#help-33
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@drifting goblet this is just straight up wrong
Is my answer correct?
Thank you
Yeah i am just confirming because my format isnβt their format
algebra lol
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Never seen a problem like this sos
( can anyone please help?
do you remember how numbers are written in other bases? 
so you know how the 1s digit in base 10 corresponds to, well, ones?
and the tens digit corresponds to tens, the hundreds corresponds to hundreds, etc?
Kinda 
it turns out that you can write this as a sum
for example, the number 123 in base 10 is really $1 \times 100 + 2 \times 10 + 3 \times 1$
higher!
you can also write that as $123 = 1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0$
higher!
does this make sense so far? 
all we're doing is writing 123 as a sum of powers of 10 instead of in its usual way
uhh
you can read this as "one hundred + two tens + 3 ones"
the plan btw is to simply convert these numbers from quaternary and quinary into decimal and then add "as normal"
mhm
lem see rw
the point here is that we can do the same thing for numbers in other bases
but instead of using powers of 10, we're gonna use powers of the base instead
for example, the number $101_{\text{two}}$ is the sum [101_{\text{two}} = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5]
higher!
as Ann said, we can just turn all our numbers into decimal form via this process and add normally 
so like if its 5 like 4,3,2,1,0 as ^4,3,2,1,0
I think Im kinda stupid at this point
yea
right
so you know that when writing a number in the ordinary way, going right to left you have a ones place, a tens place, a hundreds place and so on, yeah?
the value of each place is ten times the one before it. that is what makes our number system "base ten".
oh yea
but the fun thing is that you could conceive of numeration systems with another number as the base.
but then if i do that the answer will be over 1000 right?
for example, in base six, you would have a ones place followed by a sixes place, then a thirty-sixes place, and then a two-hundred-sixteens place, and so on. (i am stating these in familiar/decimal terms so it is a bit clunky)
we'll get back to your problem in a bit
Alright
for now i want to go over in a bit more detail about how alternative bases work
so just to show you how the same number looks in two different bases
the number 176 (decimal)
its decimal representation can be read as:
- 1 hundred
- 7 tens
- 6 ones
(and in fact this structure is reflected in English number words. "seventy" instead of "seven tens", but still)

the same number written in base six, however, would look like 452, to be read as:
- 4 "thirty-sixes"
- 5 sixes
- 2 ones
how would it be six?
wdym
like why 4" thirty-sixes"
are you asking why we're counting thirtysixes, or are you asking why we got 4 of them
why counting thirtysixes
it's six times six
oh
a thirty-six is six sixes in the same way as a hundred is ten tens
omg why are you multiplying these
no
(decimal) 4Γ36 = 144, but 5Γ36 = 180, which is too much.
if i wanted to talk about powers of six then i would
read this again. in base six, each place has value equal to six times the previous.
6**^2** = 36
oop
the "4" in "452_six" means 4Γ6^2
yes
that's 1234(quinary) spelled out, yeah
yeah that's 123(quaternary)
yeah it works out to 221
Thank youu
I think i js didn't know that kinda problem <3
thank you for helping <3
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Can someone help me to understand the conclusion of this proof? I understand the intention is to show that we have a homeomorphism from the restriction of p to U, but I don't undestand how Munkres actually... gets to the conclusion
I see that he proved surjectivity and I understand that, but he only proved injectivity for the closure so saying the closure is a homeomorphism makes sense to me, but then he just says that the open sets are also fine but how?
Restriction of a homeomorphism is a homeomorphism still
So since he's proven it for the closure, he can restrict to the interior
Okay, that makes sense... I guess I wasn't really thinking about the fact that the closure is bigger
Understandable
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Can u solve number 35 I got 25 as answer
,rccw
show your work prob
I did my work In exam this morning
Have you at least noted something or somewhat remember your procedure?
Ye
Lemme type it
Nvm I will write and send 1 min
Oh gosh I screwed up so bad
I did it wrong
In fact, well, ill suppose ill guide you back through it
Ill take it as you know your trig basics
K
First of all, you know that PQO is an isosceles with two sides = 5, other = 8
For ease of use, we will discard half the diagram and end up with something like this
Kk
okay we looking at it from a different perspective, but yeah
base is 3
Given the nature of the problem, where we asked by the tangent of the circle, we dont really care for the other "side" anyways
Do you know how to find that angle?
How base is 3 guys? My friends told the same
it's just pythagorean theorem man
Is one of the most known right triangles, but you could also find it using pythagoras, any clue how?
Wait how u find 5?!
dude....radius
is the radius of the circle, its just info they gave you
i think knowing TR is more important
OHHH
since law of cosines could really help here
not needed really
So what is TO?
#help-33 message
coming back, do you know how to find this angle?
I understood from here
i abuse cosine law so dont blame me lol
not precisely sure what this means π₯
Might be easier to just show they are similar triangles.
DUDE THEY DIDNT TEACH THAT LAW IN SCHOOL
Proportionality and similarity that's all lol
oh, then they do expect similarity
well, time to show they are similar then
cosine law is in 10th grade
It's not in my syllabus...
yeah i think its possible to do congruency
Sadge
U meant this?
Oh yeah
Yeah, i was thinking you knew a few more laws about trig, np anyways, it can be solved like this too
Oh okey
If you look at it carefully, you can see that you have two right triangles sharing a side, and they have a right angle in common.
You could take the smaller (the one in the right), rotate it, and find they are similar.
Still, you will have to find a factor of proportion. Can you see what you need to match?
AAA proves similarity.
You know the bottom side from the small one is 3.
You know the right side of the bigger one is 4, and since those are the sides that are similar, you need to find a number that makes it so:
3 * a = 4
Once you find it, you can multiply 5 by it, and youll get the answer
Similar triangles have corresponding sides. Determine which sides correspond to each other.
Yeah, thats correct.
O
Hey can u gimme a piece of advice to get better on math
I'm only getting 57/80 in this test
What kind of advice are you looking for?
To get better marks
Let me ask you this question. What do you think is the purpose of school?
To study
That's it?
Uh yeah 
Public school serves several purposes. One of those purposes is to introduce you to areas of study that you not otherwise have had access to. It's not uncommon for children to go into the same field of work as their parents. This is because parents pass on the knowledge of what they know to their children. Children grow up and go into the same field of work because it is what they are mostly knowledgable about.
Well yeah
Another purpose is to teach you the skills that you will need to learn on your own. This is an aspect which causes rural students to fail in college. High schools which do not challenge a student enough to force them to develop good studying habits hinder students in college.
O
Can confirm, rural student who struggled through college here
You should be learning how to study. Taking notes is a skill that needs to be developed.
So uh what's ur conclusion?
That's it. Develop your studying skills.
Oh okey
Everybody has different studying techniques that work for them. I have my own.
Btw can I ask smth?
Yes.
I was a web developer. In college, I had my own difficultiies related to my previous comment about rural schools. It took quite some time for me to understand why I was failing. At some point, I took a philosophy class that taught me how to ask questions. A well-thought out question gives you well-thought out answers.
Oh that's rlly cool
I wanna study law in canada in fact my dad knows a person who can arrange me for it and my seat git confirmed
My dad studied law too so I'm doing it too
Same thing u said
My one tip for learning math well. As you are solving a math problem, write out your thought process as you are solving the problem. Those notes to yourself are useful when it comes to studying. If the notes don't make sense when you re-read them, redo the notes so they do make sense.
If you need to use a formula, write out the formula. Repetition makes things easier to remember.
Oh okey but I need to tell u some harsh truths abt me
Ok.
suck it up and deal with it
That's something you will have to correct. If concentrating is problematic for you, eliminate distractions. If you have a study area, get a big cardboard box and make a wall in front of you.
no one really cares unless it makes them feel good
Sheesh it's easy for u to say lol
do you think I don't struggle with it?
Dude it almost seem impossible
Idk
U sound u unalive every distractions in ur life jeez lol
But I will try
notes isnt exactly the biggest thing but self study other than tuitions and school by your own completely is important
suffer because you tried
or suffer because you didn't try
the second option mostly leads to a net positive happiness
anyways i think this is drifting off topic
Yeahh
I don't get the mood to study in my home lol but in tution I have big fire in me lol
it doesn't matter if you like it or not you absolutely have to
delete every single game that you might be using too much
or app
Ye your right
i did that
Oh
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hello there a reason general form isnt used ?
is it not true for y' = f(at,by)
because thats a more general form
i think you misunderstand
on the previous page it covered this form
adding those extra a and b simply introduces more moving parts that are completely unnecessary
so couldnt be f(at,y)
y ' = G(ax + by) is a very specific form of ode which is not "the most general form an ode can take"
t isnt a function is it ?
i didnt say the most
in fact it's the most general form an ode that can be solved with a very specific substitution can take
y' = f(t,y) is just the most generic form you can write a first order ode in, where t is understood to be the independent variable and y is the dependent variable
oh so this consists of all the types of first order odes
yes, because the theorem is intended to apply to all first order odes
is the theorem true both ways
in other words is there a unique solution to a first order ode if and only if those conditions are met
no
the partial derivative being continuous is actually a stronger condition than strictly necessary
the less strong condition than f_y being continuous is lipschitz continuity in y. but you aren't expected to know what that means yet so they chose a condition which implies that instead
ok so if df/dy isnt continious
there can still be a unique solution
nvm read that wrong
ok so if the conditions arent met i cant make any conclusion
yes. it's likely that it will fail to exist or be unique but not necessarily guaranteed
ok
ty for your time
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hhi
Hi!
i need help
you can post your question
ill send
Post question first so bot can pin it.
,rccw
,rccw
Are you just having problem with y?
or both
lemme rephrase my problem: What's your current progress?
Please ping me when you're back.
wdym
What have you tried?
q19
i tried all prperties i know
it came in exam
today
okay, have you found either x or y?
only 110 is there
Your x is correct
and bcd is 80 in question
lemme verify y real quick
hmm
yeah, your y is incorrect
whats the ans
y is apparently half of ADC
You can figure it out, right??
You already got DCB = 80
topper of class said x is 30
sure thanks
Sure, totally legit answer
and i thought "wtf?"
π₯²
π
They're probably gaslighting you lmao
nah, you'll get half of the marks for this question
,rccw
What did you even try
y is simply 50, because DCB is 80 so the other angle must be 100, divde by 2 = 50
i added extension
If you're curious
No, you're supposed to share your work in the first place
That's not how things work in this server
can i whiteboard
you mean activity? No, it's been disabled
hmm
i added triangle pmx such that sp=px and pm bisects angle m
then sas
thats all
@midnight flower Has your question been resolved?
have you solved this? (you can ping me if you still need help. if you have done, pls close ts channel)
what is the question again
wdym
so we have sr parallel to pq right?
what do we know abt the angles if sr is parallel to pq?
that makes no sense
yes
its a ||gm?
no:β)
π
since i cant actually get a question for u right, ima say it
we have srm congruent to rmq by alternate interior angles
and bc MR bisects QRS, we have SRM congruent to MRQ
since SRM= RMQ, and QR= SRM,
we have RMQ =QRM
we have to prove pm = sp
i understand that
and?
then we have isoceles triangle qrm
ok?
do you understand ts
where?
how is it isoceles
does ts make sense to u?
then qrm isoceles by property
continue
ok
we have mq=mp bc m is midpoint of pq and ps=qr right
and qrm isoceles, so mq=qr
then we have mp=mq=qr=ps
does ts make sense to u, if not, why?
@midnight flower Has your question been resolved?
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what concept am I learning here
the prof has a thick accent, his slides are incoherent and he doesnt post book chapters
I just need to know what am I learning to know what im supposed to look up in the book and youtube
I wish there were other sections with different prof;-;
yeah calc 2
but this specific thing that im just trying to find out
and the c seems to just be a label
im not sure
hm, seems that average value is the thing
thanks
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Mean Value Theorem for Integration
(\exists c\in[a,b]:f(c)(a-b)=\int_a^bf(x)dx)
PajamaMamaLlama
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Use cos(2theta) = 1 -2sin^2(theta) to rewrite sin^2(2theta) in terms of cos(4theta)
First you can move the constant out
cant he just make sin^2theta cos^2theta into sin^2theta(1-sin^2theta)?
and then distribute
no
yeah he cano
what you do with the sin^4theta term then
reduction identity
what's the reduction identity? Google just says it's the same one I said and then you'd just end up with a cos^2 term after expanding the sin^4 term
x = sint
I'm not gonna do it
cause im lazy
but its possible
you'll get the same answer if you use my way
oic
obviously
its just that its the worst way possible
best if u like wasting ur time
@finite forum Has your question been resolved?
bro what
idk how i didnt see that
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Hello great math fellows, I need some help regarding understand the following proposition from a paper I'm reading.
If we take a non-continuous bijective additive mapping f : R β R, the natural extension g: C β C, g(Ξ± + iΞ²) = f(Ξ±) + if(Ξ²) is an additive bijection preserving orthogonality in both directions. We shall see next that this counterexample can only occur when the inner product space in the domain is oneβdimensional.
first, why the map f must be non-continuous?
Also the "natural extension" of f which is g. is it something dedicated to complex spaces?
@native minnow Has your question been resolved?
probably need the actual context of the paper to answer your question
a couple sentences usually isn't enough
@native minnow Has your question been resolved?
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This the question
Do you know the definition of displacement?
Ya
So what seems to be the problem?
Now done, it's clear
Ty
Kk
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,rccw
Intuition tells me to ||rotate triangle ADE about A onto AB so D' = B||
idk everything with x just cancels out and doesnt make sense
so 115-x?
yeah but ur right it is cancelling
There are quadrilaterls ABFE and AFCE. That may help.
^135
oh yeah
how would we use them
yeah this kills it (actually it kills both parts)
There are four inner quadrilaterals. You could make a system of equations, me thinks. π€
Or maybe not. I haven't done the mental gymnastics yet.
Give me a moment.
^
wdym by kills?
everything else is trivial from then on out
aka you basically solved the question by doing 1 thing
so like make an isosceles triangle
where is isosceles coming from
what would weo do after then?
i meant after the rotation
I'm aware of that
isnt rotation just moving it around and we already did that
lemme rephrase
what do rotations preserve
you can use your intuition based on moving things around to answer this if you want to
the lengths and angles of the shape?
correct
so you can actually fill in some more information now
fill in what you can and send a picture of what you have after that
This is what I got but idrk what to do
angles are fine
what happened to labeling the lengths
also maybe draw a nicer diagram without the scribbles because that will be helpful in not getting confused
its a rough sketch i can fix next time
ok
is it like that
o wait oops
also you can use variables or tickmarks to label equalities that you don't have concrete numerical values for
so that you can actually label everything
@cunning fiber
also you can use variables or tickmarks to label equalities that you don't have concrete numerical values for
do u mean like extra variables
what would weo d after?
you tell me
idk what to do
you can take some time to make observations
I don't expect you to get it instantly
I mean, neither did I ^ (part 1)
^ (part 2)
(observe the time gap)
oh yeah feel free to list what you see in chat
and don't worry if your observations feel "too obvious"
that's how you gain intuition/acclimate yourself (source: personal experience)
theres like a right angle
with the new angles formed
you're talking about this one right
the 25 20 and 45
thats all i hav found rn
that's fine
I have like 13 minutes before I need to leave anyway
and I'm just killing time lol
could i have a few hints
idt you need them
your first observation is already pretty good if you state it from a different perspective
90 = 45 + 45
how would we use that? to help find x
isnt that what the question is asking tho
Yeah but donβt try to get to x immediately
could we do something isosceles with the 2 a's
or the 2 1s
@cunning fiber
Don't feel bad. It took me a while to figure out what Pigeon was doing.
do u know how to do it now?
Here's a better drawing.
Think about which segments have the same length.
The caveat here is that one segment will be looked at twice.
is it AB=AD?
so its one of the lines in AFE' ?
Yes.
is it AF?
What about AF?
the segment thats looked at twice
its part of the point e in the original right
they are equal
Do you see the two triangles that extend from point A?
AFE and AFE'?
what do we do with this info?
currently taking
What do you know about triangles that have equal SAS?
oh they are congruent right
so x should be 65 degrees
Yes. I've taken geometry many moons ago and I would not have figured this out on my own either. π
It's an interesting trick to see. I'm not sure I'd recognize to do it again if I came across a similar problem.
wait do u know how we would approach part b of the question?
Without using trigonometry, I don't know.
@timber imp Has your question been resolved?
wait whats ur method with trig cause id rather just solve it than not put anything
o wait nvm with trig its easy
Right. π
<@&286206848099549185> any ideas for part b :)
AI is so wild for this explanation. It took some finessing and several corrections but it came up with a solution.
You have already seen that triangle AEF and triangle AFD' are congruent triangles.
From triangle AFE', you can show that EF is equal to FE' which is just DE + BF. This is a long-winded way of saying that EF is equal to DE + BF.
mb i was eating breakfast
isnt that already given tho
kinda
oh wait ef not e'f
The point here is that you can reframe the length of EF in terms of DE and BF.
EF = DE + BF.
Yes.
i did x+y+1-x+1-y
Yes.
ohhh i get it now
Or more formally.
EF = DE + BF
EC = 1 - DE
CF = 1 - BF
P(ECF) = EF+EC+CF
I wonder what type of psycho originally figured this one out. π
my teacher always find weird questions every now and then and I never know where its from
That whole shifting triangle ADE, I would have never thought of it.
im surprised pigeon found that out
Pigeon has probably seen this type of problem before.
ill try keep like a mental note of those types but i highy doubt ill find answers for them
Well ... we both learned something today. π
tysm for helping to explain to me it was kinda hard to understand
yw
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can i have help being walked through how to answer these questions? i think the answer is unsure without proof but im not sure and im not exactly sure how id explain it
just identify which of the sets contain [0, 4]
so, none of them?
or wait
question 14 could contain it, right?
if y starts at 4 then moves up 1
draw [0, 4] and [1, 5] on a number line and see
A set S "contains [0, 4]" means every number in [0, 4] is also in set S
in other words, [0, 4] is a subset of S
but the question says i shouldnβt use a number line, right?
i also dont remember how to write those on a number line
where does it say that
well it says only use the knowledge provided so i thought i couldnβt do anything else but think
i mean it should be obvious without a number line that 0 is in [0, 4] but not in [1, 5]
i was just suggesting using a number for you to convince yourself
well yeah i know that
The problem is asking which of these intervals contains [0,4] basically
so why do you think B could contain [0, 4] ?
So make sure every number from 0 to 4 is included
let me draw something so you understand how im thinking of this hold on
i only learned functions a few hours ago so bare with me i appreciate your patience thank you
i apologize im not that bright when it comes to math
Not really sure what the question is asking exactly
@young mantle Has your question been resolved?
me too
@young mantle Has your question been resolved?
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yo
just underlining it maybe?
Where have you gotten this from to start with
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this guy
does it
lmao it just means when you are done and want to underline the solution
its not anything deep or mathematical
fairs
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Some use boxes
some just write Q.E.D.
Its doesn't mean much aside from end of your problem
I do this but it looks more like a #
Where do i ask for help
never seen that
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Pls help with 11
i divided whole equation by 2
Iβm not seeing what that does here
u misunderstood i came for 11
no worries
πΈπππππππΒ³
yes
do u mean that it looks like a root of smth
@past thicket ?
No I just wrote it in latex so. I could see it without having to look at the paper
oh well now pls help me further
Yep just lemme think about thisβ¦
use substituion
let π₯ = 2π΄ and π¦ = 2π΅
so now we have $\sqrt{3}\sin(x)=\sin(y)$
πΈπππππππΒ³
its still same but yea it still helped to reduce my effects in writing
We can use the identity sinΒ²(π₯) = (1-cos 2π₯)/2 to transform
β3 β’ (1-cos π₯)/2 + (1-cos π¦)/2 = (-1+β3)/2
Multiply by 2 on both sides
β3 (1-cos π₯) + (1-cos π¦) = -1 + β3
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hey
can someone tell me where i went wrong
sending pics 1 sec
this is the answer
why is it 3(x-4) rather than just x - 4
this is a mess
yeah thats it
im doing a substitution worksheet
so i assumed i needed a sub
its not like the sub helped 

Like I could see your point but think about it even if you subbed there is nothing you could cancel/help simplify the integral
yeah that makes sense
thanks
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Can anyone tell me a short method without calculator?
for question 13
a number is divisible by 6 when it's divisible by 3 and 2
if a number is divisible by 2
it's obvious to see
if a number is divisible by 3
then the sum of its digit is divisible by 3
(this can be proven btw)
V is proportional to t
and i don't see what i can do to 12
12
I'm only asking for 12
well i don't see how
π
i mean, that's literally why they asked you to round to 2 decimal places lol
those numbers don't look friendly enough for a short method without using a calculator.
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We know that ||A+xb ||^2 = ||A ||^2 + x^2 ||B || ^2 +2x Aβ§B. So We know that ||A+xB||^2 - x^2||B||^2 = ||A||^2 2x Aβ§B. Then we can see that ||A+xB|| > ||A||.||A+xB || = ||A|| iff x = 0 and Aβ§B = 0. But you can't still write something that is > as >=. This leads to my issue since we know that it can only equal if the dot product is 0 but you can rewrite the inequality sign to have the equality
write \|| if you want your norm bars to show up
A+xB that's such a spoiler
do you guys have any suggestions on how to resolve this?
I'd do b by showing the contrapositive I think
but don't you get the same issue with the inequalities sign if you try to show the contrapositive?
wdym
by this I mean this statement only has equality if x = 0 and Aβ§B = 0.
I didn't get this
||A+xB|| > ||A||.||A+xB || = ||A|| iff x = 0 and Aβ§B = 0
even if x = 0 and A.B = 0, |A|.|A+xB | = |A|^2 no?
ok so we can show that||A+xB||^2 - x^2||B||^2 = ||A||^2 2x Aβ§B. so if we take away the - x^2||B||^2 . we get ||A+xB||^2 > ||A||^2 2x Aβ§B
also
|A+xB|^2 - x^2|B|^2 = |A|^2 2x Aβ§B.
did you mean
|A+xB|^2 - x^2|B|^2 = |A|^2 + 2x Aβ§B.
yes sorry
thats the only time it is equal
don't we sqrt both sides?
how do you get |A+xB| > |A|.|A+xB |
you didn't sqrt the left side
oh I get it
you didn't write the sqrt
from what I understand the equation simplifies to ||A+xB|| > ||A||. if the dot product is 0
wait do you think we can start from the top. because I think im confusing myself
let me see if I understand:
you're saying that $|A+xB|^2 - x^2|B|^2 = |A|^2 + 2x Aβ§B$ thus $|A+xB|^2 \geq |A|^2 + 2x Aβ§B \geq |A|^2$ with equality iff x = 0 and A.B = 0 \
And at the same time, $|A+xB| \geq |A|$
I think that working with the contrapositive of statement b) is easier
bloubbloub
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i.e. assume A and B are not orthogonal and find x such that |A+xB| < |A|
Well I was having trouble with was saying that I could get > with dot produc equal 0. But I think I'm going to drop that train if thought since I'm confusing myself
Isn't it all real x?
for the contrapositive you only need to show that there exists one
negation of "for all" is "there exists"
Ok I see
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i have to simplify and use conditions of existence but I'm not sure how to
(the red thing is the result im supposed to get btw)
find a common denominator
ok so 2 right then what?
I don't understand you
are you saying the common denominator is 2?
yeah
a common denominator will be a multiple of both other denominators, 2 is not what you are looking for here
oh
naturally a common denominator will be something like (x^2-2x)(x^2+2x) but there is a more reduced version. Can you see it?
a more reduced version in what sense? (sorry I'm bad at math)
so im not sure what your referring to
ah sorry its not there
aight
do you agree this is a common denominator?
i think i must learn about common denominators first before we proceed with this exercise
plus im not sure what ^ is supposed to be in this case
power
^ denotes an exponent.
x^2 is x squared
ohh
oh wait nvm
i think i understand
so just write everything together?
but does that include the upper part or just the lower part?
if you multiply the denominator by something you have to multiply the numerator by the same thing to preserve the value of the fraction
$\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+cb}{bd}$
ExpertEsquieESQUIE
so I have to multiply the first one with x^2+2x
and the second one with
x^2-2x
right?
(including the top part)
It would be easier if you factorised these quadratics first
Then you can see what is needed to create a common denominator
As long as you multiply the numerator and denominator by the same thing even a polynomial like x+2/x+2 it will still be the same, so try to use that
wait what
what did you end up with
(i have to use conditions of existence)
I ended up with this
I donβt understand what you mean
yk C.E
Now add the fractions together
you combine the fractions now, and it should simplify nicely
I think thatβs a thing your curriculum teaches idk what that means
Also I donβt see why you canβt factorise first
okay rt
ty
whats that
$x^{2}-2x=x*(x-2)$
BBMaths
Maybe factorize in us spelling
how do i combine them?
like this
ohhh
is this how its supposed to be?
i dont know either but since i already started in another way i think i might aswell finish it
||it should be times 2||
oh so it's supposed to be x^2?
So a fraction like 12/8 can be simplified to 3/2
The same applies to these polynomials
Polynomials is like x^2-2x
Using this can you turn your denominator into 4 things multiplied together instead of 2
as in x^2 or xβ’2
oh
2x
so just multiply this by 2
2x^2
you did (x^2 + 2x) + (x^2-2x)=x^2 but its actually 2x^2
$\frac{2x^{2}}{(x^{2}-2x)(x^{2}+2x)}$
BBMaths
like this guys?
No
oh nvm the bottom dosent need the extra ^2 right
The denominator was fine
ohh ok
Do you know how to factorise x^2-2x?
nope
Okay so do you know how to distribute like $(x+2)(x-1)=x^2+x-2$
BBMaths
i guess but wait what happened to -1
This is* just an example not related to the problem or do you mean in the example?
actually nevermind i dont
nevermind sorry
(example)
Wait Iβm confused now do you know distribution?
no i meant that i dont know distribution , sorry for confusing you
This question seems hard to do without knowing distribution or factorisation
Ohh im just supposed to multiply
yes
nevermind
sorry im not familiar with math lingo
so i get confused a lot
okay after i use distribution then what?
We want to do the opposite of distribution here
Factorisation is doing this in reverse
$x^2+2x \rightarrow x(x+2)$
BBMaths
There we go
We need to do this to x^2+2x and x^2-2x
Then the denominator will be fully factorised (usually better that way)
oh ok lemme try
like this?
BBMaths
like this?
did i do it right?