that the f_i are linearly dependent?
(it's not true for arbitrary choices of vectors, so there must be something special about the f_i's that you need to use)
keen pawn
because then there exist non-zero $\alpha_i$ such that $\sum_{i=1}^{n-r} \alpha_if_i=0$
grizzled pagodaBOT
A dense set
shrewd elm
(at least, you don't know yet if it is)
keen pawn
what
shrewd elm
you know that $\sum_{i=1}^{n-r}\alpha_i T(f_i) = 0$
grizzled pagodaBOT
Bungo
keen pawn
yes, we now use liearity
shrewd elm
and by linearity that means $T\left(\sum_{i=1}^{n-r}\alpha_i f_i\right) = 0$
grizzled pagodaBOT
Bungo
keen pawn
yes
shrewd elm
why does that imply that the thing inside the brackets is zero?
T isn't necessarily injective
keen pawn
We also know that T(0)=0
shrewd elm
you're on the right track
this tells you something about the sum
it's contained in what
shrewd elm
yep
keen pawn
But we know that these vectors don't map to 0_v
keen pawn
We know that \sum_{i=1}^{n-r} a_f_i doesn't lie in the span of e_1,e_2,\dots e_N
shrewd elm
the span of ker(T) is ker(T) itself, because ker(T) is a subspace
and we know the opposite actually
if T maps that sum to zero, then that sum is in ker(T), by definition
so you can write it as a linear combination of e_1 through e_r
shrewd elm
yep you are
shrewd elm
well so far you have established that
$$\sum_{i=1}^{n-r}\alpha_i f_i = \sum_{j=1}^r \beta_j e_j$$
grizzled pagodaBOT
Bungo
shrewd elm
for some coefficients alpha_i and beta_j
that's because the LHS is in ker(T) and every element of ker(T) looks like the RHS
so just rearrange that a bit
shrewd elm
and why is that?
keen pawn
e_i, f_i togetehr form a basis of $V$
grizzled pagodaBOT
A dense set
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shrewd elm
yep, and so...?
grizzled pagodaBOT
The current time for physicsrocks is 02:05 AM (IST) on Fri, 01/11/2024.
amber waspBOT
keen pawn
gn
shrewd elm
ok, gl
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Available help channel!
little hornet
Complete the equation by filling in an integer in the box so that the equation has exactly one real solution (a double root).
Can someone explain in steps how to solve it?
$3x^2 -4=$
grizzled pagodaBOT
EmprFN
flint igloo
for me two ways come to mind
the foolproof one: since 3x^2=0 only has one root (x = 0), we can subtract 4 from each side and get the equation 3x^2-4=-4
amber waspBOT
flint igloo
the second one was in fact so complicated that i couldn't even formulate it 💀
.close
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obsidian barn
Hi, I'm trying to solve a problem about a triple integral and I want to know if I'm headed to the right direction.
obsidian barn
Basically I have to calculate the triple integral of x^2+y^2+z^2 on the volume determined by the planes x=0, y=0, z=0 and x+y+z=a (with a>0). This is what I've done so far:
(I wrote f(x, y, z) instead of x^2+y^2+z^2 to fit it on the page)
patent raptor
Basically I have to calculate the triple integral of x^2+y^2+z^2 on the volume d...
I think only the bounds for z are wrong
instead from z=0 to z=a-x-y
obsidian barn
I think only the bounds for z are wrong
Oh, you are right. Thank you very much!
.close
amber waspBOT
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Available help channel!
elder fiber
amber waspBOT
elder fiber
i used the advice given to me but it didn't work out
i have one more attempt on this problem :/
this problem is very similar, and i have more attempts so i'd like to solve for this first honestly
i tried using natural log to bring t down but it did not work
my work
warm burrow
hm
elder fiber
the rest of the work not shown was done in my calculator
warm burrow
,tex .log rules
grizzled pagodaBOT
Steel
warm burrow
ok nice its here
grizzled pagodaBOT
The following error occured while calculating:
Error: Undefined function ln
warm burrow
well more exactly i got
27.45067233
but your thing calculated is basically the same
elder fiber
my calculation came out to (30/92)/0.96
warm burrow
my calculation came out to (30/92)/0.96
you cannot just cancel the logs
i see where the problem came from though
elder fiber
this one
elder fiber
did i not
warm burrow
likely not
elder fiber
hmm
warm burrow
but i dont know maybe im wron
grizzled pagodaBOT
Steel
warm burrow
i am confused here
on the step where you used ln on a bunch of things
i wouldve probably just simplified into $1.38 = (1.0875)^{4t}$
grizzled pagodaBOT
Steel
warm burrow
my math is not mathing lol
elder fiber
would i be able to do log_b x = y for that
turn it into log and simplify?
but how would that be different from log and ln
warm burrow
wait so your answer was .96 years
elder fiber
yes
warm burrow
grizzled pagodaBOT
elder fiber
ohh it was 4* the initial ln
warm burrow
yeah
elder fiber
so 4.34 rounding to the nearest hundreth
warm burrow
when you round to the nearest hundredth you just look at the thousandth
warm burrow
ok im going to sleep lol
amber waspBOT
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Available help channel!
elder fiber
amber waspBOT
elder fiber
my work
split sail
Can I grt help man
elder fiber
Can I grt help man
yes 100%, go to one of the other open help rooms and open it up with your question :)
elder fiber
yeah symplify within the parenthaseis adn then divide by the larger number on the outside
i always seem to do something wrong afterwards though
elder creek
now just do alegbra
elder fiber
cause rate is also up there to by 40
elder fiber
not what we're learning
elder fiber
yes
elder creek
thing...
elder fiber
no?
elder fiber
yeah but 40 isn't the exact same
if it's rt
cause that's 40 * rate
not just 40
a(1.08)^3.2 = 13000000 would look more like it
little widget
For the first one, you can define a function P(t) that describes the number of people after t years since 1993. Notice it starts at 13 million and after each year it increases by 8%, which can be get by multiplying by 1.08 every year, so $P(t) = 13,000,000 \cdot (1.08)^t$
since 1953 is 40 years less than 1993, you can plug in t = -40
grizzled pagodaBOT
kaue
elder fiber
elder fiber
,rotate
grizzled pagodaBOT
elder fiber
?
elder creek
ur fomrula is wrong
its not compound
it just decreases a rate of 8% a year man
thats why ur answer makes 0 sense
elder fiber
compound would mean the rate is divided by something
i dont see it being divided by anything
plus it said annually, that's the formula for that
elder creek
your formula is just P(x) = P0(1+r)^t
elder fiber
that's not what was taught
little widget
why the 0.08 on the exponent
elder creek
Changing it to 1.08 isn’t going to fix your problem
Keep trying . But I’m pretty sure it won’t
little widget
wouldn't it be a = 13mil(1.08)^-40
yes thats the population
elder fiber
598402?
elder fiber
we cracked
little widget
For the first one, you can define a function P(t) that describes the number of p...
you can use a similar approach to solve the others
elder fiber
bears 926
how would you solve THIS?
idk how you can solve for an unknown
ohhh wait
not p(a) but just x < 120
could i put 119? or are decimals needed
is it 73.1555~
little widget
could i put 119? or are decimals needed
its asking when the population is less than 120, not the number of years passed
elder fiber
it says in the year though
little widget
its asking in what year the population is less than 120
little widget
it's saying P(a) < 120, which means $11000(0.94)^a < 120$
grizzled pagodaBOT
kaue
little widget
a is the number of years since 1953
if you find 'a' you can find the year where the population is less than 120 by doing 1953 + a
elder fiber
72
no..
74
it's 73.02
i put in 72 73 and 74 into the system though and they all said it was wrong
i only have one more attempt :(
little widget
because its not asking for how many years since 1953
little widget
yeah
amber waspBOT
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Available help channel!
ember marlin
if y = 7pi^(4), then what is y'?
amber waspBOT
if y = 7pi^(4), then what is y'?
final matrix
I mean what are you taking the derivative with respect to?
because if this is with respect to x it would just be 0
quick ridge
$\frac{d}{d\pi}$
grizzled pagodaBOT
knief
ember marlin
because if this is with respect to x it would just be 0
wait can you show exactly how?
is pi just like a normal number?
final matrix
is pi just like a normal number?
pi is a constant
ember marlin
but it wouldn't be like x right?
final matrix
so 7*pi^4 is also just a constant
quick ridge
$\pi = 3.1415…$
grizzled pagodaBOT
knief
ember marlin
oh
quick ridge
,w 7(pi)^4
ember marlin
and even if its pi, i can still use the power rule on it?
grizzled pagodaBOT
quick ridge
and even if its pi, i can still use the power rule on it?
nuh uh
$\frac{d}{dx} c = 0$
grizzled pagodaBOT
knief
final matrix
and even if its pi, i can still use the power rule on it?
depends on what you're taking the derivative w.r.t
quick ridge
math is life sir
grizzled plume
no math > life
quick ridge
what’s up death
grizzled plume
understandable
quick ridge
ember marlin
the question is:
If y = 7pi^(4), then y' is...
what respect would that be asking it to
ember marlin
but there isn't an x tho
ember marlin
so it would just be 0
grizzled plume
the result would just be 0
ember marlin
right?
grizzled plume
yup
elder creek
ember marlin
ok tyty
elder creek
deriative ah 🙏😭
grizzled plume
lol
ember marlin
oh nah
ember marlin
oh yeah, cuz there isn't an x value
grizzled plume
yea
elder creek
same concept
ember marlin
ic
It says find the equation of the line tangent to:
y = (x+3)/(x^(2)+1) at x = 1, and I used the quotient rule but I keep getting -1/2 what am I doing wrong?
final matrix
ember marlin
quotient rule
final matrix
do you have any work to show?
grizzled plume
whats your working
grizzled plume
one part is wrong
ember marlin
what is?
grizzled plume
do you know how to calculate the derivative of x^2 +1
grizzled plume
uh the power rule
ember marlin
ohhh
grizzled plume
anyway @ember marlin tell me when you get the answer
ember marlin
It says find the equation of the line tangent to:
y = (x+3)/(x^(2)+1) at x = 1, ...
And for the entire problem I got y = -3/2x + 7/2
ember marlin
alright thanks so much
grizzled plume
np
ember marlin
Wait a question is asking that the derivative of the function f is given by f'(x) = -3x +4 for all x. If f(-1) = 6, what is the equation of the line tangent to the graph of f at x = -1
wouldn't that simply just be the f'(x) that it gives us
if its for all values of x
oh wait nm, I solved it
amber waspBOT
@ember marlin Has your question been resolved?
verbal arrow
Hi
idle light
Hi
verbal arrow
Can u please help me
I have some questions
About some papers I don't think they are hard I just can get to understand them
tender cipher
Just ask
amber waspBOT
@verbal arrow Has your question been resolved?
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Available help channel!
keen pawn
amber waspBOT
keen pawn
I feel like I either have to show this isn't closed under addition or doesn't have the identity
full sinew
true true true
keen pawn
It does have the identity, as $dim(null (O)) = 5$
grizzled pagodaBOT
A dense set
keen pawn
Consider $T(a,b,c,d,e) = (a,b,0 ,0,0,)$ ; $dim(null (T)) =3$. And we now consider $L(a,b,c,d,e) = (0,0,0,d,e); dim(null(T)) = 3$.It then follows that $(T+L)(a,b,c,d,e) = (a,b,0,d,e); dim(null(T+L))=0$. Thus the set isn't closed under addition and is hence not a vector space
<@&286206848099549185>
grizzled pagodaBOT
A dense set
keen pawn
Is this fine?
robust isle
yes
keen pawn
Thanks
amber waspBOT
@keen pawn Has your question been resolved?
keen pawn
.close
slate gale
hi guys, for 1b, my book says the answer is 3. i don't understand how they got 3
(btw these are the answer for 1a:
CF = 2.8
DF = 2.95
EF = 2.99)
true jackal
You can see that the nearer you go to x = 1 the more the gradient (second row of that table) approaches a value of 3
slate gale
You can see that the nearer you go to x = 1 the more the gradient (second row of...
which on is x =1?
is it the x coordinate on the points?
like point A (0,0)
true jackal
Wdym?
The x coordinates of your points are:
0
0.5
0.8
0.95
0.99
1
Therefore you can see that x goes nearer and nearer to 1
true jackal
Because at x = 1 the value 2.99 represents an approximation of the gradient, since it's the slope of the secant EF
In other words, EF is just a secant at point F (where x = 1), it's not a tangent
slate gale
In other words, EF is just a secant at point F (where x = 1), it's not a tangent
thank you so much
.close
amber waspBOT
amber waspBOT
Available help channel!
solemn escarp
i asked this yesterday but i totally forgot about it, sorry : is there any other way to know a triangle is right angles other than a^2 + b^2 = c^2
solemn escarp
this isnt a question from anywhere, i was curious
naive pivot
i asked this yesterday but i totally forgot about it, sorry : is there any other...
ngl that’s interesting
void fulcrum
If the opposing angles add up to 90 degrees
solemn escarp
well lets say we dont know any angles, just sides
turbid notch
you can also dive into more complex things like dot product
void fulcrum
If the product of gradients of the adjacent and opposite sides is -1
solemn escarp
how do we know the gradient of a side?
void fulcrum
Well
Really depends
But in most cases if you're given coordinates of the triangle
or other angles
Another way (given all 3 sides) is substituting those values into the Area of a Scalene Triangle Formula and see if it equates to 1/2 bh
If its true then the triangle is right angled
You can also experiment with vectors and dot products like the guy said before
But thats for another topic
solemn escarp
oh okay
frank zealot
or if u have ABC a triangle and J the point on segment [BC] such that JB = JA = JC (equally distant from all points of the triangle) then ABC is a right triangle in A
frank zealot
its actually a geometric rule as well
amber waspBOT
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Available help channel!
bronze basalt
if C a cercle and O its center
A and B two points in the cercles different from each other
let I midpoint of the segment AB
is AB perpendicular to OI ?
edgy oyster
if C a cercle and O its center
A and B two points in the cercles different from ...
Yes.
amber waspBOT
solar gust
AB have to be different than the diameter tho
bronze basalt
AB have to be different than the diameter tho
yeah I know
I just forgot to mention that I is different to O
amber waspBOT
Available help channel!
glad pulsar
amber waspBOT
lavish zenith
6! is da solution
glad pulsar
6! is da solution
Wrong
lavish zenith
Wrong
that is supposing you have 6 colours at your disposition
how many colours are there to use ?
glad pulsar
how many colours are there to use ?
6
lavish zenith
then 6!
glad pulsar
no
lavish zenith
oh
glad pulsar
answer is 30
lavish zenith
why not ?
glad pulsar
idk how
lavish zenith
wadasigma
glad pulsar
ye it is
glad pulsar
?
lavish zenith
cause you aint making the difference between the faces
glad pulsar
cause you aint making the difference between the faces
whaat
lavish zenith
idk
glad pulsar
samr
lavish zenith
gotta think
glad pulsar
yed
glad pulsar
first you have 1/6 chance for the first face
just choose a colour begin woth
with
it has one probability
then other has 1/5
radiant lynx
Do you need to use all the colors?
glad pulsar
Do you need to use all the colors?
ig
radiant lynx
If so take all permutations and divide by the size of its symmetry group, S_4. So $\frac{6!}{24} = 30$.
grizzled pagodaBOT
LooseEthics
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glad pulsar
If so take all permutations and divide by the size of its symmetry group, S_4. S...
what
what is symmetry group
radiant lynx
Some permutations are equivalent because you can get one from the other just by rotating the cube.
The symmetry group describes these rotations.
glad pulsar
Some permutations are equivalent because you can get one from the other just by ...
ye
glad pulsar
The symmetry group describes these rotations.
are u using circular permutations
like a necklace
glad pulsar
hint says use circular permutation
<@&286206848099549185>
amber waspBOT
@glad pulsar Has your question been resolved?
amber waspBOT
amber waspBOT
Available help channel!
upbeat gyro
amber waspBOT
upbeat gyro
In here I need to find the slope of the tangent. How would I start as I have no function to begin with
I only know I have a tangent in the coords of (1, -6)
My thought was to make a secant that starts at (1, -6) til (3, 0)
However, I'm not sure if thats recommended
bronze ore
We cannot find the exact slope of the tangent line to the parabola without knowing the equation of the parabola. The provided information only gives approximate intersection points with a line, not the parabola's equation itself. To find the slope of the tangent line at a specific point on the parabola, we would need the derivative of the parabola's equation, evaluated at that point's x-coordinate
naive pivot
you can estimate
upbeat gyro
By taking a limit?
naive pivot
yea but better
you remember the slope formula for linear functions ?