it's okay if you don't understand something
#help-36
coarse mantle
ember escarp
3x=6x?
ember escarp
oh
ember escarp
so i just multiply the whole thing
coarse mantle
isolate the x terms from the constant terms
coarse mantle
so i just multiply the whole thing
as in?
ember escarp
like depending on the ()
coarse mantle
yes
coarse mantle
you should have gotten 5x - 10 = 2x +4 right?
did you get to this?
ember escarp
yea
ember escarp
i want to know what to do nextt
coarse mantle
you have 5x on one side and 2x on another side right?
you have to bring both of them to the same side
try this:
(5x -10) - 2x = (2x + 4) -2x
do you see how we substract 2x from both sides?
ember escarp
yes
coarse mantle
can you exlain how you got this answer?
ember escarp
subtract the () then follow -2x
coarse mantle
mhm
ember escarp
or add
coarse mantle
you can't do that
parentheses don't hold much value in addition/substraction so we can get rid of them
you have
5x - 10 - 2x = 2x + 4 - 2x
do you see anything?
ember escarp
it got complex for me
coarse mantle
just move the -2x next to the 5x or 2x $5x - 2x - 10 = 2x - 2x + 4$
soft zealotBOT
Bakoles
coarse mantle
can you do 5x - 2x?
what about 2x - 2x
this is simple substraction
you can ignore the x if it bothers you
5 - 2
2 - 2
ember escarp
yea i can do those
coarse mantle
what's 5 - 2
ember escarp
3
ember escarp
0
ember escarp
yea
coarse mantle
It's 3x
you solved it
okay so let's go back to the equation
we have 3x - 10 = 4
do you know why just 4?
cause 2x - 2x = 0 so we don't write it
see how we have 3x - 10 on one side? we need x by itself
how do we do that?
get rid of the 10
how?
find a number so that that number - 10 is equal to 0
it's 10
10 - 10 = 0
since we add ten on one side, we have to add it to the other as well
so get 3x - 10 + 10 = 4 + 10
notice how we add 10 to both sides
solve
we get 3x = 14
divide both sides by 3
3x/3 = 14/3
$x = \frac{14}{3}$
soft zealotBOT
Bakoles
coarse mantle
hope you understood something
coarse mantle
🙂
ember escarp
i have like 8 more of rational equations to solve
but I think I can handle it from here?
coarse mantle
it was long but if you remember the steps it will be easier in the future
coarse mantle
okay good luck to you
ember escarp
thank u so much
coarse mantle
👍
ember escarp
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wild temple
Hello
worldly mesa
hello
wild temple
Is there anywhere I could find practice questions like the last 5
About fractional distance
worldly mesa
You can make some on your own
just choose some numbers for A and B and some fraction
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summer lynx
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summer lynx
A weird doubt but why can we not jus write the sinx^2 +cosx^2 as 1 here?
proper dagger
,rccw
soft zealotBOT
terse crypt
,rccw
soft zealotBOT
terse crypt
Simplify the trig part first
umbral hamlet
Because the trig expressions aren’t squared
terse crypt
Then do u sub
rocky tusk
A weird doubt but why can we not jus write the sinx^2 +cosx^2 as 1 here?
sin^2 x ≠ sin(x^2)
summer lynx
sin^2 x ≠ sin(x^2)
Ahh but what does the rhs mean then
umbral hamlet
First square x, then take the sin of that squared value
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flint parrot
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junior parcel
can you figure out what AC is?
flint parrot
Ok
frosty fossil
use similar triangles
flint parrot
hmm okay
junior parcel
that would work too
frosty fossil
can you see why ABC similar to DBA similar to DAC
frosty fossil
Can you find the ratio between ABC and DBA?
flint parrot
Okay
frosty fossil
what is it
frosty fossil
the ratio would be BC:AB because those are the hypotenuses of the triangles
frosty fossil
10/8 or 5/4
flint parrot
Ok
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tranquil pine
I'm doing geometry and I don't know how to solve this. I am supposed to use the Law of cosines, and then the Law of sines, but I don't know how to use the sines here
final saddleBOT
I'm doing geometry and I don't know how to solve this. I am supposed to use the ...
tired walrus
you can equate the left-hand sides of equations #1 and #2 to each other, can't you? given that they are both equal to y^2/4
so you get $x^2 - x \sqrt{3}(\sqrt{3}+1) = c^2 - c \sqrt{2}(\sqrt{3}+1)$ by the looks of it
soft zealotBOT
Ann
tranquil pine
i am not sure if this helps us
yeah; x, c are unknowns
tired walrus
you can say the areas of triangles ABD and BCD are equal. from this im pretty sure you can find x/c
tranquil pine
Why are they equal? And If they are, how to calculate with that?
tired walrus
drop a perpendicular from B onto AC and consider it as the height in the formula S = 1/2 bh, common to both triangles
the bases will then be AD and DC, and these are equal
on the other hand you can also express the areas of these triangles in terms of two sides and the angle between them (30° and 45° resp)
tranquil pine
the bases will then be AD and DC, and these are equal
When I drop a perpendicular from B, the perpendicular won't be the median since it's not an isosceles triangle. So the bases won't be AD and DC, or am I wrong?
tranquil pine
on the other hand you can also express the areas of these triangles in terms of ...
Isn't using the Law of cosines better?
tired walrus
says who
tired walrus
When I drop a perpendicular from B, the perpendicular won't be the median since ...
the perp will not be the median but the altitude dropped from B will be literally the same line segment for both triangles
maybe it is easier if i draw over your diagram, one moment
here
[ABD] = 1/2 * AD * BH and [BCD] = 1/2 * CD * BH
that's my idea here
that's my argument for why the areas are equal
tranquil pine
Thanks I see it now, it is smart. Wouldnt that imply though that BC = AB?
because AD = DC and they share DB hence if the areas are the same all the sides must be the same (thats my logic)
My diagram might be misleading in terms of shape, but every parametr is correct*
tired walrus
because AD = DC and they share DB hence if the areas are the same all the sides ...
equality of area is not grounds enough for congruence
tranquil pine
equality of area is not grounds enough for congruence
Yes, you are right my bad
I'll use that to try to solve and see how it goes
night raft
<@&268886789983436800>
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craggy plank
??
tired walrus
scammers
craggy plank
What happened
supple mantle
Hacked accounts
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normal birch
hey, trying to do a pythag theorem proof as an exercise, i know i can prove it through proportionality of similar triangles, but i can't for the life of me figure out how to prove these two angles are equal to then prove the triangles are similar. img 2 is what i currently have (its a lil rough :( ) img 3 is what im working off of
final saddleBOT
hey, trying to do a pythag theorem proof as an exercise, i know i can prove it t...
pliant shore
hey, trying to do a pythag theorem proof as an exercise, i know i can prove it t...
what's this (?) angle in terms of x?
also, what must be the angle on the circumference?
(it's a famous circle theorem)
normal birch
what's this (?) angle in terms of x?
okay i feel like its gotta be [overall angle] - x = [that angle] but i. i don't know the circle theorem. i've been trying to find it but i'm 
pliant shore
okay i feel like its gotta be [overall angle] - x = [that angle] but i. i don't ...
oh for that you don't need a circle theorem
you have a triangle and the sum of its angles is 180
normal birch
oh wait, is it 90-x? i got a lil lost in the sauce
pliant shore
oh wait, is it 90-x? i got a lil lost in the sauce
yep
normal birch
ohh okok, thought so at first
pliant shore
okay i feel like its gotta be [overall angle] - x = [that angle] but i. i don't ...
that's okay! it's called Thales' theorem
pliant shore
normal birch
THANBK YOU SOI MUCH
pliant shore
there's a picture proof like so
these two are isosceles triangles cause radii are equal
so then 2(green) + 2(red) = 180
green + red = .... ? that's how
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quick flume
In this question I have to find the set of equations and give an answer for x and y.
Here I tried to find out what the equations were in the form of y=ax+b.
I found out that the blue one is y=(3/2)x-2 and the red one is y=-x+3.
By isolating both the x and y I found out that x = 5-y and y = 11/5, making it that x = 14/5.
However this is wrong according to the book/system
final saddleBOT
In this question I have to find the set of equations and give an answer for x an...
night raft
could you show your working out
supple mantle
In this question I have to find the set of equations and give an answer for x an...
Please send us your work
quick flume
A linear set of equations consists of equations for two straight lines. The two straight lines are drawn in the coordinate system below. Use the graph to find the solution to the set of equations.
night raft
It's a bit hard to read it when you wrote it diagonally
severe canyon
,rotate 45
proper dagger
that's a thing??
severe canyon
, rotate 40
quick flume
my bad, i have to get used to writing on an ipad
severe canyon
, rotate 37
soft zealotBOT
proper dagger
first question
that blue-red pair of equations there. are those the ones you used to graph?
craggy plank
Your approach is completely doable
proper dagger
because the red eq. is y = -x + 3 but the second eq. in the og question is y = -x + 5
which is it?
quick flume
The red and blue equations is basically what I thought are the answers of those 2 equations in the graph
Which I then further used to solve for x and y
severe canyon
quick flume
Looks like a three, but it's supposed to be a 5
proper dagger
it's supposed to be a 3 tbh, looking at the red line
late bolt
This has me dead 😭
sand escarp
i was doing thyis question until i got stuck what do i do next to find X? cause X is equal to ACB right but 36 is DCB
night raft
i was doing thyis question until i got stuck what do i do next to find X? cause...
!occupied
final saddleBOT
i was doing thyis question until i got stuck what do i do next to find X? cause...
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).
quick flume
my brain is not working i notice
craggy plank
I mean, your approach is 100% correct
Cuz I’ve check your first attempt, it’s all right.
just a small hiccup
quick flume
Yeah alright now I got the right answer, thank you a lot
Think I'm digging my own grave by wanting to do it quick in my head but end up spending more time fixing it lol
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undone tiger
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worldly mesa
any other permuation will fix at least one of 2,3,4 but not all of them so it will just be a transposition which is odd
tired walrus
presumably the 4 symbols they mean are specifically 1, 2, 3 and 4
tired walrus
so beginning with 1 means fixing 1 because then σ(1) = 1
final saddleBOT
@undone tiger Has your question been resolved?
tired walrus
(123)(4)
(132)(4)
(234)(1)
(243)(1)
(134)(2)
(143)(2)
(124)(3)
(142)(3)
a lot of these do not even fix 1, i think you are seriously misinterpreting the question
they are not asking for cycle notation
undone tiger
a lot of these do not even fix 1, i think you are seriously misinterpreting the ...
I think i have to explain what I wrote
I wrote S4 all the cycles form with length 3
Actually i was discussing it two places which you don't like
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torn lotus
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torn lotus
Help plss
ripe jewel
Help plss
I can hel pyou understand what the symbols mean, but if there's a specific jargon used in neural networks I don't know them
ripe jewel
doesn't change what i said tho
are there any of them whose meaning is apparent to you?
torn lotus
no sir
ripe jewel
oh okay I looked up some terms in neural network jargon, let me just confirm
okay
here are 3 images from google image search
torn lotus
duration stayed / tips receive = satisfaction?
ripe jewel
this is what I am using to determine the jargon,I'm assuming three images that agree arent all wrong
well let's work on the indiivudal parts before the complicated parts
what's
$w_{21}^{(1)}$
soft zealotBOT
gfauxpas
torn lotus
worker?
ripe jewel
it's most similar to the first notation
torn lotus
maid?
soft zealotBOT
gfauxpas
soft zealotBOT
gfauxpas
ripe jewel
they're weights
torn lotus
hmm
ripe jewel
I believe the superscript (1) means "the first time" we use it, and the second time we're going to use it again but in a different form
$\sum$ means what it usually means in mathematics, which is ...
torn lotus
so first we use weight of consumer
soft zealotBOT
gfauxpas
ripe jewel
hint: it's in the latex symbol name
torn lotus
summation
ripe jewel
great
what do you call a summation with weights? don't think too hard
first thought that comes to mind probably is correct
torn lotus
idk
ripe jewel
weighted summation
torn lotus
yes yes
ripe jewel
probably too simple for you to think thats what i was asking for
torn lotus
true
ripe jewel
so that tells you what $\sum :$ means on th ebottom
soft zealotBOT
gfauxpas
ripe jewel
yellow circle is a special term in neural networks that you can gather from this diagram
torn lotus
hmm
ripe jewel
it's just the name of whats on this diagram on top
it's even using the same variable letter
ripe jewel
yes
and you see that unlike weights, it's added
the formula on the sheet is $z = b + \sum a_i w_i$
soft zealotBOT
gfauxpas
torn lotus
ohhh
ripe jewel
you see from here the first column of data is called the input
torn lotus
ai means?
ripe jewel
a_i is shorthand for a_1, a_2, a_3, ..., a_n, however many you need
torn lotus
ita inputs?
torn lotus
😅
ripe jewel
f has a special name, activation function
it tells you whether the neuron lights up or not
torn lotus
ohh
ripe jewel
so if the red circles are called the input layer (or just "input"? idk)
then you can tell what the yellow circle and green circle are from this
torn lotus
green output
ripe jewel
oh wait
no, there's both
there's both layer and individual
BOX A is ...
BOX B is ....
and the input layer is already labelled
torn lotus
box a is hidden layer
ripe jewel
that's box a yes
torn lotus
or calculation layer whats better?
ripe jewel
uh i dunno, the images i found on google image search say hidden layer
im not an expert buit thats what it says
anyway i think you can do them all now
now that you know f is called the activation function
torn lotus
i forgot yellow
ripe jewel
bias
turbid burrow
if this is for a class, you might want to look through the notes. neural networks have multiple names for the same things
ripe jewel
yeah but if he missed the class i cant do more than what i didd
unless you have a book or notes are from another student
torn lotus
if this is for a class, you might want to look through the notes. neural network...
i shall ask for notes for others
tysm guys
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@torn lotus Has your question been resolved?
silent falcon
The beginning is easy and the conclusion (cantor Bernstein theorem) is pretty straightforward once you’ve done the rest. However I’ve been really struggling to do the proof of "psi is a bijection onto Y"
It’s evident that its image is contained within y and its restriction to X\A is a bijection (as it is its identity function).
I just have no idea how to prove (as asked) that Y inter A is included in A inter Im(phi)
final saddleBOT
The beginning is easy and the conclusion (cantor Bernstein theorem) is pretty st...
soft zealotBOT
final saddleBOT
@silent falcon Has your question been resolved?
final saddleBOT
@silent falcon Has your question been resolved?
shadow marlin
The beginning is easy and the conclusion (cantor Bernstein theorem) is pretty st...
S'il existait un élément de $\overline{A}$ sans antécédent par $\Phi$ il existerait une partie close de $A$ plus petite que $\overline{A}$
soft zealotBOT
bloubbloub
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@silent falcon Has your question been resolved?
silent falcon
S'il existait un élément de $\overline{A}$ sans antécédent par $\Phi$ il exister...
Ah merci beaucoup chef
gentle zephyr
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dense coral
could you translate the problem? 
cosmic warren
It says find and graph the equivalence relation associated to the partition (set). How many distinct equivalence classes are there? Find a representative for each equivalence class
gentle zephyr
idk if representante = representation
drowsy epoch
,w representante
soft zealotBOT
drowsy epoch
perchance
gentle zephyr
It says find and graph the equivalence relation associated to the partition (set...
i appreciate the translation dude
cosmic warren
np 
crimson bronze
what have you tried so far?
you could start with drawing a graph like I showed for the previous exercise
gentle zephyr
first i need to find the relation before I can graph it
but this partition is a quotient set, right?
crimson bronze
yes
gentle zephyr
so we know all of this sets inside the quotient set are equivalence classes
gentle zephyr
you could start with drawing a graph like I showed for the previous exercise
i am trying
crimson bronze
yep
gentle zephyr
einen moment
crimson bronze
jawohl
gentle zephyr
you could start with drawing a graph like I showed for the previous exercise
dude the diagram is crazy now
let me re do it because the equivalence class with cardinality 4 is crazy
einen moment
crimson bronze
you could leave out the arrows going from an element to itself, might make it a bit clearer
gentle zephyr
you could leave out the arrows going from an element to itself, might make it a ...
here are the equivalence classes
we can find the equivalence relation from this, no?
oh I forgot 5R5
gentle zephyr
you could leave out the arrows going from an element to itself, might make it a ...
this should be it
we can find the equivalence relation from this right?
crimson bronze
yep
gentle zephyr
or what should we do from here onwards?
can we find the representative of each equivalence class?
crimson bronze
yes to both
gentle zephyr
how?
I think we first need to find the equivalence relation in order to find the representative of each equivalence class, no?
@crimson bronze
crimson bronze
you don't need to, but you can if you want
gentle zephyr
how?
crimson bronze
btw, we usually say a representative, because there may be more than one
so let's start with finding a representative
gentle zephyr
how?
crimson bronze
what even is it? 
gentle zephyr
its the first coordinate
crimson bronze
no, there's no concept of "first" in an equivalence class, or in a set in general
gentle zephyr
its hard to explain, its tied to the definition of equivalence class
crimson bronze
it's not that hard, you should see if you can find a definition
that should always be step one: what is the definition
gentle zephyr
the equivalence class of $x \in A$ is the subset of A, that contains all the elements $y \in A$ related with x, it is denoted $\bar{x}$
soft zealotBOT
Renato
gentle zephyr
$\bar{4} = \bar{8} = \bar{10} = \bar{9} = {4,8,9,10}$
soft zealotBOT
Renato
gentle zephyr
that should *always* be step one: what is the definition
what is the representative dude
crimson bronze
are you sure you don't have a definition in your book?
gentle zephyr
of representative?
crimson bronze
yeah
gentle zephyr
or equivalence class?
crimson bronze
representative
gentle zephyr
representative
i think a possible representative of some equivalence class are whichever of the elements themselves of each equivalence class, am I making any sense?
$\bar{4} = \bar{8} = \bar{10} = \bar{9} = {4,8,9,10}$
soft zealotBOT
Renato
gentle zephyr
like in this case, any element in the equivalence class of 4, are possible equivalence class representatives
crimson bronze
yep
gentle zephyr
En cada clase
podemos elegir un representante, es decir un elemento en la clase que
“representa” la clase
translation: In each class we can choose an representative, in other words, an element from the class that represents the equivalence class
gentle zephyr
im triggered there is no definition of it
same happened with quotient set
like no definition whatsoever
or like, in between lines
crimson bronze
translation: In each class we can choose an representative, in other words, an ...
to put it another way: a representative of an equivalence class is just any element of the equivalence class
yeah, this book seems a bit poorly structured
gentle zephyr
btw, we usually say *a* representative, because there may be more than one
just, 1, 2, 4, 5
crimson bronze
yep, those are representatives for each equivalence class
gentle zephyr
do I still need to name all the element of the equivalence relation?
whats the cardinality of the equivalence relation?
crimson bronze
if you want, but it'll be a bit tedious, because there are 30 elements in the relation, if I counted correctly
soft zealotBOT
Result:
15
I think it's 2^2 + 3^2 + 4^2 + 1^2
for the equivalence class {4, 8, 9, 10}, there is an arrow from each element to each other element, so 4 + 4 + 4 + 4 = 4*4 = 4^2
gentle zephyr
like, the cardinality of each pairwise disjoint set in the quotient set to the power of 2
crimson bronze
yeah, I guess you can also think of it as the cardinality of the cartesian product {4, 8, 9, 10} x {4, 8, 9, 10}
gentle zephyr
what?
the cardinality of an equivalence relation R on A, is the cardinality of AxA?
crimson bronze
no, we're looking at each equivalence class separately, then adding them up
{4, 8, 9, 10} is an equivalence class, and it is fully connected: it has arrows between every element. Therefore the arrows are the entirety of {4, 8, 9, 10} x {4, 8, 9, 10}
crimson bronze
there are only 28 elements in that relation, you're missing 2
I'll give you a hint: it's also missing in your drawing, it's in the equivalence class {4, 8, 9, 10}
gentle zephyr
(8,9),(9,8)
gentle zephyr
I'll give you a hint: it's also missing in your drawing, it's in the equivalence...
I appreciate the hint dude
crimson bronze
yep
gentle zephyr
.solved
tranquil pine
got 172
umbral hamlet
What have you tried so far
tranquil pine
cutting it
umbral hamlet
Could you elaborate on that
tranquil pine
got 42
umbral hamlet
yup
umbral hamlet
u can probably figure this out, but if you get stuck I can give a hint
umbral hamlet
nice
umbral hamlet
draw a picture
tranquil pine
mhmm
its a diagonal/side
Then AC = √2s if s is the side
nvm
2/√5 i think
no 2
AC is sqrt(2)
nvm i got 2
verify>
umbral hamlet
That looks annoying
I guess not really
You can use ratios to find most of the areas
tranquil pine
hmm
tranquil pine
You can use ratios to find most of the areas
18?
tranquil pine
lol
umbral hamlet
erm okay I’ll just do it
umbral hamlet
(3,3) = 45/2
(1,3) = 35/2
(1,1) = 28/3
x = 63/28 * 12 = 27
tranquil pine
alr back
umbral hamlet
I might have made a mistake
umbral hamlet
doing this on my phone in my head 🥀
but it’s just comparing ratios
what was your process
tranquil pine
idk lol
messy work
lemme draw
wait
i might be wrong
be = 20
af = 12
cg = 15
OOOH
DH is 21
kk, its IS 27
mb bro
umbral hamlet
woo
umbral hamlet
Whoa there’s such a cool visualization for this actually now that I think about it
Oh well doesn’t matter now
tranquil pine
lol
tranquil pine
ok
28
i got 28
if the sides of 5 and 1
wait i did smth wrng
nvm
verify?
lemme draw
tranquil pine
I would set variables to the outer segments
better way
5+1+2+2+1+3+2+1+5+3+1+2
set long side x and short side y
x+y=5
x-y=sqrt(15)
got 2root(5)
a;r 2,2
similar triangles?
umbral hamlet
better way
did you just guess, im not sure how else to do this without setting variables
tranquil pine
did you just guess, im not sure how else to do this without setting variables
ye lol
well
it cant be 6 and cant be 4
so HAS to be 5
umbral hamlet
oh u assumed theyre all integers lol
umbral hamlet
the altitude from P to AM has to 1/5 the height of the square
umbral hamlet
i think so
tranquil pine
kk
correct yay
i did this and got an answer wait
5 i think
used pythag, cordinate geo and got EF root(5)
am i right?
umbral hamlet
these are escalating
tranquil pine
its reminding me of olympiad math
😫😫😫😫
umbral hamlet
analytic is so gross
tranquil pine
i dont remember any of the geo tricks
is it 5? i mena im not sure
umbral hamlet
is there really no synthetic solution to that
tranquil pine
its reminding me of olympiad math
i might be him
i did tha
and realized it was 4 and 7
lol
i might have accedently added a sqare root
shell condor
this is last jst so u know
Dude What is this
umbral hamlet
why did the 7 4 become 8 5
tranquil pine
why did the 7 4 become 8 5
mb
umbral hamlet
oh that isnt bad though i think
let me get my ipad these are not mental mathable anymore
tranquil pine
ye
tranquil pine
let me get my ipad these are not mental mathable anymore
fr
i gtg to karate class in 8 mins
shell condor
Eh I might have something
tranquil pine
i did this and got 8+4root(3)
shell condor
Is there any way to prove that angle(PAB) and angle(QAD) are equal
umbral hamlet
i like this but it still seems annoying to work with
tranquil pine
yeye
tranquil pine
..
shell condor
4×cos(15°)?
umbral hamlet
Is there any way to prove that angle(PAB) and angle(QAD) are equal
right triangle congruence
tranquil pine
mhmm
shell condor
Root(6) + root(2) IG
Cos(15°) = (sqrt(3) + 1)/2(sqrt(2)
4 cos(15°) = sqrt(6) + sqrt(2)
Can you check if this is correct?
sqrt(2)×(sqrt(3) + 1)
3.8637 should be the answer
tranquil pine
waht da
shell condor
Yeah I'm not sure, I used my head only
umbral hamlet
i think 4 sqrt3 + 8?
cuz
we want [4cos(15)]^2
2cos^2 (15) - 1 = cos(30)
= sqrt(3) / 2
shell condor
we want [4cos(15)]^2
Yeah but why, why should we square this
shell condor
Ahhhhhh missed that, I thought we were finding the side
Mb
Yeah you're correct
So 3.8637²
tranquil pine
wut the
shell condor
14.93 sq. units
edgy mauve
shell condor
as a number?
In the sense?
shell condor
i think 4 sqrt3 + 8?
This is in root form
tranquil pine
i think 4 sqrt3 + 8?
got that
shell condor
10th grade contest geometry
💀, that's pretty tough for 10th grade TBH
tranquil pine
💀, that's pretty tough for 10th grade TBH
there 5th graders in calss
shell condor
there 5th graders in calss
Damn, for olympiads IG
tranquil pine
ye
shell condor
One sec
umbral hamlet
sqrt(9 + x^2) + x = 5
9 + x^2 = x^2 - 10x + 25
10x = 16
x = 8/5
white area = 2 * 8/5 * 3 = 48/5
total area would be 30 = 150/5
51/5 is shaded?
final saddleBOT
final saddleBOT
Available help channel!
tawny snow
final saddleBOT
tawny snow
Simplify the following expression, list restricted values.
Algebra 2 review of Pre-Calculus
.close
final saddleBOT
tawny snow
.reopen
final saddleBOT
✅
young bobcat
uh let me see
so a fraction will be undefined when either
- the numerator is undefined
- the denominator is undefined
- the denominator is 0
since the numerator and denominators are polynomials in this case, they will always be defined
thus, we only need to worry about when one of the denominators is 0
so specifically, the restricted points are when either $x=0,x^2+1=0,$ or $x^3+x=0$
soft zealotBOT
Arnavutköy
young bobcat
@tawny snow does that make sense?
tawny snow
I need to simplify it.
young bobcat
oh i was doing the restricted values part first
uhh sure we can simplify it first
ok so i see you have expanded the denominator $x^3+x$ good
soft zealotBOT
Arnavutköy
young bobcat
now you shouldn't multiply the whole sum by that new fraction
instead, you simply want to result in a common denominator
this implies multiplying different fractions by different values
in this case, our common denominator would be $x(x^2+1)$
soft zealotBOT
Arnavutköy
tawny snow
Yep
young bobcat
thus, for the $\frac{2}{x^2+1}$ fraction, we can multiply by $\frac{x}{x}$
soft zealotBOT
Arnavutköy
young bobcat
and for the $-\frac{1}{x}$, we would multiply by $\frac{x^2+1}{x^2+1}$
soft zealotBOT
Arnavutköy
tawny snow
I’ll try
young bobcat
so we would be calculating $\left(-\frac{1}{x}\right)\left(\frac{x^2+1}{x^2+1}\right)+\left(\frac{2}{x^2+1}\right)\left(\frac{x}{x}\right)+\frac{1}{x(x^2+1)}$
soft zealotBOT
Arnavutköy
young bobcat
specifically we evalulate out all of the products and can then sum the resultant fractions just by taking the numerators, as they all share a common denominator
tawny snow
Then multiply the whole thing by x^3 + x / x^3 +x?
young bobcat
Then multiply the whole thing by x^3 + x / x^3 +x?
no need to do that
we don't need to multiply by anything
we will already have a common numerator and denominator
tawny snow
So add numerator together?
young bobcat
yep
lets first see what we get then
once we have a singular fraction, it will be much easier to work with
tawny snow
So it’s ( x^2 +2x +1 ) / ( x^3 + x )?
young bobcat
uhh you must have made a calculation mistake. perhaps you forgot the negative term in front of the 1/x?
young bobcat
( -x^2 + 2x ) / ( x^3 + x )?
yes, that looks correct
tawny snow
It’s wrong
young bobcat
can we simplify the sum now?
young bobcat
It’s wrong
its not the final answer, but i mean it is a correct reresentation
can you factor (-x^2+2x)?
tawny snow
No?
young bobcat
ok so the -1 powers are implicit fractions right
so first, i want you to combine the implicit fractions within the numerator into one fraction with a common denominator
i want you to then do a similar thing for the denominator
lets first start with that
tawny snow
Did it
young bobcat
ok what did you get
soft zealotBOT
Arnavutköy
young bobcat
you can only combine numerators when denominators are equal
it doesn't work around the other way
you cannot combined denominators when the numerators are equal
tawny snow
They’re under the same fraction?
young bobcat
They’re under the same fraction?
no the numerators are same but denominators are different
you cannot just combine the denominators in this case
if the denominators were both the same, then you can combine the numerators. but not vice versa.
tawny snow
So what am I supposed to do?
young bobcat
So what am I supposed to do?
multiply by a 'unity sandwich' as you did in the previous problem, to create a common denominator
tawny snow
There’s only one denominator?
young bobcat
in this case, it would be $\left(\frac{1}{x+1}\right)\left(\frac{x}{x}\right)-\left(\frac{1}{x}\right)\left(\frac{x+1}{x+1}\right)$
soft zealotBOT
Arnavutköy
young bobcat
There’s only one denominator?
no in this case x+1 and x are both denominators
tawny snow
Ignoring the blue writing, there is one denominator.
young bobcat
Ignoring the blue writing, there is one denominator.
one denominator in which case? are you talking about just the implied fractions in the numerator, the implied fractions in the denominator, or the whole fraction overall?
tawny snow
In the black original question, there is one denominator in the whole original question.
young bobcat
In the black original question, there is one denominator in the whole original q...
well the -1 are implied denominators
$(x+1)^{-1}=\frac{1}{x+1}$
soft zealotBOT
Arnavutköy
young bobcat
$x^{-1}=1/x$
soft zealotBOT
Arnavutköy
tawny snow
I know that
young bobcat
I know that
so we get rid of the exponents and we can rewrite this as $\frac{\frac{1}{x+1}-\frac{1}{x}}{1+\frac{1}{x+1}}$
soft zealotBOT
Arnavutköy
young bobcat
now you need to do two things:
- simplify $\frac{1}{x+1}-\frac{1}{x}$
- simplify $1+\frac{1}{x+1}$
tawny snow
K
soft zealotBOT
Arnavutköy
tawny snow
And how do I simplify? ( x+1 ) / ( x+1 )?
young bobcat
And how do I simplify? ( x+1 ) / ( x+1 )?
multiply 1/x+1 by x/x in numerator. mmultiply 1/x by x+1/x+1 in numerator. multiply 1 by x+1x+1 in denominator. multiply 1/x+1 by nothing in denominator.
young bobcat
uh sorry i gtg now
or like in a few min likely
i think i have given you enough to solve the problem
only other thing you need to know is that $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$
soft zealotBOT
Arnavutköy
proper dagger
where's the original problem
tawny snow
This
gentle zephyr
final saddleBOT
sudden comet
Can you translate?
gentle zephyr
turbid burrow
do you know the definition of an inner product (dot product)
and also norm of vector
and do you know their geometric intuitive descriptions
gentle zephyr
and do you know their geometric intuitive descriptions
a.b = |a|.|b|.cos(θ)
fair bison
a.b = |a|.|b|.cos(θ)
that's a formula, not an intuition
fathom tundra
First step is to look when |V•W| can be zero, can you get an equality then?
turbid burrow
$a\cdot b = \sum_{k=1}^n{a_k b_k}$
soft zealotBOT
walltile
turbid burrow
would be the dot product
but for this problem, you probably don't need that
do you know how to prove cauchy schwartz inequality?
bc this problem should be solvable without using a specific inner product
but it might be easier to use the definition of dot product. bc it should be equivalent for all inner products
fathom tundra
I think the proof of the theorem kinda solves the whole problem if he's familiar with it
turbid burrow
$|x|=\sqrt{x\cdot x}$
turbid burrow
I think the proof of the theorem kinda solves the whole problem if he's familiar...
it would, but you would need to think about the implication to solve the problem
fathom tundra
it would, but you would need to think about the implication to solve the problem
Yeah I don't know how to show his work in that case lol
soft zealotBOT
walltile
turbid burrow
is the definition of norm
rewrite the cauchy schwartz inequality using the definition of dot product and norm (with a dot product) and you should see when they'd be equal
gentle zephyr
rewrite the cauchy schwartz inequality using the definition of dot product and n...
we dont know the number of coordinates V has
V = (v1, ... , vn)
gentle zephyr
θ=0
fathom tundra
we dont know the number of coordinates V has
The Schwartz's-Cauchy inequity holds for all scalar products and norms induced by them so you shouldn't use the standard dot product so this doesn't matter
gentle zephyr
and what does that say about u and v
idk
turbid burrow
what is theta
gentle zephyr
angle between them
gentle zephyr
and what does that say about u and v
parallel
terse crypt
in other words, u and v are scalar multiples of each other
This
gentle zephyr
how to prove it
terse crypt
@gentle zephyr the question asks “is there any”
Showing existence of one case is enough
gentle zephyr
the answer is "yes" then?
terse crypt
Yes
terse crypt
“Yes, consider ${\theta = 0}$. [Explanation]”
soft zealotBOT
k
gentle zephyr
how do I explain it
terse crypt
Show that ${|V \cdot W| = ||V||||W||}$
soft zealotBOT
k
gentle zephyr
Show that ${|V \cdot W| = ||V||||W||}$
wdym
gentle zephyr
Show that ${|V \cdot W| = ||V||||W||}$
v,w ∈ R^n
gentle zephyr
Show that ${|V \cdot W| = ||V||||W||}$
help please
terse crypt
What do u think u’re supposed to do here
gentle zephyr
how do I prove it
terse crypt
What’s ur plan
gentle zephyr
Show that ${|V \cdot W| = ||V||||W||}$
how?
turbid burrow
a fast way is to state that u=kv where k is a scalar
and use the dot product formula in the problem statement
turbid burrow
a.b = |a|.|b|.cos(θ)
this formula
terse crypt
Yes
atomic tinsel
Can someone help me?
final saddleBOT
Can someone help me?
late bolt
Can someone help me?
Tension acts on the other weight pulling it up
The force on the left weight is T - friction - weight component
And on the right it’s weight - T
U then progress by using the fact that they will have the same acceleration
atomic tinsel
yeah but i got a negative value
late bolt
yeah but i got a negative value
Can I see ur working
atomic tinsel
wait
late bolt
,, \frac{T-80}{8} = \frac{70-T}{7}
soft zealotBOT
AnitaG
atomic tinsel
atomic tinsel
So, is this problem wrong?
late bolt
Possibly our assumption that friction is 40N is wrong
Since that is the max friction
late bolt
So, is this problem wrong?
Maybe that too, not sure
Also the friction cannot be acting down the slope if the acceleration is negative
Scratch that it can it’s just atypical
Iirc the formula is MAX friction = mu * R
late bolt
Iirc the formula is MAX friction = mu * R
So maybe the friction isn’t limiting in this case
atomic tinsel
if we assume 7kg go up?
late bolt
if we assume 7kg go up?
But that’s not what the question asks
atomic tinsel
So maybe the friction isn’t limiting in this case
is this same as 1/sqr(3)
late bolt
is this same as 1/sqr(3)
No that’s the coefficient, mu
atomic tinsel
is they equal
late bolt
Iirc the friction is mu R only if the opposing force is greater to or equal than mu R
late bolt
is they equal
Nah u have to multiply that by the reaction force, as u have done
atomic tinsel
i mean the question give miu = this
late bolt
Yeah no that’s correct
But maybe friction isn’t limiting in this scenario
U got an MS by any chance?
Or maybe it is still going down
But that would require the particles to not be stationary in the first place
atomic tinsel
U got an MS by any chance?
i am a/l🙂
atomic tinsel
What does that mean
advanc level
atomic tinsel
no
late bolt
Oh damn
atomic tinsel
may be fricition wrong here?
late bolt
Yeah the friction could potentially be less but idk how we’d find it if it was
atomic tinsel
can you tell me is the one in photo =1/sqr(3)
late bolt
Too many unknowns
late bolt
can you tell me is the one in photo =1/sqr(3)
Yeah that’s the coefficient of friction
atomic tinsel
Yeah that’s the coefficient of friction
ok
late bolt
F_max = that * normal reaction force
atomic tinsel
but the all questions asking about 7kg going down you see?
late bolt
But maybe friction is not limiting, though we’d need the other stuff like tension for example to find the friction
atomic tinsel
and thengoing up
late bolt
but the all questions asking about 7kg going down you see?
Yeah
Does it say anything about initial velocity?
atomic tinsel
Does it say anything about initial velocity?
nope
late bolt
The only way I see it going down is if it has some initial velocity also going down
atomic tinsel
Even if the 7-kg goes down, will it go up again after hitting the ground?
late bolt
Even if the 7-kg goes down, will it go up again after hitting the ground?
Maybe bc the string will go slack and the tension goes away
I’ll be honest my memory’s not so fresh on this type of question
atomic tinsel
the question is weird one😑
late bolt
Look anyways I wish u the best of luck but I’m officially baffled as well
Ima go have breakfast
atomic tinsel
haha, thanks enjoy your breakfast 😄
final saddleBOT
@atomic tinsel Has your question been resolved?
jade fable
You guys think you are good?
final saddleBOT
You guys think you are good?
jade fable
11a-10c=15
drowsy epoch
Nah thats tough