complete the square
#help-17
finite hatch
finite hatch
lmao i think i got it now but thx :D
quick vigil
complete the square
That isn't always that easy to do
quick vigil
Alr
finite hatch
I’m confused with the brackets
can I just put double brackets
oh
wait
I think I had a bracket error
oh nvm
i frogot the 16 on the bototm
oopssssssssss
ok noted thx guyz i understand completing the square now
tis quite easy\
and i will nbever usie the quadratic formula
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vocal sleetBOT
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median gate
coral ibex
How to do? (Highschool lvl easy)
vocal sleetBOT
How to do? (Highschool lvl easy)
gentle thicket
How to do? (Highschool lvl easy)
,rotate
twin meteorBOT
steady lake
Integral calculus
I am stuck in the yellow highlighted line. I have no prior understanding of where 1/10 comes from.
The second picture is from where I cant work it anymore
gentle thicket
**Integral calculus**
I am stuck in the yellow highlighted line. I have no prior...
!occupied
vocal sleetBOT
**Integral calculus**
I am stuck in the yellow highlighted line. I have no prior...
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).
steady lake
Thank you so much
gentle thicket
!status
vocal sleetBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
split wind
**Integral calculus**
I am stuck in the yellow highlighted line. I have no prior...
hello there Raiven, i bet during your typing progress someone already claimed this channel, you'll have to copy your question and paste it on an unoccupied channel
steady lake
hello there Raiven, i bet during your typing progress someone already claimed th...
okay thank you!
proven whale
hi!
full ferry
How to do? (Highschool lvl easy)
do you know the composition of an alpha particle?
coral ibex
do you know the composition of an alpha particle?
Theres 2 protons and neutrons?
full ferry
Theres 2 protons and neutrons?
yes
and the uranium loses 22 protons so how many alpha particles are emitted?
coral ibex
11
full ferry
correct
coral ibex
Uh not sure
full ferry
a beta particle has a charge of +1 electrons
coral ibex
10 beta particles, idk why I am stuckhere
coral ibex
Ah okok
vocal sleetBOT
@coral ibex Has your question been resolved?
thorny galleon
i dont get whats going on in D
vocal sleetBOT
i dont get whats going on in D
thorny galleon
1/2 (7-2) 10 was the area of the triangle inside
bitter pilot
They subtract from the triangle area the little parabola area
Where they integrate from B to C
halcyon forum
help i cant visualise this word problem 😞
vocal sleetBOT
help i cant visualise this word problem 😞
halcyon forum
uhh yess
grim lotus
can you imagine the equator
halcyon forum
yes
grim lotus
forget the equator
imagine a circle going through the north and south pole
vertically
halcyon forum
yes..
grim lotus
draw a line from the centre to a point on the circle
i mean basically you wanna focus on this circle alone
the radius of this circle is the same as of the sphere
halcyon forum
rough sketch but—
mild flower
drawing not to scale
grim lotus
yes
halcyon forum
OHHHH
so just to specify whats the question trying to say and whats the 5°N to 8°N latitude??
grim lotus
drawing not to scale
should it be 1 degree
grim lotus
ah
grim lotus
so just to specify whats the question trying to say and whats the 5°N to 8°N lat...
so imagine the angle in the middle is 5 degrees and whip it around the sphere
halcyon forum
and then?
halcyon forum
ouu so like across the circle? like horizontal?
mild flower
so just to specify whats the question trying to say and whats the 5°N to 8°N lat...
0ºN would be at the equator, 5ºN would be, well, 5 degrees north of that. 8ºN would be 3 further degrees north of that. The north pole would be 90ºN
grim lotus
np
mild flower
halcyon forum
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vocal sleetBOT
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minor dagger
Hi! Would anyone be able to tell me what I would have to search to find more information on solving problems like this one? I am trying to find a video that goes through this type of problem but lack the terminology to find one. Thanks!
vocal sleetBOT
Hi! Would anyone be able to tell me what I would have to search to find more inf...
twin horizon
Hi! Would anyone be able to tell me what I would have to search to find more inf...
Express all the terms as products of prime factors
dark rune
simplifying algebraic expressions?
minor dagger
awesome thank you so much i was able to find something!!!! ^_^
dark rune
Express all the terms as products of prime factors
huh?
vocal sleetBOT
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Available help channel!
night zephyr
How would I go about proving that (2^35)(2^33-1)+1 is composite? I can't seem to factor it into some sort of (a-b)(a+b)=a^2-b^2 and I'm not sure how else to approach it
vocal sleetBOT
How would I go about proving that `(2^35)(2^33-1)+1` is composite? I can't seem ...
dark rune
How would I go about proving that `(2^35)(2^33-1)+1` is composite? I can't seem ...
just multiply it out
vocal sleetBOT
@night zephyr Has your question been resolved?
vocal sleetBOT
@night zephyr Has your question been resolved?
vocal sleetBOT
@night zephyr Has your question been resolved?
white palm
How would I go about proving that `(2^35)(2^33-1)+1` is composite? I can't seem ...
factoring is the way
multiplying it out is also part of it
dense eagle
How would I go about proving that `(2^35)(2^33-1)+1` is composite? I can't seem ...
note that 2^35 is very close to 2^33
in fact, 2^35 = 4(2^33)
if we set x=2^33, see if you can do something with that
solution: ||2^35(2^33-1)+1 = 4x(x-1)+1 = (2x-1)^2||
ruby hill
So I have two 3x3 matrices and their cross product (no matter the order) is an identity matrix. What can I say about their relationship?
vocal sleetBOT
So I have two 3x3 matrices and their cross product (no matter the order) is an i...
flat whale
So I have two 3x3 matrices and their cross product (no matter the order) is an i...
Do you know what inverse is
And you probably don't mean cross product
Just product
vocal sleetBOT
vocal sleetBOT
Available help channel!
night zephyr
solution: ||2^35(2^33-1)+1 = 4x(x-1)+1 = (2x-1)^2||
thanks
night zephyr
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median sun
for all n natural number, n/4n-1 <= 1, is true isn't it? because: n < 4n then n + 1 <= 4n
vocal sleetBOT
for all n natural number, n/4n-1 <= 1, is true isn't it? because: n < 4n then n ...
median sun
hehe char
so, if I'm proving that the sequence n^2 - n / 4n^2 - 1 converges to 1/4, this arguments are correct?
| (n^2 - n / 4n^2 - 1 ) - 1/4 | = | - (4n - 1/ 16n^2 - 4)| = 4n - 1/ 16n^2 - 4
since 0 < 16n - 4 < 16n^2 - 4
(4n - 1/ 16n^2 - 4) < (4n - 1 / 16n - 4) < (4n/16n-4) = n/4n-1 < 1, for all n natural number
since 1 > 0 I can conclude that the sequence converge to 1/4, right?
dull bear
so, if I'm proving that the sequence n^2 - n / 4n^2 - 1 converges to 1/4, this a...
Hmmm, not really 
if you're trying to do it formally, remember that you wanna show that, given any distance, you can make the distance between (n^2 - n) / (4n^2 - 1) and 1/4 smaller than that distance
At the moment, that only shows that the distance between them is at most 1, but not arbitarily small 
dull bear
Mm so this form that a classmate made is the correct way?
Yep, cause then you can make that 1/n as small as you want to 
median sun
Okay, thank you, also
by intuition I affirm that n^6 - 5n^3 - 2n -1 > n^3, but now my classmate is making me doubt about it, that inequality is wrong?
dull bear
Well, from some point on, it will definitely be true, but it needn't be true for all values of n (indeed, it isn't true for e.g. n = 1, left hand side is -7, right is 1)
median sun
Mmm and if it's no true for all values of n, then the proof is wrong
dull bear
Mmm and if it's no true for all values of n, then the proof is wrong
Which proof? 
median sun
dull bear
It's easy to work around, you don't need it true for all n, but from some n on
,w solve n^6 - 5n^3 - 2n -1 > n^3,
twin meteorBOT
dull bear
It's true for all integers from 2, it seems, so just make your N to be at least 2, in additon to any other conditions you have on it 
median sun
So If in my proof I write that N >= 2 all the other procedures are correct? can I always make that N >= x, for some x, and the definition of converge continuos being valid?
dull bear
So If in my proof I write that N >= 2 all the other procedures are correct? can ...
Yep, if you choose your N to be larger than both 2 and 1/epsilon, you're happy and needn't worry
median sun
Yep, if you choose your N to be larger than both 2 and 1/epsilon, you're happy a...
just write that N have to larger that 2 and 1/epsilon, it's not necessarily write something like M = max{2, 1/epsilon}?
dull bear
just write that N have to larger that 2 and 1/epsilon, it's not necessarily writ...
If you do, remember that $\frac1{\epsilon}$ may not be an integer, so e.g. $\max{2, \ceil{\frac1{\epsilon}} + 1}$ would be a good choice instead, but otherwise
twin meteorBOT
@dull bear
analog scaffold
Hello, can you explain me why floor(1/epsilon) + 1?
median sun
Hello, can you explain me why floor(1/epsilon) + 1?
he is my classmate lol
dull bear
Hello, can you explain me why floor(1/epsilon) + 1?
If 1/eps is an integer, then you won't have the strict inequality
dull bear
I mean, is that the ceiling or the floor function applied to 1/epsilon?
Ceiling 
Misinterpreted the question, it's the ceiling (though the floor works too, I believe)
analog scaffold
good, but then why is ceil(1/epsilon) + 1 being used in the max function? I can understand the 2 but not that other number
our teacher wants us to proof more rigourously than andrew wiles did with fermat's theorem, so we shoudnt take a single stuff by granted lol
dull bear
good, but then why is ceil(1/epsilon) + 1 being used in the max function? I can ...
Means that you get the 1/N < epsilon part
median sun
If you do, remember that $\frac1{\epsilon}$ may not be an integer, so e.g. $\max...
So, the sketch of the proof will be something like this?
Proof that n/n^6 - 5*n^3 - 2n -1 converge to zero
For all n > 2, we have that n^6 - 5n^3 - 2n -1 > n^3 > n^2, then
(n / n^6 - 5n^3 - 2n -1 ) < (n / n^2) < 1/n
since n >= N, 1 / n <= 1/N < epsilon
I'm lost with that n > 2 and n >= N and the maximum 
dull bear
So, the sketch of the proof will be something like this?
Proof that n/n^6 - 5*n^...
Yep, that's it, the aim now, is that you want to find an N, such that you get the 1/N < epsilon
The n > 2 allows you to make the claim that (n / n^6 - 5*n^3 - 2n -1 ) < (n / n^2) < 1/n, so you need to make sure that you have N > 2 so that you can use it
median sun
sure transitivity n >= N > 2
analog scaffold
but by the definition of convergence we suppose that n >= N, which means that 1/n <=1/N. And then we can use the archimedean property by saying that for epilson > 0 there is N natural such that 1/N < epsilon, but since 1/n <= 1/N by transitivity we have that 1/n < epsilon
dull bear
Oh if you're allowed to use the Archimedean property, then yea you can literally just say that too, choose N to be larger than both 2 and 1/epsilon in that case
median sun
The n > 2 allows you to make the claim that (n / n^6 - 5*n^3 - 2n -1 ) < (n / ...
But, as my classmate mentioned, our teacher is very rigorous, so how can I justify that claiming N>2 won't affect the definition of convergence? :c
dull bear
But, as my classmate mentioned, our teacher is very rigorous, so how can I justi...
Because you're allowed to pick the N however you want to, you can't choose epsilon, but you can choose N such that it gets things that make your life easier!
median sun
Because you're allowed to pick the N however you want to, you can't choose epsil...
Mm it's because the definition said that exists N such that for all n, n >= N?
analog scaffold
So the proof will be as follows
For all n > 2, we have that n^6 - 5n^3 - 2n -1 > n^3 > n^2, then
(n / n^6 - 5*n^3 - 2n -1 ) < (n / n^2) < 1/n
By the Archimedean property, for epsilon > 0 there is N natural such that 1/N < epsilon. By definition of convergence, suppose that n >= N, then 1/n <= 1/N and by transitivity 1/n< epsilon. Again using transitivity we have that (n / n^6 - 5*n^3 - 2n -1 ) < 1/n, for n >=2
is that correct?
I mean we already found N which would be 2 I guess, but I think there might be something missing
dull bear
Mm it's because the definition said that **exists N** such that for all n, n >= ...
Yep, some N exists such that you can whatever n >= N you choose, that happens - you just need to find a particular N, or justify why one exists
median sun
I mean we already found N which would be 2 I guess, but I think there might be s...
jajajaja lo que provoca Sigi con tanta perfección :')
analog scaffold
sigifredo me esta volviendo loco xd
median sun
Yep, some N exists such that you can whatever n >= N you choose, that happens - ...
oooh I think that finally I understand! thank you sooo much char <<33 hahaha sorry for being two people bothering you :')
analog scaffold
You guys are the bomb
dull bear
I mean we already found N which would be 2 I guess, but I think there might be s...
N needs to be both at least 2, so you can have the (n / n^6 - 5n^3 - 2n -1 ) < (n / n^2) ,
then you need N to also be the one such that 1/N < epsilon too,
so pick the larger of those two
analog scaffold
got it
dull bear
oooh I think that finally I understand! thank you sooo much char <<33 hahaha sor...
You're never a bother 
and lovely to meet you too @analog scaffold 
analog scaffold
pleasure is mine mate
median sun
You're never a bother <:LoveYou:672804816497541130><:LoveYou:672804816497541130>...
Again thank you <3
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vocal sleetBOT
What to do
flat whale
What to do
Area of rectangle is width times length
vocal sleetBOT
@spare hound Has your question been resolved?
spare hound
spare hound
north sky
you just simply sub them as it shown in the bracket
spare hound
Ok
spare hound
you just simply sub them as it shown in the bracket
What is sqrt2x times 5x-4
vocal sleetBOT
@spare hound Has your question been resolved?
tame river
Let me see here,
vocal sleetBOT
Let me see here,
lost lake
(8x+5)^2/3 is just (8x+5)^-1/3 • (8x+5)
Since 1-1/3=2/3
So you can factor out (8x+5)^-1/3
tame river
-7x (-24x-15)^-1
lost lake
(8x+5)^1/3 • (-7x-3(8x+5))
(8x+5)^1/3 • (-31x-15)
lost lake
i get -7x -3(8x+5)^-1
That -1 should be +1 tho
tame river
i get -7x -3(8x+5)^-1
4x + 5(x-2) = 5x-10 = 4x+5x-10 = 9x-10 [prev problem]
-7x-24x-15
-31x-15
${\frac{-1}{3} - \frac{2}{3}}
${\frac{-1}{3} - \frac{2}{3}}$
twin meteorBOT
MutinyHasMail4liw
tame river
${\frac{-3}{3}$
twin meteorBOT
MutinyHasMail4liw
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
tame river
-3/3 = -1
silk osprey
${\frac{-3}{3}$
you don’t need the braces around the \frac
tame river
you don’t need the braces around the \frac
oh ok
silk osprey
it’s $\frac{-3}{3}$
twin meteorBOT
knief
vast shale
Or just -1?
silk osprey
yes but i was showing him the latex
vast shale
Ok
tame river
(8x+5)^1/3 • (-7x-3(8x+5))
(8x+5)^1/3 • (-31x-15)
is (8x+5)^1/3 • (-31x-15) the correct answer?
lost lake
I think so
vocal sleetBOT
timid rapids
<@&286206848099549185>
surreal basin
work = integral force dx, so the question wants you to use the midpoint to estimate integrating f(x) from 2 to 34
timid rapids
mhm
surreal basin
n = 4 means you should be considering 4 intervals
can you name those 4 intervals
(the first one is [2, 10]
surreal basin
yep
timid rapids
cool so how would i set up the integral
surreal basin
anyways midpoint rule then has you estimate the area for each interval
for [2, 10], midpoint rule estimates it as a rectangle with width 8 and height f(6)
timid rapids
oh i see
surreal basin
it chooses the x-coordinate in the middle of the two ends to use for the height
surreal basin
yes
timid rapids
then take the average?
surreal basin
no
timid rapids
so then what
surreal basin
calculate the four rectangle areas first
surreal basin
@timid rapids did you calculate them?
timid rapids
nah im just wondering what the next step is
my bad im working on a seperate problem right now but i was planning to js do that problem later with ur steps
surreal basin
lmk when youre good to get back to this problem
you said that once you have the four rectangle areas, you average them
whats the reasoning behind that
timid rapids
you said that once you have the four rectangle areas, you average them
whats the...
becuase its telling me to estimate on the whole interval so when i have 4 parts of it i just take the average and that should be applicable to the whole interval
surreal basin
put it this way
if you walk 16 meters,
you can split that into you walking 4 meters four times
timid rapids
right
surreal basin
becuase its telling me to estimate on the whole interval so when i have 4 parts ...
you are now saying that if you find out you walked 4 meters four times, you average that to get that you walked 4 meters total
(4 + 4 + 4 + 4) / 4
you couldve just done 4 + 4 + 4 + 4
you dont need to say (4 + 4 + 4 + 4) / 4 * 4, calculate the average then multiply by the number of intervals (which is what Im assuming youre doing, which you did not bother to say)
timid rapids
you couldve just done 4 + 4 + 4 + 4
oh lol
surreal basin
so once you have the four rectangle areas, what would you do to find the integral
surreal basin
you can split that into you walking 4 meters four times
youre missing a crucial hint
what's the first rectangle area?
timid rapids
48
timid rapids
uh no
surreal basin
whats the height of the first rectangle
timid rapids
6?
surreal basin
and whys that
timid rapids
cuz 7+5/2 = 6
surreal basin
bro tried a wild guess and got surprised when its wildly wrong
timid rapids
lolll
surreal basin
for [2, 10], midpoint rule estimates it as a rectangle with width 8 and height f...
go look here
I mention "f(6)"
and (2 + 10) / 2 = 6... what could it all mean 🤔
try taking another guess at what midpoint rule would do with this
timid rapids
so since 6 is the x coordinate we plug 6 into f(x) to get the height f(6) and then multiply the two to get the area?
surreal basin
youre leaving out where the 6 comes from
timid rapids
it comes from the average of 2 and 10
surreal basin
why 2 and 10, then?
timid rapids
because thats the interval we are evaluating
surreal basin
thats good
timid rapids
and theres also 3 more intervals too
surreal basin
when youre multiplying "the two", f(6) is one of them, whats the other
timid rapids
the other is 6
surreal basin
6 * f(6)?
timid rapids
oh sorry 8
timid rapids
cuz its the distance between 10 and 2
surreal basin
yep
timid rapids
so the area would actually be 47.2
timid rapids
right
surreal basin
thats how midpoint rule would estimate for an interval
with that, you have four rectangles
timid rapids
right
surreal basin
they will each give you an area
as a result, you have an area that estimates each of [2, 10], [10, 18], [18, 26], and [26, 34]
how then do you estimate the area of [2, 34]
surreal basin
youre missing the hint again
surreal basin
if you walk 16 meters,
look back here
when I split up the walking into 4, 4, 4, 4
I combined them back together by adding them
timid rapids
right
surreal basin
because thats how I walked in each of them
and so when I combine them
thats how I walked in total
surreal basin
as a result, you have an area that estimates each of [2, 10], [10, 18], [18, 26]...
youve got an area for consecutive stretches of the same function
surreal basin
how then do you estimate the area of [2, 34]
how do you combine them
timid rapids
add them up then divide by 4
surreal basin
why divide?
timid rapids
cuz theres 4 intervals
surreal basin
youre not finding the average
surreal basin
np
whats important is that you essentially got here on your own
next time you wont need to go through this again
haughty rover
Find all values of z such that $z^3 = -8i$
vocal sleetBOT
Find all values of z such that $z^3 = -8i$
twin meteorBOT
pixel
haughty rover
$z = (-8i)^{\frac{1}{3}} \ \ \text{Convert to Polar form} \ r = 8 \ \theta = tan^{-1}(\frac{-8}{0}) = \text{undefined}$
sly sierra
twin meteorBOT
pixel
sly sierra
draw a picture to see what the angle is
sly sierra
plot the point z = -8i, what angle does it make with the origin?
haughty rover
plot the point z = -8i, what angle does it make with the origin?
its not z = -8i tho
z^3 = -8i
make any didfference?
sly sierra
sure but you start with z in polar form right?
sly sierra
sorry, let's call it w
w = -8i
you're looking for the 1/3 roots of w
so put w in polar form
twin meteorBOT
pixel
sly sierra
$w = 8e^{i\theta}$
twin meteorBOT
Bungo
sly sierra
what is theta?
haughty rover
90?
sly sierra
no
haughty rover
oh wait
sly sierra
how do you write your polar forms?
yes, 3pi/2
ok so magnitude is 8 and angle is 270 or 3pi/2
so how do you find the cube roots from that?
sly sierra
all good, you'll probably eventually learn the e^(i n theta) notation and why it is used
haughty rover
$z = 2cis(\frac{\pi}{2})$
twin meteorBOT
pixel
sly sierra
yea that's one of them
to get the others you can write:
$w = 8cis(\frac{3\pi}{2} + 2\pi n)$
twin meteorBOT
Bungo
sly sierra
since adding multiples of n doesn't change the value of cis
then when you divide by 3...
haughty rover
**Bungo**
oh yeah that form just means that like
the value repeats every 2pi?
if that makes sense
sly sierra
yea
so when you take cube roots, that becomes $2cis(\frac{\pi}{2} + \frac{2\pi}{3}n)$
twin meteorBOT
Bungo
sly sierra
which tells you all three roots (take n=0,1,2)
haughty rover
wouldnt the value just repeat itself everytime cause going 2pi is just going in a circle
sly sierra
but after dividing by 3 you have 2pi/3, not 2pi
haughty rover
ohhh ya
sly sierra
so you're adding multiples of 2pi/3 to pi/2
sly sierra
it will repeat in a cycle with period 3
haughty rover
which changes it
sly sierra
yep
shadow oak
What is area of line AD when there's triangle ABC while AB=AC and point D is over line AB while BD =5 ,FC=13
vocal sleetBOT
What is area of line AD when there's triangle ABC while AB=AC and point D is ove...
@shadow oak Has your question been resolved?
sacred ruin
What is area of line AD when there's triangle ABC while AB=AC and point D is ove...
where the point F located
shadow oak
Extended line of AC
vocal sleetBOT
@shadow oak Has your question been resolved?
ancient sorrel
picture?
graceful aspen
vocal sleetBOT
pine hull
What's the doubt
pine hull
Okk
pine hull
i have no idea what any of this means
Well literal equation are ones where multiple variables are involved
graceful aspen
okay
opaque minnow
hi
graceful aspen
okay so it’s a bunch of different equations
and the answer will have a number of letters
so if i’m trying to find the answer would i need to solve all of the equations
opaque minnow
I am guessing the answer will be more than one
i thinks so
but it's kind of a silly exercise hahaha
pine hull
so if i’m trying to find the answer would i need to solve all of the equations
No, they are not asking for solution
As I can see
graceful aspen
genuinely what
okay so what are they asking for
i promise i’m smart
just not in math
opaque minnow
bro
pine hull
okay so what are they asking for
Can you show another question
graceful aspen
Can you read?
it doesn’t make sense
graceful aspen
wait hold on this is a bad question
opaque minnow
dont over think
pine hull
Well this is easy, they are asking first step . For an example
x+2 = 3
First step will be Subtracting 2 both side
graceful aspen
oh okay so it’s 6-4
pine hull
Yeah combine same term as it says for an example
2x= x + 8
Subtracting x both side
2x-x = x+8 -x
x = 8
graceful aspen
so it’s 2kx =8
pine hull
the second one
pine hull
Solve the equation, two kx terms are there so can you add them?
graceful aspen
slow down
sorry
oh okay
so the first step has already been done
so i can add the common variables
and it would be 5kx=10
graceful aspen
okay so then would i have to divide?
pine hull
How do you just keep x on left side ?
graceful aspen
so i have to divide 5k
pine hull
Yep
graceful aspen
?
pine hull
Use / always in variable
graceful aspen
oh i thoight it was because you divide it and then you do something with the extra letter on the other side
how did it because 10/5k
become*
oh wait is it 2
x=2k
okay so it’s not that either
pine hull
how did it because 10/5k
10/5k meaning
10/5 * 1/k
2 * 1/k
graceful aspen
i’m not good at fractions i’m ngl to you 😭😭💀
pine hull
Well that's going to be difficult then 😅
To get x value you need to divide 5k both side so
5kx/5k = 10/5k
x = 10/5k```a denominator?😭
graceful aspen
yes
pine hull
10/5 is what?
What will be this value?
pine hull
Yes so numerator becomes 2
pine hull
10/5k can’t be the final answer
What is left in denominator
pine hull
Yup
graceful aspen
so x=2/k
pine hull
pine hull
No prob
pine hull
Sure
pine hull
can you add the like terms
Shortcut is actually you can move terms from left to right or right to left by changing signs, that's how I used to do it (don't do this)
Kids nowadays do it like subtracting or adding terms on both side
For an example
So you will subtract 2x both side
2x+ y -2x = 3x-2x
y = x```You need to subtract or add something both side so that left side you get y term and right side you get a term
graceful aspen
so you switch signs after you add or subtract
civic otter
No
pine hull
so you switch signs after you add or subtract
Ignore the switch part I said, just follow the example
graceful aspen
okay okay
pine hull
4y-3a = 13a
What can you add or subtract both sides , so that you remain with only 4y at the left side ?
pine hull
Well that's one way to do it , but you can think of removing a term from left end instead of y term
pine hull
4y-3a = 13a
What can you add or subtract both sides , so that you remain with on...
@graceful aspen
graceful aspen
can you do -9
civic otter
If I take away 3 apples, how many should you bring to restore the original number?
pine hull
a is also there
graceful aspen
okay okay
so my goal is just to get two variables on the same side so i can add them
pine hull
No, your goal is to keep the 2 variables on opposite side
pine hull
But you have one -3a variable also on left side
graceful aspen
so it’s 4y=16a
pine hull
Yep
pine hull
pine hull
Next one is same
graceful aspen
okay wait i’ll try and solve it on my own
pine hull
You have px term on left side that you probably want to remove
pine hull
okay wait i’ll try and solve it on my own
Sure
graceful aspen
pine hull
does it just because of
7px-3px will be something in terms of px
pine hull
You are doing
12n + 3px - 3px = 7px - 3px right?
graceful aspen
yes
pine hull
Soo 3px-3px will be what?
pine hull
Yep
graceful aspen
then why did you say earlier where did ir go
graceful aspen
4px
pine hull
Its alright
graceful aspen
so it’s not 4px?
graceful aspen
So 12n = 7px-3px
after this
it’s this
pine hull
so it’s not 4px?
It is 4px
graceful aspen
imagine the n is after the 12
pine hull
12n = 4px
Now you need answer as x = something
graceful aspen
how do i get it over
pine hull
How can you get x = something
What do you need to divide or multiply both side ???
graceful aspen
let me think
pine hull
12n = 4px
graceful aspen
divide by 4p
pine hull
Yep
graceful aspen
so 3n/p=x
pine hull
Yes
graceful aspen
how do i get the x on the other side?
pine hull
1 = x is same as x = 1
graceful aspen
so i can just move it
civic otter
1 = x is same as x = 1
Computer scientists: 🤔
pine hull
graceful aspen
pine hull
Your welcome
graceful aspen
i didn’t actually think i would learn anything
pine hull
I am glad if you did
graceful aspen
have a good rest or your morning/day
pine hull
have a good rest or your morning/day
Make sure to do .close if you are done
graceful aspen
.close
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plush bloom
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plush bloom
I'm stuck :(
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@plush bloom Has your question been resolved?
plush bloom
<@&286206848099549185>
shadow urchin
Hmm...
plush bloom
We only need to find the similarity ratio between YBZ and XDZ though...
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plush bloom
plush bloom
tf is that diagram
ikr 
plush bloom
somebody help
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inner osprey
how did you get the numbers written in purple?
vocal sleetBOT
@plush bloom Has your question been resolved?
plush bloom
how did you get the numbers written in purple?
Powers of points and Ptolemy's
inner osprey
they're incorrect
muted anchor
These are incorrect too
There's an accurate version of the figure if it helps, note that BZ is not actually tangent to that circle, it's just very close (assuming there's no numerical error going on, but it says it's about 1° so I doubt it)
dusky ravine
Powers of points and Ptolemy's
No
It's triangle TYD is directly an interior
And BZ is tangent
vocal sleetBOT
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plush bloom
Is this your drawing
muted anchor
Yes
plush bloom
And BZ is tangent
Why…?
plush bloom
anyone know why they're tangent ;-;
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hexed pagoda
the square root of x to the power of 4 is x² = Ix²I why not -x² because -x² . -x² = x to the power of 4
civic otter
the square root of x to the power of 4 is x² = Ix²I why not -x² because -x² . -x...
!occupied
vocal sleetBOT
the square root of x to the power of 4 is x² = Ix²I why not -x² because -x² . -x...
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).
hexed pagoda
and now
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plush bloom
unsolved for 3 days 
vast shale
Okay, let me say somethings, even if it might be obvious to you.
Or actually I have to first ask: How did you determine that the angles marked with double line at Y and D are equal, and the angles marked with single line at B and X are equal?
Okay: Assuming your angles are correct, then the triangles YBZ and XDZ, whose area you are supposed to compare, are similar.
So the ratio of area(YBZ)/area(XDZ), which equals p/q as a simplified fraction, is the square of the ratio of corresponding sides of those two triangles. In other words,
p/q = (BY/DX)^2 = (BZ/ZX)^2 = (YZ/DZ)^2
I hope this helps.
plush bloom
Quadrilaterals TYZD and TBZX are cyclic.
I've been trying to find YZ/DZ, but I've been stuck on it for five days.
|| @vast shale ||
vast shale
Ok I understood the angle thing, thanks. So because TYZD and TBZX are cyclic quadrilaterals, ∠TDZ and ∠ZYB add up to 180 degrees, but ∠TDZ and ∠ZDX also add up to 180 degrees, implying
∠ZDX = ∠ZYB,
and an analogous argument can be used to prove
∠YBZ = ∠DXZ.
I am slowly catching on to what you understood days ago 😅 but who knows, I may yet be able to solve it
Can you explain how you got the lengths?
Just wondering if Herons formula for the area of a triangle, together with similarity of triangles YZB, ZXD, YXT together with the given lengths 6,7,8 might be enough to solve it.
But I think that will not work.
Ok, I have got a kind of idea what one could do, but I have not worked out the details.
Step 1: Because in the triangle BDT, all lengths are known, it should be possible to determine all angles. using these, the angles in the similar triangles YZB, ZXD and YXT should also be determinable, so also the ratio of the sides there.
Step 2: Use previous step together with the fact that TX = DX + 7 and TY = BY + 6 to find two equations which also contain scaling factors of the triangles.
Then maybe divide one equation by the other, or something like that, to get p/q or something like that.
Besides all these partial ideas, the basic thought I want to share is that the lengths 6,7,8 all need to enter the calculation
gloomy sapphire
wait, I replied to the wrong message
gloomy sapphire
Ok, here is the corrected version @plush bloom:
Use circumradius formula to get length TM, set r1 = 1/2 TM (for later)
Use law of cosines to get angles T, D, and B in triangle TBD
Angles DTM and BTM are half of angle T, since the line connecting a vertex to the incenter bisects the angle
Use law of cosines to get lengths MD and BM in triangles TDM and TBM respectively
Set s1 = 1/2 MD, s2 = 1/2 BM
Use law of sines to get angles TDM and TBM in triangles TDM and TBM respectively
angle TDM + angle TBM = angle BMD
pi - angle T - angle TDM = angle TYD
pi - angle T - angle TBM = angle TXB
Use law of sines to get lengths YT, TD, and DY in triangle TYD, and length TX in triangle TXB
Use circumradius formula to get circumradius of triangle TBX (r2), and circumradius of triangle TDY (r3)
Let O1 to be the circumcenter of triangle TBD, O2 to be that of triangle TBX, and O3 to be that of triangle TYD.
O1O2 = sqrt(r1^2 - s1^2) + sqrt(r2^2 - s1^2)
O1O3 = sqrt(r1^2 - s2^2) + sqrt(r3^2 - s2^2)
Angle YO1X must be equivalent to pi - angle BMD because YO1 and XO1 are perpendicular to BM and DM respectively.
Use law of cosines to get length O2O3 in triangle O1O2O3
Use Heron's formula to get the area of O2O3Z, then divide double the area by O2O3 to get the height of the triangle.
Double the height to get MZ
Use law of cosines to get angle MO2Z, then halve to get angle MXZ, add to angle TXB to get angle DXZ
TX - TD = DX
Double angle DXZ to get angle DO2Z
Use law of cosines to get distance DZ in triangle DZO2
Use law of cosines to get angle MO3Z, then halve to get angle MYZ
DY - MD = YM
Use law of sines to get angle ZYM; it will be an ambiguous case, but we know the angle ZYM will be obtuse, so that allows us to solve for the correct angle
pi - angle MYZ - angle YZM = angle YMZ
Use law of sines to get length YZ
Now you have all the lengths and angles to solve for the area of the triangles
0.5 sin(angle YZM) * YM * YZ -> area of YBZ
0.5 sin(angle DXZ) * DX * DZ -> area of XDZ
A bit trickier than I initially thought, there's a bunch of variables so I could have forgotten to explain how to explicitly solve for one of them or something, but I should have covered all of the major variables
I think it should be correct, but let me know if you spot any problems with it
plush bloom
Whoa, thank you so much, I'll take a look later.
gloomy sapphire
yeah np
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stable crystal
inner osprey
flat whale
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vast shale
no way
glossy cypress
surreal basin
.reopen
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mental egret
i got the answer for this as it having no solution from augmented matrix method (rank(A|B) != rank A) which is the correct answer .however i tried using cramer's rule by taking the 4th equation as 0's equation but it will give the result as infinite solutions because all formed determinants are singular matrix as their 4th row will always be 0 . Is there something i did wrong ?
vocal sleetBOT
i got the answer for this as it having no solution from augmented matrix method ...
surreal basin
i got the answer for this as it having no solution from augmented matrix method ...
- you found out theres no solutions
- you try to find solutions anyway using cramer's rule
- it breaks
- hmmmm
in general the formed determinants are 0 when the matrix isnt full rank
so cramer's rule will only work on full-rank matrices
mental egret
in general the formed determinants are 0 when the matrix isnt full rank
so cramers rule can only be used when the number of given equations equals number of variables ?
mental egret
the word is "full rank"
how is that different from what i said ?
subtle summit
e.g. the concept of "rank" accounts for redundant equations (if you know x + y = 1, then the equation 2x + 2y = 2 doesn't add anything)
mental egret
you can have 4 equations without full rank
wait , are you saying cramer's rule can't solve anything other than unique soln ? becuase full rank means its determinant is non zero
mental egret
e.g. the concept of "rank" accounts for redundant equations (if you know x + y =...
so linearly independent
surreal basin
wait , are you saying cramer's rule can't solve anything other than unique soln ...
yes
surreal basin
hopefully you read the textbook before calling it out
torn cobalt
vocal sleetBOT
viral copper
- 1 - 1 in numerator, factorize, and split
torn cobalt
+ 1 - 1 in numerator, factorize, and split
????
also can u teach me how to solve that with a gdc
surreal basin
absolutely ananas
viral copper
eh actually now that I think about it I don't think that's as useful as I originally thought
I don't get what you mean by GDC here
vocal sleetBOT
@torn cobalt Has your question been resolved?
torn cobalt
I don't get what you mean by GDC here
cuz i take math ai hl so i kinda just wanna rely on my calc lol
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rustic talon
need help on finding the parabola's vertex cause in here theres only y², typically there's equations like
y²+5y-2x=9 something like that where there's two terms with the same variable
vocal sleetBOT
need help on finding the parabola's vertex cause in here theres only y², typical...
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forest maple
and i know that P(X≥2) = (5C2)⋅0.7^(2)⋅0.3^(3) + (5C3)⋅0.7^(3)⋅0.3^(2) + (5C4)⋅0.7^(4)⋅0.3 + (5C5)⋅0.7^(5) hence P(X=2) = (5C2)⋅0.7^(2)⋅0.3^(3)
vocal sleetBOT
and i know that P(X≥2) = (5C2)⋅0.7^(2)⋅0.3^(3) + (5C3)⋅0.7^(3)⋅0.3^(2) + (5C4)⋅0...
forest maple
those values are: P(X≥2) = 48461/50000 and P(X=2) = 1323/10000
at this point im just confused how to come to this ratio answer
arent you supposed to do frac A divided by frac B to find the ratio?
i swear that does not yield any of the multiple choice answers
this should be a fairly reliable exam paper from what i know
unless you have to subtract the probability of P(X=2) from P(X≥2) before doing the ratio division?
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magic ether
hi
magic ether
I am trying to find the shortest possible lenght for vector v = [3t, t-2]
I want to find this without using any tools
|v| = sqrt((3t)^2 + (t-2)^2)
how do i find derivative of this?
Do I have to set derivative v'(t) = 0 to find minimal / maximum point
and use second derivative to find if its minimal or maxium
?
strong vortex
Sqrt function is increasing, so minimising the inside of the sqrt is sufficient
magic ether
hmm, what dou you mean? that it always increases? idk still how to do this
strong vortex
Instead of differentiating |v|, differentiate |v|^2
strong vortex
Because the smaller the input of sqrt, the smaller the output
So minimising the input minimises the output
strong vortex
That gives v = [0, -2]
Which means |v|=2
There might be a value of t that gives a smaller |v| without using a minimisation process
magic ether
Because the smaller the input of sqrt, the smaller the output
i still dont understand that part lol that they removed sqrt or differneitate lenght^2
confusing
heavy yoke
increasing functions preserve inequality. if f(x) is increasing, then
f(x) > f(y) means x > y
strong vortex
The x axis is like |v|^2
You want to be as left as a possible on the x axis, which is the same as minimising |v|^2
The y axis is then |v|
magic ether
You want to be as left as a possible on the x axis, which is the same as minimis...
idk what you mean by |v|^2
magic ether
yeah i dont understand that
strong vortex
We are minimising sqrt(3t)^2 + (t-2)^2)
If we look at sqrtx (the graph), we say the smaller x is, the smaller sqrtx is
magic ether
yeah
strong vortex
If we wanted to minimise sqrtx
Instead of differentiating, we cld say let’s make x as small as possible
strong vortex
When I say sqrtx, I mean for some generic function, not related to ur question itself
In ur question, there isn’t sqrtx, but sqrt((3t)^2 + (t-2)^2)
magic ether
yeah
strong vortex
We are just seeing what happens to sqrt, as the thing inside it changes
So an different problem, if I asked u how to minimise sqrtx, where x is some random variable, what would x need to be and why?
magic ether
0
strong vortex
And why do u know it’s 0, and some rlly big number not on the graph, say x=100000000000
magic ether
because the function i saw graph of
and sqrt 0 is lowest you can go
cant do -1 etc..
sqrt(1) = 1
sqrt(0) = 0
strong vortex
Yh, so u look at the graph, and see that the leftmost point that is possible must surely be the smallest
magic ether
yeah
strong vortex
So u did the minimisation problem, without taking a derivative, but instead by simply minimising x (taking the leftmost point)
When asked to minimise sqrtx
strong vortex
The thing is, for sqrt((3t)^2 + (t-2)^2)
U want to make (3t)^2 + (t-2)^2 as small as possible
The same way u took the smallest value of x u cld
magic ether
cld?
strong vortex
Could
strong vortex
But ur not minimising sqrt(t)
sqrt((3t)^2 + (t-2)^2) is what ur trying to minimise
strong vortex
|v|=sqrt((3t)^2 + (t-2)^2) = sqrt4
magic ether
yeahh
strong vortex
Now let’s say (3t)^2 + (t-2)^2= 9, which is a bit bigger than 4
Then |v|=3
Notice that |v| increases when I increase (3t)^2 + (t-2)^2
magic ether
yeah
strong vortex
That’s how the sqrt function works, if I make the input bigger(4 —> 9), then the output gets bigger (2–> 3)
So instead of minimising the output numbers (2,3 etc), we try to minimise the input numbers (4,9)
magic ether
so would t = sqrt(1/5)?
strong vortex
No, t=1/5 is the value that minimises (3t)^2 + (t-2)^2
You would need to substitute that into the right expression
(3x1/5)^2 + (1/5-2)^2
And then take the sqrt of it, to find |v|
magic ether
so you find t that minimises for without sqrt because that is what minimized max with sqrt?
this confusing lol
strong vortex
Yh, the easiest way to think abt this is, minimising |v|^2 minimises |v|
If u can convince urself this fact, then the rest is just calculus/derivatives etc
magic ether
i think i understand it
sqrt(x) is always increasing, the lower the x value the more it minimizes
reason removed sqrt part was to find easier how small can get x value
if you think of x value being 10t^2 - 4t + 4
the lowest value of 10t^2 - 4t + 4 leads to lowest y value of sqrt(10t^2 -4t+4) maybe i gave bad explination
strong vortex
Yes, this reasoning is correct
magic ether
thanks i understand it more now 🙂
strong vortex
Good luck
magic ether
.close
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lyric moon
Hello I've been trying to solve this problem for like an hour, would appreciate if someone can help me with it
vocal sleetBOT
Hello I've been trying to solve this problem for like an hour, would appreciate ...
lime gorge
lyric moon
Should I redo it or just tell you?
I'm stuck at y=n(1+sinx/cosx)*n-1 * (1+sinx/(cosx)*2)
lyric moon
Yea so now the problem is the *n-1
lyric moon
I'll just type it out