#help-0
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kjmkty
ohhh
you see the difference
its not the answer
ye
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hihii i need help w the hence part onwards
my fren got a sequence of natural logs of even no. but I didn’t get that
<@&286206848099549185>
do you know the principle behind telescoping sums?
Where's ln(n)?
ln(n-1) should also get cancelled out, no?
You have a -ln(n-1) and + ln(n-1) in the telescope
Oh wait no I'm blind why am I having brain moments
Show me how you got from the first line to the second
im not sure why you decided to calculate the sum like that. Try to split the whole sum up to three seperate sum, sum of ln(r), -ln(r+1) and ln(r+2)
and i assume you know that sum of ln(r) from 1 to n is ln(n!), you can try applying this the same method of how you would calculate the sum i mentioned
@wanton tusk Has your question been resolved?
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Hello, I have an assessment that my teacher will grade for homework, so could someone please check through all the work throughly and tell me if there is anything wrong?
@glacial dagger Has your question been resolved?
hihii I found a mistake in q4
u can’t divide by 1/2 with a + still there
it needs to be a cf in order to divide on both sides
yepp! other than some presentation errors
but idk how it is where ur from thoo
but the ans shd be fine from what I’ve seen :))
someone else can try checking too bec I might’ve missed smth out
@glacial dagger Has your question been resolved?
<@&286206848099549185> could someone else check again?
@glacial dagger Has your question been resolved?
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Ahem
Why did i get differnt brackets although i used quadritc formula isnt is supposed to work for every binomial?
When the leading coefficient isn't 1 then you need to multiply by it to the (x-a)(x-b) for it to be equal to the quadratic expression
What does (x-a)and (x-b) refer to?
Oh wait
Like this right?
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this -49 should be +33 where did i go wrong?
@main pagoda Has your question been resolved?
Oh yea thanks
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any astrophysicists here 
A Molniya satellite orbit is an elliptical orbit around the Earth with a high eccentricity and a large inclination relative to the equator plane. Seen a satellite on such a Molniyabaan very
Moving slowly around its apogee (the furthest point from the orbit), it spends a great deal of time over the Northern Hemisphere, which is useful for telecommunications between areas at high latitudes north.
The job is limited by the following wishes:
- we want the satellite to go through its apogee twice a day. The orbital period is therefore half a sidereal day (you may equate the duration of a sidereal day to 24 hours for simplicity).
- we want the satellite to stay 600 km above the Earth's surface at perigee (closest point of orbit) so as not to suffer from friction with the atmosphere.
To ask:
- Calculate the semimajor axis a and eccentricity e of this satellite orbit.
- How many hours has a satellite on this orbit been above the Northern Hemisphere (per circle,
of course)?
i did 1. using kepler third law and for e basically 600km=a*(1-e)
i dont know how to do second
G universal constant, M mass of earth, e eccentricity, a semi major axis, P period and v velocity
answer to 2 is supposed to be 10 hours i came 3 seconds xd
isnt this maths server
maths is a subset of astrophysics!
so is physics?
i thought of using
$v=\sqrt{\frac{(GM)(1-e)}{a(1+e)}$
and
Fucktalogist
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
$P=\frac{2\pi*a}{v}$
but didnt help
Fucktalogist
G universal constant, M mass of earth, e eccentricity, a semi major axis, P period and v velocity
answer to 2 is supposed to be 10 hours i came 3 seconds xd
n0 
@limpid spade Has your question been resolved?
@limpid spade Has your question been resolved?
@limpid spade i would ask in the physics server in #old-network
they’ll probably be able to help more
im banned there for some reason xd and i already asked in astronomy server
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A factory tests 100 light bulbs for defects. The probability that a bulb is defective is 0,02. The
occurrences of defects among the light bulbs are mutually independent events,
Calculate the probability that exactly two are defective given that the number of defective bulbs is two or fewer.
@dry osprey Has your question been resolved?
this isnt anything school related for me, my friends a math major and he told me he would only play if i could answer this math problem
i dont even know the first step
since we are talking theoretical probability
the probability that a light bulb is defective=0,02
literally means that if you have 100 bulbs 2 of them will be defective 100% of the time
but that last statement is putting me kinda off
i think there lies the twist
That's not true. The probability of getting heads on a coin flip is 0.5, but that doesn't mean you'll always have one head and one tails when flipping twice
theoretically yes
but practically no
Theoretically no
i flipped a coin 15 times once and never got head 😭
there are 2 kinds of probabilities
you are wrong
Theoretically, you can expect two defective bulbs in 100 bulbs
That does not mean you will "get two defective bulbs 100% of the time." That's nonsense
so its saying that there are confirmed 2 defective lightbulbs but i just need to determine the probability of that happening?
but you were already given the proba of that happening
It's saying that you know for a fact there are no more than 2 defective light bulbs. There could be 2, there could be 1, or there could be 0
oh i see
It's asking "what are the odds you have 2?"
There's Bayes formula for calculating the odds of A given B, or you can note that the probability of getting 2 given there's 0, 1, or 2, is the same as the probability of getting 2 divided by the probability of getting 0, 1, or 2
I wouldn't suggest giving up. It's not a hard problem
Think of it intuitively: the probability of something happening is (the number of outcomes we want)/(total number of outcomes)
For example, the odds of getting heads on a coin flip is 0.5, because there's 1 way to get heads and 2 total outcomes (heads and tails). Thus, the probability is 1/2
This is assuming each outcome is equally likely
Alternatively, you can think of it like (the odds of getting the outcomes we want)/(the odds of getting all possible outcomes). This isn't the best way to word it properly, but for most cases, the denominator of that one is 1.
In our case though, all possibilities are "0 defective bulbs," "one defective bulb," or "2 defective bulbs." So the denominator would be the odds we have one of those cases
I'm probably not explaining this as well as I'd like
no no its ok im just really bad at understanding ive never been too good at math
so we know its 2 or below
So, what are the odds of getting two defective bulbs in general?
That's the odds of getting one defective bulb in a set of... one bulb
ohhhh
We want the odds of getting exactly 2 defective bulbs in 100
so then i would need it to be 4/10000?
Not quite
That's the odds of getting 2 defective heads in 2 bulbs
There's actually a formula to calculate "the odds of getting exactly x successes in n tries."
so just increasing the denominator doesnt really change much?
It's nCx * p^x * (1 - p)^(n - x)
You multiplied (2/100)(2/100), right? That only applies if you're looking at only two bulbs and you want to know the odds both are defective
Also, p here is the odds of getting a success, i.e. the odds of getting a defective bulb
wait so then theoretically i could do this 100 times and get it?
No. That would tell you the odds of getting 100 defective bulbs
You're on the right track though
We can think of it as the odds of getting 2 defective bulbs AND 98 non-defective bulbs
So instead, you could multiply the odds of getting a defective bulb two times, times the odds of getting a normal bulb 98 times
Well, almost
But that'll get us closer
Are you following?
i think so
So, what do you get after multiplying those probabilities? Feel free to leave it as a decimal, it's probably pretty small
i think im lost again
would the odds of getting a defective bulb two times be 4/10000 like before?
No, that's the odds of getting a normal bulb once
The odds of a defective bulb is 0.02
So the odds of a normal bulb is 0.98
But just as you multiplied 0.02 twice in order to get the odds of two defective bulbs, what can you do to get the odds of getting 98 normal bulbs?
oh would you have to multiply it 98 times?
Yes
Or just raise it to the 98 power
And then multiply that with 0.0004
It's probably a very very small number
,calc (0.02)^2 * (0.98)^98
Result:
5.5235133650459e-5
damn
Looks right to me
oh nice
Not as small as I thought
i thought i was wrong
Now, we're not done yet
right
That's the odds of a particular combination of bulbs. I.e., that's the odds of specifically having 2 defective bulbs, then 98 normal bulbs
But we could have 98 normal bulbs and 2 defective bulbs, or 1 defective bulb, 90 normal bulbs, another defective bulb, and 8 more defective bulbs
We need to multiply that by the total number of ways you can order two defective bulbs and 98 normal bulbs
so how would i got about doing this
Do you know how to find the number of ways to order objects?
Combinations, permutations, that kinda thing?
i knew having fewer the 2 possible defective bulbs was possible but i didnt put into consideration the number of normal bulbs
i dont tbh
Well, here's how we can do it. Let's say I only had 5 normal bulbs and 2 defective bulbs
That's a total of 7 bulbs, right?
Do you know how many ways you can order 7 objects?
do you mean like averages?
No
Let's say I had the numbers 1, 2, 3, 4, and 5
I can order those 12345, or 54321, or 13254, or 15234, etc
How many ways can I order those 5 numbers?
ohhh so like how many different orientations of those 5 numbers can there be?
im trying to do it in my head but im assuming theres a better way that im not aware of
Consider this. How many numbers can go into the first spot?
5
Good. Now, once a number is in the first spot, how many can go in the second?
exactly
The total number of combinations is then 5 * 4 * 3 * 2 * 1
This is denoted 5!, or "5 factorial"
Yes, though we need to be careful
We have duplicates. We can swap two normal bulbs and have the same ordering
Because the 5 normal bulbs can be ordered in any way, we need to divide by the number of ways to order 5 bulbs
Similarly, we also need to divide by the number of ways to order 2 bulbs due to the two defective bulbs
so then for the defective bulbs we would just divide by 2?
we would divide by 120 for the normal ones right?
Yep
We can generalize this idea. The number of ways to organize m normal bulbs and n defective bulbs is (m + n)!/(m! * n!)
So in our original example, how many ways are there to organize 98 normal bulbs and 2 defective bulbs?
4950?
this probability right?
Yep
i got 0.273394935
,calc 4950 * (0.02)^2 * (0.98)^98
Result:
0.27341391156977
nice
Now we need to know the odds of getting two or less defective bulbs
We can find that by finding the odds of getting 2 bulbs (already done) + the odds of getting 1 bulb + the odds of getting 0 bulbs
So, what are the odds of getting exactly one defective bulb?
It's the same principle we've done already
so one would be 2/100?
Sorry, I meant one defective bulb out of 100
No
damn
Remember what we did to find the odds of getting exactly 2 defective bulbs out of 100 bulbs?
We're doing the same thing, but with 1 defective bulb
It was (2/100)² times (98/100)^98
And then times the number of ways to order the bulbs
Yep
The odds are still 0.02 and 0.98
i think i confused myself and changed it from 2/100 to 1/100
lemme retry
0.00270652154
is this any closer
Yes
But now you have to multiply by the number of ways you can order one defective bulb and 99 normal bulbs
What's 499950? The odds? Or the number of ways you can order them?
It's incorrect either way
danb
How'd you get 499950?
0.270652154?
Yep
Lastly, we need the odds of getting zero defective bulbs in 100 bulbs
Aka, the odds of getting 100 normal bulbs
would this just make is (2/100)⁰ x (98/100)¹⁰⁰?
0.13261955589
Yep
That's the odds of having 2 defective bulbs or fewer in 100 bulbs
For our final answer, what's the odds of getting exactly 2 defective bulbs in 100 bulbs divided by the odds of having 2 bulbs or fewer in 100 bulbs?
The odds of getting exactly 2 bulbs out of 100 bulbs isn't 0.00005...
.
Well, you had the disadvantage of not having a lot of prerequisite knowledge
Answering the question requires knowledge of permutations, combinations, probability, etc
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In an experimental study, are the number of treatment combinations equal to the number of experimental units?
Look at the (0, -5, 0)
are we suposed to take whatever numbers we want
cuz the exercise says
Take a point of whichever u want from one flat
idk if i translated good
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Excuse me, I speak Spanish but I did not understand that question and they told me to join so they could help me
Im assuming they want to know which theorem is being described in the image?
then you justify why it is that one?
What he asks me is : According to each graph, indicate to which notable product it belongs and why, make three examples:
<@&286206848099549185>
@alpine sable Has your question been resolved?
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hey, is there a easy way to calculate 14/29?
What does an "easy way" mean?
I mean if you think of getting 14/29 = 0.482758621
Then there are various speed calculation methods out there
@tame hound Has your question been resolved?
Such as?
Yeah
@tame hound Has your question been resolved?
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Hello! Can someone help me with this question please
I tried equating the x of both of them and getting equations in terms of t and s
but i cant seem to get the right answer
You can also set the y and/or z coordinates equal
,w -2t=-9+5s, 1+2t=2+3s
do i plug it into the simultaneous equation? because then I get x as 0 but thats wrong
Back into the original equations
L1 or L2
You should get (-4, 5, 6)
Either, the equality you set at the start means they'll be the same
Cause you're solving for intersection after all
np
if i have another question, do i have to close this and reopen the channel? or do i just send it here
nicee okay im working on this one now
i followed this: https://math.stackexchange.com/questions/2686606/equation-of-a-plane-passing-through-3-points
my AB was (1,-3,0) and my AC was (1,-2,0)
and my cross product was (0,0,1)
i think i did that here
so I got z + c = 0 -> 1 + c = 0 -> c = -1
and then the final equation became z - 1 = 0
but thats wrong :(
yeah its these
i went back through my steps a couple of times but i cant figure it out
ahhh
oh shit
goddamnit it worked
i spent a solid 2 hours on this
thank you so much lol
yesss
i have one last question!!
for this one
i equated the x, y and z for L1 and L2
and for x, i got t as 32
14+6t = -18+7t -> 7t-6t = 14+18 -> t = 32
so i plugged that back into the x but it looks like its wrong
You prolly need to compute the coordinates as a single number
However, they intersect if and only if t is the same for the x, y, and z coordinates
,w 20+6t=-14+8t
,w 20+6t, t=32
,w -14+8t, t=32
They prolly shouldn't have used the same parameter/scalar, but I don't think that was the intention based on what I got above
oh wait yeah youre right i think
i got the answer
i just changed the variable in the second set of equations
so i used s instead of the t they gave
and i solved it to find t and s, and plugged it back in
thank you so much @rustic coral @alpine sable !!
life savers
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Excersise 1
,rotate
Isolate the y variable in the linear equation
Then replace y in the circle
@wispy cave
I’m this far I am new to this topic trying to figure out on my own before cover in school
(like this?)
oh
Hence the ≠ symbol
-5
You sure?
Multiply
Umbraleviathan
Shouldn't be too hard
Yeah
that the awnser
Well no
Use quadratic formula
hihii
hello
u can’t have a negative inside a sq rt
i stupid
me too HAHA
no square number will be negative
oooooooooooooooooooooooooooooooooooooooooooh
in the real world
so for the formula
the negative is actually outside the sq rt
a= 10,, b= 30,, c = 20
oh so i went away ahead of myself
u shd get (-30 +- rt(30^2 -4(10)(20)) / 2(10)
yepp or u can just factorise ^^
this method is easier imo
i got it
yay
not i replace the x
with enough practice u can HAHA
now
or u can just do a quick side working
wait what’s the question
as in did he ask why not replace the x instead?
this
oo
yes can
yes can take our common factor 10 to make it easier for u to see
for some reason i has route - 3
but issok this method works too
and did not like that
cus could not devide
divide
so if i were to do q2 would it be the same thing
have pic?
yeshh,, similar
i like how the other guy got fed up with me lol
HAHAA
my teachers got fed up w me a lot too
slow learner 😭
issok @alpine sable won’t give up on u
lol
HAHAH
I did that too when I first learnt this topic
imaginary world no?
ooooo icic
wait I think I’m learning that soon
mm
sounds imaginary like my good grades
nopes HAHA what’s that
I’m in jc
u irish
Junior college
jc in ireland is the junior cert
It’s like pre-uni
HAHAHAH u got accepted into the so called better uni
u know there’s a lot of drama there rn
idk what’s happening to NUS
OMG
ITS SO HOT
Istg it’s like 40 degrees here everyday 😭
@wispy cave are u oki
just factorising now
Yaya
yuh
doesn’t that give imaginary roots
yuh
wait did u manage to prove that it’s a tangent
y=-3
Yaya
y=-3 or y=-3
so ull only get one answer
it’s the same HAHA
1 awnser
no need to write 2 times :))
but there is 2 of them
it’s basically (y+3)^2
yuh
so y= -3
yuh
just write one time is good
k
ure actually suppose to prove first HAHA
how
u know u made x = 3y +10 right
then you subbed into the circle eqn
yuh
then after u do that
ure left with 10y^2 +60y +90
when u get this
use the discriminant
b^2 -4ac
a = 10,, b= 60 and c = 90
what do u get
i am lost
ouh what happened
oh
basically if we want to see whether a line cuts a curve at 2 points, 1 point(tangent) or none at all
yayayay
we use it for this
nope
so that means i proved it
no u need to explain
^^
it proves that (-3,1) is a tanget
if i am wrong i am so sorry
u lost too many brain cells cus of me
lol
basically, since the discriminant is 0,, the line hence only touches the circle at one point and is hence a tangent to it
making it a tangent
yes
so how would i fraise that
so (1, -3) is the pt of intersection
like this
k
shd be enough
instead of discriminant u can just say since b^2-4ac=0 to save time HAHA
the bad thing about this is this might be the last time u struggle teaching an idiot math but now im starting to pay a guy to do it
woah
when i should be doing it for free
ure paying a guy?
HAHA ohh u mean tuition?
we call them grinds
i learned this from a canadian that shared there cigarrette with me
……
yup
wait how old r u
17
wait
i dont smoke often
HAHAH WHAT
we can drink at 18
omg nooo don’t smoke
but i get into clubs at 17 and drink sometimes
i knoiw
i only do it when im super drunk
But it is a well known fact that smoking and drinking makes you atleast 90% cooler
dont tell an irish teen not to drink
not too much pls
i am lightweigh
I will
and there’s nothing u can do abt it
hmph
95%*
it is expensive as fuck todrink
that would kill u
WHY
Just let the man be, we are supposed to be helping the lad with math
do u have that
my dad got it for me when he went on holidays
finee u do u 🕊
is this just common in Ireland
the fact that ur father bought it for u 🥲
but i told him what i drink on a night out and he saw it in a shop for cheap
and he got for me
i know that
u americans and canadians are lightweights
on nights out i would drink vodka blacks and this with a fizzy apple drink called cidona
There are like a handful of places on earth where it's 21. What are the odds that we have one here that's not from the US the biggest one of the handful
ohh
but both my parents are irish
I think legal age for drinking is 18 tho
yepp here too
where is singapore
Asia
Then why did you say you need to be 21? That's weird
that’s for smoking
Singapore is a city-state. It's not part of another country, it's a country made of a single city
yaaa it’s so nice here
is it rain
who is that?
cus i like the rain
no it’s the sun
but recently it’s been raining more often
nvm
so it’s been quite nice
huh
go to North Korea
what does that even mean
Even less places on earth where smoking is age-gated at 21 btw 😄
18/21
the uk are trying to put up the age limit every year for the next 15 years
so its 18 now
and next year 19
20
driving at 18
etc
not many students obey the rules though
nah i aint a heavy smoker i would just have 1 or 2 tokes and thats me done
it is very sickening
I’ve seen students in uniform smoke in the open
of course
u said women are ooooof
what else wld I expect
🧐
oh
I still don’t take my words back
@wispy cave
yo
close
how
how i do that
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thx
now do .reopen
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now do .close
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ok I’m sry HAHA
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The bot is about to get mad at you
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no what’s that
I remember I studied history for 2 years
This evolved from random bullshit to straight nazi shit real quick
it’s heinrich we’re talking to
i have to get up for school in 6 hours
yuh the fucking ukranians take all the accomidation
ATB!
we took so many of them in and they do nothing
???? HAHAHA what
they get free money and housing and they treat people like shit
but Singapore celebrates it on 1 May
and sg is not a communist country
????
I don’t understand those terms 😅
no it’s not HAHAH
unless..
ya not me
but we have governments
im a girl
it looks wrong, but this is actually completely correct.
it was chosen by the marxist international socialist congress
anyway this is completely out of context so bye
fishballs
I am@not knowledgeable in this area at all
he the guy from squid game
since you’re so good at math, solve the riemann hypothesis
nvm
prove it
bye bye
@wispy cave it’s ur time to shine
yuh
comeback from rt -3
here is an artical written about me
in this case,, the storm@was ur rt-3
and my solvings of this equation
the rainbow will be u proving the Riemann hyp
no
the first paragraph talks about how little we know about it
10011001000101011010110100101
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What does this have to do with math lol
And no, communism doesn't work, nor does capitalism
What we need to for YOU to give ME all your money
Trust me, it'll work
Just give me your life savings
All of it
Anyways @wanton tusk close the channel
it’s not me tho
It's yours
wha-
Closed by @wanton tusk
Use .reopen if this was a mistake.
sryy!
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need help
can yall tell explain what the answer to this is
why do you think that?
happy to help
You’ve got to take up your own help channel to ask the question
@safe flume Has your question been resolved?
Closed by @safe flume
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$a+\frac{b}{c}=
a*\frac{c}{c}+\frac{b}{c}=
\frac{ac}{c}+\frac{b}{c}=
\frac{ac+b}{c}$
fedeCreeper
@alpine sable
@alpine sable Has your question been resolved?
I will never learn how to use that bot