#help-41
1 messages · Page 6 of 1
Yup maybe u can contact a proff and ask them for help Or study from yt Or somewhere else
My prof from Uni doesn't know how to help sadly
Did ask him once a year ago about it, but I was too occupied to keep up with Uni math to learn on my own
dam u can try using yt and if ur stuck anywhere the server is always here for u
and me too
Thank you for that
Luckily as long as it's numbers I manage to do them with the correct tutorial

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is this solvable?
Yes
@quartz crescent Has your question been resolved?
@quartz crescent Has your question been resolved?
can any1 help?
Connect vertices with the midpoint
like from the corner to the intersection?
take your time
Yes
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let’s do this again
I thought the last guy did a good job lol 😨
what was bro even yappin about
did not help one bit he just started spewing words that I didn’t even understand what was going on
smth like complete the square
can anyone help me figure out what I'm doing wrong
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yeah but the steps I just don’t get those
that’s literally all I’m asking
what do you not get?
Sorry I forgot
u want to change the original equation into a form like (x+a)^2 + b
@rare hatch Has your question been resolved?
Cause vertex form
Wow so helpful
my fault
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$\mathcal{(L}_s)^{-1} (\frac{s^2 + s +1}{s^2 + 4s +5})$
Triaxyz
I will explain my process but I got $\delta (t) -3e^{-2t}\cos(t) -4e^{-2t}\sin(t)$ which symbo says is wrong
Triaxyz
convert $\frac{s^2 + s +1}{s^2 + 4s +5}$ into $\frac{s^2 + s +1+3s+4-3s-4}{(s+2)^2+1}$
Triaxyz
which becomes $\frac{(s+2)^2+1}{(s+2)^2+1} - \frac{3s}{(s+2)^2+1} - \frac{4}{(s+2)^2+1}$
Triaxyz
all s variables have to be replaced by s-a for you to use that
so 3(s+2-2) which turns into +6 on that last term
and 6-4 = 2 which is the coefficient that symbo says
alright I got it
thank you
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I’m not sure if I plugged In the number correctly. Can someone please tell me if I have or not? And if I’m going in the right direction?
clicky with the linky?
^ porn
pls I just want help with my hw
<@&268886789983436800> 18+ (nsfw)
(sorry for ping)
(i turned off reply ping to that message)
<@&268886789983436800>
So can anyone help me with this
I rewrote it bc I think the first time I wrote it wrong
<@&286206848099549185>
@glad schooner Has your question been resolved?
<@&286206848099549185>
why are you using f(x)?
What am I supposed to plug in for y then 💔
because I thought f(x) and y is the same thing
you don't plug anything in for y
no, y = f(x) is very common notation, but you can't assume it here unless you're told to do so
So I just keep it as Y? Like this?
Oh okay thank you!
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Hello guys, quick question:
How can I write
$\dots + \frac{-7}{2} + \frac{-5}{2} + \frac{-3}{2} + \frac{-1}{2} + \frac{1}{2} + \frac{3}{2} + \frac{5}{2} + \frac{7}{2} + \dots$ as a series?
Akkaman
why do you want to write that as a series
Because I need it to solve a problem. 😄
I'm just unsure about how to write it as a series with the sigma symbol.
bruh isn't it just 0
it is
Well...
not really
Yeah it is, I am looking at the reciprocals of them, so $\frac{1}{\frac{-7}{2}}$ etc...
Akkaman
sane answer dropped
It has something to do with another problem.
I really just need to know how to write it without the dots but rather with the sigma symbol. 🥲
you better show the another problem then, this is not an expression you wanna play with
why though?
Wait.
is $\int_{-\infty}^{\infty} x dx = 0$?
rbit
In which order should I even be summing it?
erm, upto infinity can't say
You can't unless you have a specific order for every term, and even then it won't necessarily make any sense
I guess that since it's both directional, I can pick any order.
So 1/2 + 3/2 - 1/2 + 5/2 + 7/2 - 3/2 + 9/2 + 11/2 - 5/2....
whoops - that's not 0
the cauchy principal value of it is 0, but just applying the basic definitions it will diverge
Yeah, really this can evaluate to any rational number or -inf or +inf
Can't it evaluate to any real number actually
well to answer the question though
$\sum^\infty_{n=1} \frac{1}{(2n-1)^2}$ would be the series over the squared reciprocals of all positive odd numbers, right?
Akkaman
All the terms are of the form n+1/2 (n integer), so I wouldn't say so, but maybe the fact it can evaluate to +inf and -inf means it can, idk
There are series, all of whose terms are rational, that converge to pi
Yes but they all have different denominators
And you are trying to compute this?
$\sum_{k=-\infty}^{\infty} \frac{2k+1}{2} = \sum_{k \in \mathbb{Z}} \frac{2k+1}{2}$ something like this, but it will really just diverge
rbit
Oh, you're right. This series is actually completely divergent, no matter how you rearange it
its partial sums will oscillate
yes
this surely converges, although it doesnt seem to be elementary
If I multiply that by $\frac{1}{2}$, would I not just get the series over the squared reciprocals of $\frac{1}{2} + \frac{3}{2} + \frac{5}{2} + \dots$?
Akkaman
that's another interesting point, does it necessarily oscillate or can you find an order such that it doesn't? I guess since the terms are growing absolutely on either side, you actually can't find such an order, so yeah it can't even evaluate to a specific number
Yeah it converges to $\frac{\pi^2}{8}$.
Akkaman
(It always oscillates or diverges completely, because all terms have magnitudes >= 1/2. So the partial sums must change by at least 1/2 at every "iteration")
yeah, this sounds right
except instead of + you should use ,
Yeah sorry. 😄
and the series would be multiplied by 2 if im not mistaken
$2\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^{2}}=\sum_{n=1}^{\infty}\frac{2}{\left(2n-1\right)^{2}}$
MæthIsAlwaysRight
Anyway, like @pseudo crescent already said, you should just state the original problem instead
he did 💀
what's the goal of multiplying it by 2 btw?
I must be blind
There should be a mathematical operation to go from $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots$ to $1, 2, 3, \dots$, right?
Akkaman
Yes, +1/2
I think you are doing quite weird things here though
Is there a way with multiplication?
No
no
Yeah the topic is weird. 😂
do you know value of zeta(2)?
there is a simple way to get value of this from that
That's the topic. ^^
anyway, what would you do now?
are you trying to transform this to zeta function or sth?
But my proof is not a typical algebraic proof but rather geometrical / physical.
All pairs of consecutive terms in both sequences differ by 1. If you multiply one sequence by some number (apart from 1), that will change and it cannot give you the other sequence
well.. expect weird result then
I'm still confused, so you do not have proven zeta(2) yet, or do you? And now you want to prove the value for the odd squares?
Explaining the entire proof here would be too time consuming, that's why I'm trying to break it down to the most important things. 🥲
I wanted to know how to get from $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2}, \dots$ to $1, 2, 3, 4, \dots$ by some mathemical operation.
Akkaman
add + 1/2 to each term
isnt the real goal here getting from sum of 1/(2n-1)^2 to the sum of 1/n^2?
No no.
you can also multiply by 2, index shift and divide by 2
multiply by 2, add 1, divide by 2
just a warning: I don't think you can prove this thing "geometrically" by just playing around with numbers
I think I made a mistake at some point.
This feels so directionless
also proving the value for zeta(2) is really not easy at all
I know an elementary proof but it's a long journey
I have already done it algrebraicly and geometrically.
But I wanted to modify the geometrical proof and now I am stuck. 😂
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how can this be proven in some other way
its unclear how one would reach this choice of q
one would reach this choice of q after hours of trying stuff
its not that crazy. you want that p^2-2 in there quite naturally
yes but that doesnt narrow down the choices for you so that you can reach this choice of q by trying some choices
because there are still too many choices that involve p^2-2
so there isnt a way to find these unless you gain more experience and gain this intuition naturally ?
I mean thats how proofs generally go
thats right i understand your point
is there another way to prove this so that i can gain more insight
its also not like this is the only choice of q that will work
you would maybe start with q=p-(p^2-2), then square it and check what q^2-2 is. it doesnt work (presumably, otherwise they would have chosen that), so you maybe decide to use p-(p^2-2)/something. for that something you might decide it could be some multiple of p, so maybe kp. you try it out, maybe it works, maybe not. if it doesnt you might change it to kp+n
and then you are there
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Hi, I've got a question I'm confused by... it's a really simple one but I have two different answers for it
A container will hold 106 ounces of grape juice. How many 6(5/8) ounce glasses of grape juice can be served from this container?
the picture shown is my working
the answer in the textbook says it is 16 (glasses)
but mathway says it is 11 (1/24)
a 106-ounce container being served into glasses that are 6(5/8) ounce
you did a mistake in second step
lol
53/5
it is a kind of magic tric k
:magician:
i couldn't work it out but as soon as i asked it...
some kind of quantum magic 🧙
thanks guys
what u guys doing?
what does that supposed to mean/
i came to this help-41/ap button saw u guys here what's up
alright
hey 77^2 i have a qn what's 1 and 2/5 power 3
what?
i'm just curios of what way you will use
I didn't understand what your question even is?
1+2/5?
yess
{7/5}^3
like the mixed fraction
A container will hold 106 ounces of grape juice. How many 6(5/8) ounce glasses of grape juice can be served from this container?
hi btw sorry lol
god that's 10
hi guys
wassup
How is this 11 ;-;
im facing a problem when i try to solve through my calculator it keeps saying syntax error
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try typing it out on a keyboard, use a calculator online as well. it might be easier on a keyboard
yes, 11^2 = 121
now what happened here
yeah for 13 it'd be 143
!status quo
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
- i guess
i'm trying to say 530 / 53 = 10 ap
i'm sorry, i'm not thinking clearly at the minute lol -_-
Yea
i will check with mathway again as well
it happens to the best of us no worries
16 is the answer that was given in the textbook, but i had the wrong answer (and working as well)
Well in ur working out u got 11 and here im showing u why its 16
that's good, at least the tetbook is right lol 
are you a spammer jassbery?
Sadly
you did 53/5
it was supposed to be 53/8
So are you all good now or still stuck ?
so the answer should be 2
oh you are right as well
d'oh!
Yeah i noticed that baha
i hate when people don't care to calculate things
what are you talking about btw?
what how did i not see that good call 77
i mean 10
@austere jungle so all good now :))?
Yay!!
horay : )
yayyyyyyyy
yes, thank you all
yeah idk, i'm glad i clicked to show messages because i thought it might be a message from one of the bots at first
hope you guys all have a great day btw, thanks again
bro thinks life is all sunshine and rainbows
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Could I get help with this question please
I am not sure where to start
Do you know what a basis is?
Linearly Independent Set that Spans the space
And do you know what the notation [(2,5)]_B means?
If I remember correctly it means that it is 2u + 5v where u = (1, 2) and v = (2, 3)
Is [(2,5)]_B different from (2,5)_B?
I don't know what (2,5)_B means
This is what is in my textbook
It's almost the same except [(2,5)]_B is vertical, I guess
Ignore this. I get it
I don't get this though
(2,5) = au+bv, like here
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Hello
i’m a little confused on what i’m supposed to do for this problem
can anyone help me with it
and what does it mean by use any method
g(x) is also a line
do you know how to find the equation of a line given two points?
I assume there's also a tool in the e-tool thing that allows you to form a line with two points
no not really
we learned abt it in class today but i didn’t understand it
don’t you like set them equal?
oh wrd pmo
???
but how do you find the two points
that’s what i’m really worried abt
any two points define the line
it doesn't matter which points you choose on g(x)
you can also notice that there's a constant difference in the g(x) values and a constant difference in the x values. this also indicates its a line
so like -3, 21 and -1 13
yes
you can use point-slope form
y-(y1)=m(x-x1)
where (x1,y1) is a point on the line
and m is the slope
you know how to find slope/gradient right
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I don't get what it means by "increasing" and "decreasing"
increasing means slope/derivative is positive
In this section we will discuss what the first derivative of a function can tell us about the graph of a function. The first derivative will allow us to identify the relative (or local) minimum and maximum values of a function and where a function will be increasing and decreasing. We will also give the First Derivative test which will allow u...
Could that be related to the first derivative test maybe
related, sure. but you're already given dP/dt so just solve for when the right side is positive/negative/zero
equilibrium means dP/dt equals what value?
zero
If you do that you just get P=0 or P=230
But if I remember those are called critical points of the function
correct
if dp/dt is zero at only those two points, then you can determine the sign around those points
,w plot x(1-x/230)
your sign looks wrong for p < 0
I made a typo so it should be p less than 0 means it's decreasing
so do you know how to do your problem then after reading the link i sent here?
I don't know how I would phrase it, could I say for any p<0 the population is decreasing
does p < 0 have physical meaning?
Most likely not
do you understand why?
You can't have negative population
right
do you understand how to answer your questions now?
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guys trying to solve this with L'H
i did someting wrong
but idk how
(e^1/x -1) derived
is chain rule right
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Hello! I understand you need to find the limit, right?
try pinging them, they've closed the help channel so they might not see your message
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How to find left and right side limit
@compact wagon Has your question been resolved?
"Follow your finger from the left/right, what does it look like you're getting closer and closer to?"
i know thr right hand limit at x=0 but why does the left hand limit equal x=2 when there is ajump discontinuity
@compact wagon Has your question been resolved?
@compact wagon Has your question been resolved?
@compact wagon Has your question been resolved?
at x = what?
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Is this correct?
,w sqrt(3x+12)=sqrt(x+8)
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Guys, what should it be like? Is it number one or two?
I’m confused on how we’re supposed to connect the chords
Bcs i’m getting a gradient of 0 for the chord CB in the 2nd drawing
number 2
that's correct
so that means the perpendicular bisector will be a vertical line
since its slope is undefined
actually notice that you have a right angle at B, due to the coordinates
so the centre is just the midpoint of CA
remember your circle theorems from IG
Ohhhh yeahhh
How do i know if no 2 is the correct one in case CB and BA are not perpendicular to each other?
well actually, for three points on a circle
you just need 2 out of the 3 perpendicular bisectors, so drawing 1 isn't wrong
it's just messier to work out
Ohhh
Okk i’ll try to do it first
Just finished my As level on this yr may June got 73 on my p1 I'm pretty sure this has a formula
Thank you so much for your help
Ohhh yess
Just replace x and y and u will get 3 unknowns and 3 equations
yes so radius is|| sqrt((7 - 4)^2 + (9 - 5)^2)||
no need lol
@woeful temple
sorry for interrupting but i think x and y are supposed to stay as x and y and others to be substituted by number values so as to be an equation?
Thank youu for this!
that is correct, x and y are variables
h, k, r are constants which depend on the circle
If we know h, k, r we can sub those in and get the equation of the circle
ah k
i was thinking of the word that should be used to describe "x and y are supposed to stay as x and y"

But we don't know h, k, r so don't we have to substitute x and y so that we can get h, k, r?
Btw we can also find the centre from the midpoint of BC right?
woooops I meant CA
CA is the diameter
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can someone explain venn diagram union and intersection im really confused
Union means or
Intersection means and
so for example, if we take a union b
we're saying a or b
in a venn diagram, it would be both the circles including the middle/overlapping section
however, if we say a intersection b
we're saying a and b
so you would only take the middle/overlapping section of the venn diagram
so does a or b mean both
and a and b means like
whatever is the same
because i get confused
okay thank you i get so confused
ight allgs
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what can i even do here
do you have the original version
this is the original
yea ik
thing is aabb is a 4 digit perfect square where a and b are its digits
so we need to find the number
hm
calculators arent allowed btw
yea ik 😭
ignore the GIF part , its the greatest integer function
the question is basically asking this
I kinda got some method but it still involves some calculations
some significant calculation*
whats the method?
if you write the number in the form of 1000a+100a+10b+b
you can pair the a and b things together
oh i see
erm
like the whole number
a multiple of 11 and also a square number
yea
so like 100a+b=11k^2
like this
so that both 11 makes 11^2
but there could be a more efficient method
maybe
most of the questions on this test had a lot of trial and error involved tho so this may be correct
erm probably,
at least with this method you don't have to check all the numbers
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In how many ways can 4 boys including John and Matthew, and 4 girls, including Sally, be arranges in a line if:
(ii) Girls and bous alternate and John and Sally stay together
@sharp depot Has your question been resolved?
<@&286206848099549185>
is 252 the answer?
@sharp depot Has your question been resolved?
<@&286206848099549185>
for this
first we will make 8 spaces
and then we will make a block of john and sally
This is the first hint
@sharp depot
sure
then iget confused on how to rearrange the other around the block
while ensuring girl abnd boys alternate
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HOw is 81% not right? I'm sick of this problem, please help
show how you got 81%
,calc (2 * 81 ^ 2 + 50) / (81 * (100 - 81))
Result:
8.5588044184535
yea that's not 30
i did 2x^2 + 50/x(100-x) = 30
like here
Result:
31.421985815603
,calc (2 * 93 ^ 2 + 50) / (93 * (100 - 93))
Result:
26.648233486943
,calc (2 * 93.9 ^ 2 + 50) / (93.9 * (100 - 93.9))
Result:
30.874177272648
,calc (2 * 93.8 ^ 2 + 50) / (93.8 * (100 - 93.8))
Result:
30.344040167824
,calc (2 * 93.79 ^ 2 + 50) / (93.79 * (100 - 93.79))
Result:
30.291965519296
,calc (2 * 93.78 ^ 2 + 50) / (93.78 * (100 - 93.78))
Result:
30.240058315316
,calc (2 * 93.779 ^ 2 + 50) / (93.779 * (100 - 93.779))
Result:
30.23487677328
,calc (2 * 93.65 ^ 2 + 50) / (93.65 * (100 - 93.65))
Result:
29.580142177903
,calc (2 * 93.68 ^ 2 + 50) / (93.68 * (100 - 93.68))
Result:
29.730020862835
,calc (2 * 93.69 ^ 2 + 50) / (93.69 * (100 - 93.69))
Result:
29.780297129201
,calc (2 * 93.71 ^ 2 + 50) / (93.71 * (100 - 93.71))
Result:
29.88132925056
,calc (2 * 93.73 ^ 2 + 50) / (93.73 * (100 - 93.73))
Result:
29.98300592271
,calc (2 * 93.74 ^ 2 + 50) / (93.74 * (100 - 93.74))
Result:
30.034087895893
,calc (2 * 94 ^ 2 + 50) / (94 * (100 - 94))
Result:
31.421985815603
@pallid canopy so if im rounding its 94%\
,calc (2 * 93 ^ 2 + 50) / (93 * (100 - 93))
Result:
26.648233486943
,calc (2 * 93.5 ^ 2 + 50) / (93.5 * (100 - 93.5))
Result:
28.851501439737
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not sure how to approach both questions
According to google itself you should use:
Cauchy-Schwarz Inequality
AM-GM Inequality
for 7 am-gm should work i think
hmm i think im missing something
i got (a+1/a)^2 + (b+1/b)^2 >= 8
did i apply AM-GM incorrectly?
how did you get >= 8
oh nvm i misunderstood
that lets a,b=1, which isnt possible with a+b=1 (i also did the same thing lmfao)
hmm
am i stupid or does the inequality not hold
can both be true
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hi
so in my class we are getting groups of 4 that we will be using for some time and i want to calculate the probability of getting paired with a certain person
A for ease of use
The groups will have 4 people and there's 26 people in total
That's 6 groups with a remainder of 2
So 7 groups where 2 of them will have 3 rather than 4
Now I could just go 1/7=14.29%
but
now that the groups arent equally sized, that introduces a whole other can of worms that i dont know how to deal with
that's what i need help with
do you want to know how to do it?
Yes
itll just be around 16% im like 90% sure
oh wait 2 groups of 3
in that case itll be more like 10-11%
i assuming there would be 1 group of 2 but that sounds kinda dumb now that i think about it
say you are A and the certain person is B
label the groups to be groups are
a,b,c,d,e,f,g where f and g are the 3 groups
try finding the P(A in a)×P(B in a)×ammount of groups of 4+P(A in f)×P(B in f)×ammount of groups of 3
im pretty sure
wait nk
there
we multiply by ammount of 4 so we take into consideration goinf into a,b,c,d,e
so here, P(A in a) is 4/26, ill let you do the rest
Hm
P(A in a) = 4/26
P(B in a) = 4/26
So 5(4/26)(4/26)
=80/676
=20/169
P(A in f)=3/26
P(B in f)=3/26
So 2(3/26)(3/26)
=18/676
=9/338
So 20/169+9/338
180/57122
=0.03%
Oh wait
I multiplied
=14.5%
Is that right?
.close
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can someone explain the difference between or, and because i'm confused in probability when it says or n and what does it want me to do
hm so i'm confused
so idk how to tell the difference and how to always get the right answer
or = either or both sides are true
and = both sides are true
false or false = false
false or true = true
true or true = true
false and false = false
false and true = false
true and true = true
ohhh okay
or just understand it with simple english
you have to eat fries or burger
you have to eat fries and bruger
ohhh so or limits you to one category
The first can easily be confused with xor here
No, that's xor
what's xor
false xor false = false
false xor true = true
true xor true = false
[;
or means that either side is true (including both being true), xor means that exactly one side is true
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did 1, need help with 2
they probably meant express with a rather than A.
Here is a hint: what is PQ:AD? can you, through parallel properties, get some equalities?
No they meant exactly to express with A
The answer says 2/3a
I just have no idea how to get to it
PQ is a straight line
So it should be PQ= a+2a/2
that would be 3a/2
yeah i mean they probably meant "a" instead of the capitalized one
try to follow my hint
Oh yeah
Hm
I mean
you can prove that some triangles are similar? maybe
oh yeah ofc!
but do you know some parallel lines ->equal ratio properties?
Wait no i dont
welp lets prove it then
Wait hold on what do you mean by this, can you give an examle
if DE is parallel to BC
Yes
then AD:AB=DE:BC=AE:AC
yeah lol
since PQ is parallel to AD, can you transform the ratio PQ:AD to something else?
👍
similarly if the triangles are like this we have some ratio properties
Yeah thats another variation
ok, MP/PA seems hard to compute, can you translate it to something easier?
notice BM is parallel to AN
Right
Lets see
AN/BM=AP/PM=NP/BP
Its the same
Exept AN is a
and BM is 0.5a
Lets see
If it helps
should be MP/AM
wait one sec
why AN/ND?
did you mean PQ/AD?
it should be MP/AM and MQ/MD
Ok hold on ill write it down rq
welp, since the proof is easy, you can just try to derive it on the spot if you forget
Ok so i still have no idea
i wrote it down but theres only ad that we know
yeah thats why we wanna transform PQ/AD to some ratio we might know
we see that PQ/AD=MP/AM
also you see that AP/PM is in very similar form here
Oh shit
and AN/BM have a known value: 0.5a/0.25a=2
Wait wait what
AN is a
BM is 0.5a
isnt it
Oh right
AN/BM=2
Yeah
so AP/PM=2.... note the similarity between AP/PM and MP/AM here...
maybe you can try to transform one to the other?
hint: write everything as multiples of PM
Ok ok
what is AP in terms of PM?
you will need to use AP/PM=2
Yeah i get it
the length of AP in terms of the length of PM
since AP/PM=2 we have AP is 2PM
Arent AP=PM
I had an explanation to 1at was that DM and AM meet at the midpoint M and CN and BN meet as midpoint N and because they intersect, I think it would form a straight line that’s parallels to the base AD
Exactly thats why i got confused
what straight line is formed? PQ you mean?
yes
shouldnt have P turned into a midpoint
And Q too
because does it just turn into a straight line without them being midpoints
as long as P and Q have the same height above AD, then they are parallel?
PQ being paralle to AD doesn't tell us a lot
i think a lil bit more rigor needs to be added
you want to put PQ parallel to AD into ratio equalities
what would imply PQ being parallel to AD?
MP/AP=MQ/QD
Oh wait
that gives us that
MQ/QD is 2
correct
correct
and this indeed proves AD parallel to PQ rigorously
i think your argument for why AP=PM is maybe just getting a lil bit carried away by some wrong intuition
I understand that it makes no sense but i had no other explanation
things regarding intuition is hard to help with, for now my suggestion is to not trust anything unless you know the full rigorous proof.
you build intuition through those proofs and example
yes that AP=PM
while thats not the case
But is PQ really a straight line
wdym by that? PQ is defined to be a line that is straight
My explanation to 1 didnt really state that P and Q are midpoints
I just stated how it could be a staright line
i don't understand what you mean by straight line here
u can't have a wiggly line connecting PQ by definition
Ah ill try to explain it to you
I dont really know the meaning in english
but ill find it rq
Google translate says it means "Means section"
do you know what that is?
not rlly
ok then ill try to find an accurate one
midsegment
Right
I remembered
midsegment is the name
midsegment of?
my explanation in 1 was to prove that PQ is a midsegment
PQ is a midsegment of AMD iff P and Q are midpoints
Exactly
Thats why i was confused when AP/PM was 2
that means i also proved 1 incorrectly
💀
the midsegment is defined for a triangle
but you are "extending" this to the intersection of two triangles
which doesn't quite work
Yeah
I dont know how to prove that PQ is parallel to AD
this is sort of a scenario where you really have to ask yourself whether the "intuition" you have is correct
i thought it made sense
Can you help me prove 1 then continue to 2
parrallel <-> same ratio
if you want to prove it, try to use similar triangle
ok lemme state it formally
How did we get MP/AP=MQ/QD before proving its parallel
oh im just saying we want to show that
MQ/QD is just 1/2 as you suggested b4
^try and see why
you might've gotten the variable mixed up look at the graph again
MQ/QD is MC/ND
Yes
0.5a/a
it is 1/2
it is 1/2
so they are equal!
just to make sure, you know why the equal ratio implies parallel right?
1/2 is much easier to work with
Yes
ok good!
ill use the 1/2
so you see the ratio isn't really 1, but 1/2
We said it was 2 then 1 and now 1/2
which sort of proves that your intuition originally was wrong
True
for the 2 we had the fraction flipped
its correct
Ok ill try to put it in other ratios
we said AP/PM=2
see if i can get somethin
but PM/AP=1/2 is still correct
yes
👍 PM/AP=1/2 is a known ratio now, and you can try to make it in some way equal to PQ/AD
(they are not equal but you can do something to the fractions)
if we use thales on triangle AMD and use a varation we get MP/MA=MQ/MD=PQ/AD
AM=2PM
Or am i dumb
MD=2MQ
AP=2PM not AM
Right
from this ratio
but its an easy fix
MQ=2QD
how many PM is AM?
not really AM is bigger than PM
MP is x AP is 2x
hint: AM=AP+PM
YES!
PQ=3AD?????
so what is this?
No wait
1/3
CORRECT!
and you are done!
AD is 2a isnt it?
It is
Wait uhhh
Thats crazy
thank you so much
I have more problems that idk how to do but woudve helped me for a hour so i wont bother you
no worries you got that PQ is 2a/3 right?
no worries
ill prob open a new help channel in a couple of hours since the deadline is tomorrow....
I got 5 more problems
Thank you again ill close this channel now
.close
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idk how to being
summation of F_x and summation of F_y = 0?
can someone solve it and send me the answer, i know its not allowed, but i just need to see the working of one question, so i can attempt the rest on my own
@high berry Has your question been resolved?
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hi
can someonehelp me
Ok.

that why you gotta put it formally