the same for the other intervals
#help-33
lilac siren
if none of them give you errors (like dividing by zero), then the inequalities and the critical values are all correct
gtg, hope that helped 🪼
still temple
thanks
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jade valley
hi i have two problems i'm stuck on! having to do with critical values and regression, i'll link the questions below
marsh citrusBOT
hi i have two problems i'm stuck on! having to do with critical values and regre...
jade valley
just help with ti84 calculator if needed with these questions
marsh citrusBOT
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limber hawk
I am confused on the exterior angle inequality theorem
marsh citrusBOT
I am confused on the exterior angle inequality theorem
night mica
could you post it please?
marsh citrusBOT
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limber hawk
<@&286206848099549185>
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solar heron
Measures less than m of angle 8
Measures greater than m of angle 3
I might be able to help, but I'm not sure how to think about what you posted
idle fractal
What do you want help with Idk what you want me to do with that picture
limber hawk
Measures less than m of angle 8
Measures greater than m of angle 3
^
it’s fine, just help me understand the exterior angle inequality theorem using that picture I just poster
Posted*
like examples
oh shoot wait that’s my bad I meant measure of angle 8 and e
3*
jaunty estuary
hi! im tryign to learn parameterizing surfaces and i am a bit confused as to what exactly i type into geogebra here
marsh citrusBOT
hi! im tryign to learn parameterizing surfaces and i am a bit confused as to wha...
long osprey
Surface(expression for x coord in terms of parameter vairables, y coord, z coord, name of variable 1, start of variable 1, etc.)
so in this case Surface(x,y,z, u, -1, 1, v, -1, 1)
with x,y,z in terms of u,v
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stark egret
Hi, Can someone please help me?
midnight obsidian
Hi, Can someone please help me?
is there anything specific you have an issue with
should be pretty straightforward once you factor out sec^2 (x) from the left
stark egret
is there anything specific you have an issue with
Im just lost
midnight obsidian
Im just lost
$\sec^4{(x)} - \sec^2{(x)} = \sec^2{(x)}[\sec^2{(x)} - 1]$
elfin berryBOT
nebula40
midnight obsidian
does this make sense to you
stark egret
Nope
marsh citrusBOT
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solar heron
Hey @stark egret if you want to join us on stream I can help guide you through the question
marsh citrusBOT
elfin berryBOT
Mycobacterium
silent marten
if f(x) is a function that is decreasing and is contineous can i say that the integral from n to + infinity f(x) dx is larger then the serie from n+1 to + infinity of f(n)?
marsh citrusBOT
if f(x) is a function that is decreasing and is contineous can i say that the in...
devout mauve
well, neither has to exist (meaning finite limit)
but this is called the integral test
to check whether the series converges
solar heron
It will definitely be dependent on the series
so sometimes the series will be larger, sometimes it will be smaller
But I think the general idea (from my intuition) is that if the sum of the rectangles is smaller than a convergent integral, then the series converges. If the sum of the rectangles is larger than a divergent integral, then the series diverges.
silent marten
to check whether the series converges
But this is to check if the integral and the series and the ingegral have the same nature right ? Here i am looking for way to prove that the sum from n+1 to infinity of n e^(-(n^2)) is smaller then the integral from n to + infinity of x e^(-(x^2))
solar heron
But this is to check if the integral and the series and the ingegral have the sa...
You are correct, you can only compare them if they have the same nature (conv or div)
devout mauve
well you can always compare them. its just not always useful
silent marten
Ok but for example why did you choose this 2 examples you did you know that the first one is bigger and the second one is smaller ?
devout mauve
if the integral converges then so does the series
if the series diverges then so does the integral
silent marten
Yeah right but for example if the integral converges and the series convergest too , which one is bigger ?
The integral or the Serie , thats my question
devout mauve
I mean you just have to draw the rectangles under the graph
instead of above
then just like the intuition says, the integral is bigger
silent marten
Thats the only method ? 😭 😭 😭
devout mauve
of course you shouldnt really use the word bigger if they diverge but whatever
well if you draw them above then the integral is smaller
from your wording of the question it makes much more sense to draw them below. because the series starts at n+1
solar heron
Thats the only method ? 😭 😭 😭
integral test should work as long as you know f(x) is continuous, positive and decreasing
silent marten
But that only tells me about the nature , lets say both are convergent how do you know that one is bigger ? Do we always need to draw the graph ?
solar heron
But that only tells me about the nature , lets say both are convergent how do yo...
I think the idea here is that you can always use EITHER right or left endpoints. So ONE of those is always an underestimate of the area (and this underestimate will match up with the series)
So I would say its more about convincing yourself that the integral converges or diverges
In your example above I evaluated:
to kind of convince myself that the integral converged
solar heron
I feel like I need more information about what N is
marsh citrusBOT
@silent marten Has your question been resolved?
silent marten
Sorry i am not that good at English, but i tried to translate the goal of the question and my work
I am seeing your stream now XD
And somehow i need to use the first inequality ( i think)
humble jewel
marsh citrusBOT
frail orbit
!status
marsh citrusBOT
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
frail orbit
i'm reading it more as 'whats it bounded below by'
stoic saddle
then they should write it that way
ffs
"which of the following is a lower bound for (sum)"
or "forall n in N, (sum) ≥ ..."
not "the min value" lmao
marsh citrusBOT
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ivory turtle
i'm reading it more as 'whats it bounded below by'
Bounded below by 1, done
frail orbit
well out of the options
idk
AH
integral test!
$\int_{1}^{N} \f{1}{x^p},dx \leq \sum_{n=1}^{N}\f{1}{n^p}$
@humble jewel
elfin berryBOT
🫎 A Certain User(Moosey) 🫎
stoic saddle
Bounded below by 1, done
this
stoic saddle
well out of the options
you know there is a "none of the above" option right
stoic saddle
i'm tired
tired or not, you've attempted to do the book author's job for them.
which is NOT something a student should ever do.
marsh citrusBOT
@humble jewel Has your question been resolved?
humble jewel
So option D?
humble jewel
<@1186149220160381102>
I am not understanding what is happening here?
@stoic saddle
Please teach me step by step if possible what's going here
If you have a little time
tough hazel
Hi
past frigate
Try finding f'(x) first
past frigate
Did you find it?
tough hazel
I dont know how to get value of p
tough hazel
Did you find it?
Yes
past frigate
You could compare coefficients on both sides after finding f'(x)
past frigate
Right so
ivory turtle
you can simplify the first term once more
past frigate
That is $x^{-1/2} + lnx/2x^{1/2}$
elfin berryBOT
Lorentz
tough hazel
ivory turtle
$\frac{a}{b}$
elfin berryBOT
Flappie
ivory turtle
technically yes, but you can simplify more
tough hazel
That is $x^{-1/2} + lnx/2x^{1/2}$
Like this??
past frigate
And you need it in form of this, so just compare the terms on both sides
You might wanna expand the x^p(...) Term
ivory turtle
$\frac{1}{\sqrt{x}}+\frac{\mathbf{ln}(x)}{2\sqrt{x}}$
elfin berryBOT
Flappie
past frigate
Yes but seperate the left side into 2 fractions
tough hazel
**Flappie**
Like this??
past frigate
And now compare the terms on both sides
We have x^(-1/2) and x^p
x^p g(x) and (x^(-1/2))* (lnx)/2
Find any similarities?
past frigate
Indeed
tough hazel
And g(x) is lnx/2
past frigate
Ye
tough hazel
Thanks
past frigate
Np
past frigate
I'm not sure, to find another way I need to think about it
tough hazel
Anyway thanks for helping
past frigate
Np
twilit totem
marsh citrusBOT
orchid oracle
Most basic definite integral ever; what am i doing wrong
marsh citrusBOT
Most basic definite integral ever; what am i doing wrong
woven eagle
how to solve the mode of this problem?
woven eagle
help, plss
ivory turtle
do you know what the mode is?
stoic saddle
do you know what the mode is?
^ @woven eagle
woven eagle
the value that occurs most frequently
ivory turtle
so, which of these values appears most frequently?
hint: that what is being shown here is that 12-15 appears 4 times, 16-19 6 times, etc
woven eagle
ohh, so it's 16-19..
ivory turtle
you sure?
stoic saddle
which of these score ranges has the highest frequency next to it?
woven eagle
nvm, 28-31
ivory turtle
there you go
sour relic
how isnt this symmetric?
marsh citrusBOT
how isnt this symmetric?
distant peak
is it equal to <y,x> ?
sour relic
isnt $<y,x> := y_1 x_1 - y_2 x _ 1 - y_1 x_2 + 3y_2 x_2$?
elfin berryBOT
Fractalogist
sour relic
or is it $:= y_1 x_1 - y_1 x_2 - y_2 x_1 + 3x_2y_2$
distant peak
it is still y2 x2 yes
elfin berryBOT
Fractalogist
sour relic
then ofc not symmetric
distant peak
it is tho
sour relic
**Fractalogist**
so its this?
ivory turtle
it looks symmetric, idk why it wouldnt be
distant peak
sour relic
ivory turtle
XD
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visual cedar
guys i don't see why from above eqution to 1.54. classical mechanics by goldstein
marsh citrusBOT
guys i don't see why from above eqution to 1.54. classical mechanics by goldstei...
visual cedar
i don't understand the subscript on grad (del) operator, what does that mean?
main idol
i don't understand the subscript on grad (del) operator, what does that mean?
With respect to the ith variable
Or whatever variable force is indexed by
marsh citrusBOT
@visual cedar Has your question been resolved?
visual cedar
Or whatever variable force is indexed by
i still don't get it. $\grad{V}=(\pdv{V}{x_1}...\pdv{V}{x_n})$ right? what does subscript here mean? only $\pdv{V}{x_j}$? Could you write down expanded version of this equation?
elfin berryBOT
yehuihe
main idol
i still don't get it. $\grad{V}=(\pdv{V}{x_1}...\pdv{V}{x_n})$ right? what does ...
F is a vector. V is a scalar
Find a relationship between F and V
marsh citrusBOT
@visual cedar Has your question been resolved?
visual cedar
Find a relationship between F and V
you meant $F=-\grad{V}$? that's for conservative fields and is assumed in here. I know this. but the very right side of expression how does this translate to 1.54
elfin berryBOT
yehuihe
main idol
**yehuihe**
F is a vector. Do you know what F_i is?
visual cedar
$F_i=-\grad_i{V}$
elfin berryBOT
yehuihe
visual cedar
so i guess it's a summation of $-\pdv_1{V}\pdv{r_1}{q_j}...-\pdv_n{V}\pdv{r_n}{q_j}$
elfin berryBOT
yehuihe
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main idol
Follows from the definition of total derivative
Called The Generalized Chain Rule there
visual cedar
ok sure i know generalized chain rule and it surely looks like it. Thanks ill take over from here
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frank eagle
Consider the series $\sum_{k=1}^\infty(-1)^kk$. I want to compute its Cesaro sum and its second Cesaro sum, but I get stuck doing the second iteration. So the sum reads $$-1+2-3+4-5+6-7+\ldots,$$ with partial sums $$-1,1,-2,2,-3,3,-4,4,\ldots.$$ The means are simply $-1,0,-2/3,0,-3/5,0,-4/7,\ldots$. Now I want to compute the mean of the means. I get $$-1,-1/2,-5/9,-5/12,-34/75,\ldots.$$I'm lost here, since I do not see a clear pattern, and it doesn't look like it converges to $-1/4$, which it should. Appreciate any help.
marsh citrusBOT
Consider the series $\sum_{k=1}^\infty(-1)^kk$. I want to compute its Cesaro sum...
elfin berryBOT
Philip
frank eagle
Please ping me if you'd like to help.
marsh citrusBOT
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vernal citrus
Hi all. I've been working on this question for a fair while now..
marsh citrusBOT
Hi all. I've been working on this question for a fair while now..
vernal citrus
I've managed to do parts A, B and C fairly easily. I'm struggling on part D.
Say A = Aaron, R = Rhys, J = James, D = Daniel
A) p(A) x p(R) x p(J) = 0.024
B) I found all the probabilities within the venn diagram i drew and done 1 - (all) to get the answer of 0.336
C) Simply used my venn diagram to get 0.084
D) I'm struggling here. I've been doing a lot of stuff. First, I try using P(D|J) to get P(D and J) as 0.18
Then for P(D|J') i done 0.25 x 0.8 to get P(D and not J) which I got as 0.2
But then when it comes to finding "1 or fewer are late" i struggle a lot. I've tried adding together the probabilities of more than 1 being late. I also tried finding individual probabilities of i.e P(Aaron and not anyone else) + P(Rhys and not anyone else) etc
But I never seem to get to the right answer
Any guidance would be appreciated. Cheers
<@&286206848099549185>
:/
shell ether
:/
vernal citrus
The question is that i'm not sure where to go after finding the probabilities in part D as every way i've gone (as said above) results in an incorrect answer?
gilded prairie
you got three cases in the last question, when james is late, when someone else is late besides daniel and james and when someone else is late
wait i`m sorry the last case is when daniel is late
vernal citrus
where's the "when someone else is late besides daniel and james"
gilded prairie
well that depends on who`s late i guess so you got four cases, when aaron is late you use the probality of him being late which is 3/10=0.3, when rhys is late you use 4/10=0.4
vernal citrus
its asking for "one or less being late", so surely you cant use those probabilities as they contain the probabilities of i.e aaron and reece?
gilded prairie
these probabilities are of them independently arriving late as it is said in the question
gilded prairie
But then when it comes to finding "1 or fewer are late" i struggle a lot. I've t...
the problem is that you are considering more than one being late, you just have to consider the cases of one or fewer arriving late
vernal citrus
is that not "P(Aaron and not anyone else) + P(Rhys and not anyone else) + P(...) + P(none late)" ?
vernal citrus
I simply used the values from the Venn Diagram P(Aaron and not anyone else) being 0.144, P(Reece and not anyone else) being 0.224, etc
gilded prairie
hmm, try multiplying the chances instead of using the venn diagram, if you cannot do that, just say and i teach you
vernal citrus
The mark scheme answer for the question is 0.6204 (mark scheme doesnt show any workings) , so I don't think multiplying is going to help in this situation as it will result in quite a small value. I tried doing P(Aaron) * P(Not Aaron) + P(Rhys) * P(Not Rhys) but i think that was very wrong lol
gilded prairie
hmm, ill try doing it and see what i got, be back in 3 minutes
vernal citrus
thank you!
vernal citrus
yes sure let me find my paper
when doing P(aaron and not others) etc i got 0.77
0.63 without the "none"
gilded prairie
ok that is not right
vernal citrus
that was using a P(D) of 0.2
gilded prairie
ok let me kinda explain what you got wrong, i dont know if can do this through text on discord but lets go
this question involves each case that someone of the five people arrives late at the party, right?
so lets look at each case carefully this time
lets say that aaron is the one arriving late at the party
considering that aaron has 0.7 of not arriving late
we can say that he has a 7/10 chance of arriving on time, agreed?
vernal citrus
yes
gilded prairie
ok, so based on that we can assume that he has a chance of 3/10 of arriving late
vernal citrus
yes
gilded prairie
considering that in this case aaron is the only arriving late, we multiply 2/10 with the chances of the others not arriving late
3/10
my mistake, sorry
so in this case it'll be: aaron (3/10) * rhys (6/10) * james(8/10)
and considering that james is not arriving late the chances for daniel will be 7.5/10
so the first case will be aaron (3/10) * rhys (6/10) * james(8/10) * daniel(7.5/10)
gilded prairie
yes, exactly
you just need to remember the cases, and add all of them in the final
remember that there's five cases
try it and let me see what you got
vernal citrus
yes trying it now
gilded prairie
pin me when you finish, i'll go drink some water
vernal citrus
i done:
P(A) * P(not R) * P(not J) * P(not D)
- P(not A) * P(R) * P(not J) * P(not D)
- P(not A) * P(not R) * P(J) * P(not D)
- P(not A) * P(not R) * P(not J) * P(D)
= 0.612 ?
i think i messed up when figuring out the P(J) * P(not D) i can try again
gilded prairie
yeah i almost messed up in that part too, remember that the chances of daniel changes when james is late
vernal citrus
i'm now getting 0.5364 if i do 0.2 * 0.1 for P(J) * P(not D)
hold on let me just write exactly what i'm doing
0.3 * 0.6 * 0.8 * 0.75 (A is late)
- 0.7 * 0.4 * 0.8 * 0.75 ( R is late)
- 0.7 * 0.6 * 0.2 * 0.1 (J is late)
- 0.7 * 0.6 * 0.8 * 0.25 (D is late)
gilded prairie
the numbers checks out, you're just forgeting one case
vernal citrus
oh when none are late
gilded prairie
i'm now getting 0.5364 if i do 0.2 * 0.1 for P(J) * P(not D)
btw, there's something wrong with this, even with this numbers, it should not be 0.5364
vernal citrus
ohhh i've managed to get the right answer
gilded prairie
yes, congrats!!
vernal citrus
0.3 * 0.6 * 0.8 * 0.75 (A is late)
+ 0.7 * 0.4 * 0.8 * 0.75 ( R is late)
+ 0.7 *...
i done this + (0.7x0.6x0.8x0.75) = 0.6204 😃
thank you so much for your help, i appreciate it a lot
gilded prairie
anytime
btw, just remember that the venn diagram can be really usefull to visualize problems, but sometimes it can get in the way
vernal citrus
yes definitely has its benefits and drawbacks
gilded prairie
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subtle swan
find all possible values of a^2+b^2+c^2
someone told me to add the first two equations and minus the last one two times
but how exactly would i get the answer
marsh citrusBOT
find all possible values of a^2+b^2+c^2
someone told me to add the first two eq...
@subtle swan Has your question been resolved?
marsh citrusBOT
@subtle swan Has your question been resolved?
still temple
you will get $(a^2-3a)^2+2(b^2-2b)^2+4(c^2-c)^2=0$
elfin berryBOT
Maria
subtle swan
you will get $(a^2-3a)^2+2(b^2-2b)^2+4(c^2-c)^2=0$
and then?
still temple
when sum of squares equals 0?
marsh citrusBOT
subtle swan
.reopen
marsh citrusBOT
✅
subtle swan
when sum of squares equals 0?
if they're all zeros?
still temple
if they're all zeros?
yes
subtle swan
ok ty
subtle swan
but check to make sure i didn't make any typos here
yes i got the same thing
ty 4 the help
marsh citrusBOT
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Available help channel!
charred pelican
marsh citrusBOT
charred pelican
how do i solve these equations for i and i1
i got i from the second one, which is i1*R+E/2R
how
isnt it DE
i cant group em man
i1 just dosent separate
🤦
not from a coaching
☠️
wild
novel juniper
what's teh equation you're trying to solve again?
charred pelican
what's teh equation you're trying to solve again?
solving for i and i1
charred pelican
what's teh equation you're trying to solve again?
two variables two equations so
charred pelican
i got i from the second one, which is i1*R+E/2R
i got this
marsh citrusBOT
@charred pelican Has your question been resolved?
forest orchid
marsh citrusBOT
proper zodiac
Consider the cases x > 0 and x < 0 separately
hidden plaza
What are these options…
lucid zenith
for real why is -8 an option bro
lucid zenith
algebrically ?
in general, the limit of a function involving |x-a| at a causes problems
forest orchid
Consider the cases x > 0 and x < 0 separately
Ok
i got -3 and 3
so DNE
were these good number choices? 0.1 and -0.1
for x>0, 0.1, 0.11 and x<0, -0.1, -0.11, -0.111
forest orchid
in general, the limit of a function involving |x-a| at a causes problems
how do i recognize the |x-a| problems
Like the one in this
Lol
i Might be Dumb guys
marsh citrusBOT
@forest orchid Has your question been resolved?
vale panther
Am I tripping or its just basic algebra
marsh citrusBOT
Am I tripping or its just basic algebra
lucid zenith
it is
still temple
probably to hint at what to do for the following part(s)
marsh citrusBOT
@vale panther Has your question been resolved?
vale panther
I think its just telling us that Xn and Xn+m for any m in integer form a field?
wild tiger
Hi, I have a small problem. I have two possible answers or solution approaches for the T test for independent samples. One I got from the book, the other from my lecture. Are both correct or have I made a mistake? Shouldn't the result be the same? Or is one approach completely wrong? If both are correct, which one is more accurate and what is the difference? Thanks for your help 🙂
$n_1=9 // n_2=9//\bar x_1=102,33//\bar x_2=105,89//s_1^2=187,50//s_2^2= 190,86$
$H_0:\mu_1=\mu_2 // H_2: \mu_1\ne \mu_2$
-> Number 1
$S_p^2=\frac {(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}=\frac {S_1^2n_1+S_2^2n_2} {n_1+n_2-2}=\frac {187,50 \cdot 9 + 190,86 \cdot 9} {9+9-2} =212,8275\
S_{\bar x_1-\bar x_2}=\sqrt {\frac {\hat \sigma_p^2} {n_1}+\frac{\hat \sigma_p^2} {n_2}}=6,88\
t_{df}=\frac {\bar x_1 -\bar x 2} {S{\bar x_1-\bar x_2}}=-0.517$
Number 2
$S_p^2=\frac {S_1^2-S_2^2} 2= 189,18 \
S_{\bar x_1 -\bar x_2}=\sqrt {S_p^2(\frac 1 n_1 + \frac 1 n_2} )=6,48 \
t_{df}=\frac {\bar x_1 -\bar x 2} {S{\bar x_1-\bar x_2}}=-0,55$
i am really lost so thanks for any help
marsh citrusBOT
Hi, I have a small problem. I have two possible answers or solution approaches f...
elfin berryBOT
isa
blazing pulsar
You should just read your lecture notes carefully and see what the difference in assumptions are being used between each test, but I've never seen anything that looks like number 2 before, (S_1^2 - S_2^2)/2 doesn't really make sense because it looks like a biased estimator to me and there isn't enough context to see what they are trying to do.
wild tiger
You should just read your lecture notes carefully and see what the difference in...
thats the weird part, the assumptions are both the same (in name) i know one here is \hat\sigma_p^2 and the other is S_p^2 but they are both named the same. i m confused with the same step where u say it doesnt make sense
wait for the second one i can send the whole part of what they do
elfin berryBOT
isa
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wild tiger
that is the original example by the book
and i thought i could give it a try and calculate it the way my prof did in the lecture, which is seen in number 1
<@&286206848099549185>
blazing pulsar
Your work is annoying to read because it is full of typos in the notation. That document did not say to calculate (S_1^2 - S_2^2)/2, and you should have hopefully noticed that was an issue when I mentioned it instead of dumping copypasta out.
The work you said your professor wrote doesn't even make sense, you write that (n_1 - 1)S_1^2 + (n_2-1)S_2^2 = n_1S_1^2 + n_1S_1^2, this is just not true. One side has a finite population correction factor the other does not. Then you write a formula where S is now a function of sigma, which isn't even included in the data.
marsh citrusBOT
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quartz hearth
help me im going to break down
marsh citrusBOT
help me im going to break down
quartz hearth
where did they get the modulus things i generally dont understand the problem at all TT
frail orbit
which step do you not understand?
modulus thing is just the absolute value, and that's just makes negative inputs positive, and does nothing to positives and zero.
you can also think of it as modulus, but imaginary part zero.
i.e. |x|=sqrt(x^2)
marsh citrusBOT
@quartz hearth Has your question been resolved?
marsh citrusBOT
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leaden berry
e^(x/2-[x/2])
marsh citrusBOT
e^(x/2-[x/2])
leaden berry
How to find without geogebra that this function repeats intervals?
stoic saddle
you are missing a closing parenthesis.
leaden berry
By putting random numbers 0 to 1,2,3
stoic saddle
does [...] mean floor?
leaden berry
Yes
leaden berry
{x/2} is decimal
X=0
It will be 0
at 1 it will be 0.5
But at 2 it will be 1
After 2 it will repeat itself
Nice
.close
marsh citrusBOT
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lean imp
marsh citrusBOT
lean imp
so im trying to do the root test
because it seems to be the way this is done in analysis, ive always struggled with it
when i put this in absolute values and i raise it to the 1/n power
do i also bring the 1/n to the bottom? so that the demoninator becomes 9(n^2+1)?
im unsure how i distribute the 1/n to the denominator essentially
nm i figured it out
.close
cerulean oxide
is this funtion differentialble when x<0?
when the root becomes lets say -8?
main idol
you should find out if the function is even defined at x=-8 first
cerulean oxide
it isnt
but how do i show it isnt
or rather let me ask a better question:
"For all C > 0, find a continuous function y : [0, ∞) → [0, ∞)
with y(0) = C which is differentiable p ̊a (0, ∞) and satisfies the differential equation (1) here: y'(x)=4 sqrt (yx)" is the question itself.
So far i've:
differential equation by seperable variables -> sovle the intergral -> find C (C=4) -> found y -> is positive for all real numbers, due to $$ \sqrt(x^3 $$ dont need to think of if the numerator is 0.
Now i believe i have to show that y(x) is a differentiable and continuous function
elfin berryBOT
// mav
cerulean oxide
<@&286206848099549185> have i understood the question corrctly
marsh citrusBOT
@cerulean oxide Has your question been resolved?
marsh citrusBOT
@cerulean oxide Has your question been resolved?
cerulean oxide
<@&286206848099549185>
marsh citrusBOT
@cerulean oxide Has your question been resolved?
hidden cobalt
marsh citrusBOT
hidden cobalt
@still temple Help pls
solemn plover
What's the answer
main idol
I have no clue
Find the roots to eliminate some options
solemn plover
It's d
hidden cobalt
Find the roots to eliminate some options
by looking at it i can see -4 , 0 , 6
main idol
And observe there's a double root at 0
solemn plover
See where the roots are
solemn plover
Do you have desmos
Forget it
I'll show
For a single root, the graph cuts clean through it
hidden cobalt
Do you have desmos
yes
solemn plover
For even powered root it's roughly the shape in the second picture
hidden cobalt
For a single root, the graph cuts clean through it
well i know that
hidden cobalt
For third powered root the graph is like the third picture
🗿
still temple
root at -4, 0 , 6, get each function, and set x to one of those 3 numbers, all of them should be equal to 0
if all 3 numbers make the polynomial equal to 0 then thats the correct oen
hidden cobalt
Is there something wrong
no I don't understand 😭
solemn plover
Do you understand how the graph cuts the X axis for each power of root
hidden cobalt
I have never seen these graphs
hidden cobalt
Do you understand how the graph cuts the X axis for each power of root
yes
hidden cobalt
yes
solemn plover
It cuts clean at x=-4
hidden cobalt
yes
solemn plover
Therefore one factor is (x+4)
hidden cobalt
ok
solemn plover
At x=0
hidden cobalt
yes
hidden cobalt
Therefore we can assume x² is a factor
ok
solemn plover
At X=6
hidden cobalt
is (x-6)
solemn plover
hidden cobalt
oh ok
hidden cobalt
well yeah
solemn plover
x⁴ and x² have strikingly similar graphs
hidden cobalt
x⁴ and x² have strikingly similar graphs
fr?
hidden cobalt
You can only assume from given options
If I had to eliminate, it would be B, C, and A
solemn plover
Why not e
solemn plover
Does the graph cut x=0 like a triple root?
hidden cobalt
Does the graph cut x=0 like a triple root?
Oh shit
my bad
I didn't see that
thats it?
solemn plover
Yes
hidden cobalt
dang
hidden cobalt
Remember what I said
yes look at the roots and assume
solemn plover
Highly recommended you to mess around with desmos
tender oyster
can you prove a square is the only type of quadrilateral with four lines of symmetry?
marsh citrusBOT
can you prove a square is the only type of quadrilateral with four lines of symm...
tender oyster
10
solemn plover
Let's try this
We have 2 lines of symmetry through the centre of any side
For a quadrilateral
tender oyster
We have 2 lines of symmetry through the centre of any side
wdym
solemn plover
We have 2 lines of symmetry through the centre of any side
Prove that it's possible only for a rectangle
Then prove for diagonals
marsh citrusBOT
@tender oyster Has your question been resolved?
leaden berry
marsh citrusBOT
still beacon
you have a 1/4 chance of guessing correct
leaden berry
What??
static quarry
why would you ping moderators
leaden berry
I guess they are trolling
leaden berry
Gambling?
still beacon
how am I trolling?
static quarry
look at columns 1 and 3, what linear combination of them would give 0 in rows 1 and 4?
static quarry
there's no determinant here, it's not a square matrix
static quarry
no not true
all matrices have a rank, square or not
rank = # of linearly independent columns
(or rows)
leaden berry
Yes they have but their rank is considered only equal to square
static quarry
Yes they have but their rank is considered only equal to square
no
static quarry
you know the rank is at least 2 since cols 1 and 3 are linearly independent
leaden berry
Rank has many definitions
static quarry
tell me your definition
next raft
number of leading ones in the reduced row form is one such definition
static quarry
ok
so that's one option, you can row reduce
that will be a lot of work though, due to the variables
maybe working with the rows is the most straightforward:
obviously rows 1 and 4 are linearly independent
so the rank is at least 2
now show that row 2 can't be expressed as a linear combination of the other rows (or if it can, then row 3 can't)
so then the rank is at least 3
(and therefore exactly 3)
leaden berry
My question is if we put r=-2 we can get a matrix 2×2 which is non zero determinant but we need to he sure for rank 3 first
next raft
number of leading ones in the reduced row form is one such definition
I'm not Harry! you should have waited for him to respond
leaden berry
If rank 3 is Determinant 0 then it will have rank 2 otherwise rank 3
static quarry
If rank 3 is Determinant 0 then it will have rank 2 otherwise rank 3
this is not a square matrix, you can't take determinants
leaden berry
Let me try now with row reduction
static quarry
I'm not Harry! you should have waited for him to respond
oh sorry, i didn't notice that ..
leaden berry
Rank is 3
I got one row is 0 others are non zero and echelon form
Wait
Check option 2
B is correct
It gives us rank 2 i check it with row reduction
@static quarry
static quarry
yes
another way to see it, if you plug in r=2 and s=1 then the 2nd column becomes zero
so the rank is at most 2 in that case
and cols 1 and 3 are LI so the rank is always at least 2
marsh citrusBOT
Available help channel!
midnight cipher
the answer should be 25/17 but idk why its wrong in the part i marked
midnight cipher
sorry for the bad quality
old glacier
im having headache ngl
midnight cipher
ik
old glacier
is that rodrygo?
midnight cipher
who? the one in my pfp?
old glacier
ye
midnight cipher
nah, its alex valera
old glacier
who
marsh citrusBOT
marsh citrusBOT
Available help channel!
still temple
marsh citrusBOT
still temple
can someone please help me with this
which one is displacement, which one is velocity and which one is acceleration
night mica
artemetra try to not give the full answer
no body cares abut ak 47 of course ...but you could be banned by some wierd staff
becouse of you just give the full answ
vernal forge
wdym
still temple
is that displacement, velocity, or acceleration
unborn condor
define displacement for me
night mica
define displacement for me
marsh citrusBOT
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Available help channel!
cerulean oxide
@glass silo
marsh citrusBOT
<@788085606483361802>
cerulean oxide
third question if you have time
night mica
chartbit is the best here for the "third question"
cerulean oxide
chartbit is the best here for the "third question"
(he knows it already)
glass silo
Let me fetch it for context
cerulean oxide
"For all C > 0, find a number K > 0 and a continuous function y from [0,K] -> [0,inft), with y(0) = C. This function must be differentiable on (0,K) and satisfy the given differential equation (2). For the given C, what is the largest possible value of K?"
glass silo
Differential equation being $y'(x) = (y(x))^2$ or?
elfin berryBOT
@glass silo
cerulean oxide
yes
glass silo
In that case, you're happy with starting it off?
cerulean oxide
can what i seperate tho
just y' = y^2
unless dx counts as one
so you get integral one dx = 1 = x
why can i view dx as 1
glass silo
Well, more that when you separate, you have 1 on the right hand side
cerulean oxide
yh but why does it work to replace dy and dx with 1
glass silo
Well I mean you're not really "replacing" dy and dx with 1, it's just that from $y' = y^2$, you get $\frac{y'}{y^2} = 1$ and from there when you integrate you end up with $\int y^{-2} \dd y = \int 1 \dd x$
elfin berryBOT
@glass silo
glass silo
More that you divide both sides of y' = y^2 by y^2
glass silo
the solution assumedly would only be valid for certain values of x
glass silo
Your solution at the end I believe should be basically $y = \frac1{C - x}$, but of course, initial conditions and stuff
elfin berryBOT
@glass silo
cerulean oxide
<@788085606483361802>
C-C=0?
glass silo
but then it isnt defined
That's where the [0, K] part comes in
The function only needs to be defined (and the other stuff) on [0, K]
cerulean oxide
The function only needs to be defined (and the other stuff) on [0, K]
how do i even know my C is more than C, or i that something i ought ot suppose
glass silo
\catthink think you should actually basically get that $y = \frac{C}{C - x}$ as the solution in terms of the inital condition $C$ they give you?
elfin berryBOT
@glass silo
cerulean oxide
\catthink think you should actually basically get that $y = \frac{C}{C - x}$ as ...
why is C in the top as well
?
glass silo
why is C in the top as well
Because remember that you need to label your integration constant differently - then when you find it in terms of this C, you should get there
So like your one should be $x= -\frac1y + C_{int}$, then $y(0) = C$ gives you $0 = -\frac1C + C_{int}$ and $C_{int} = -\frac1C$
elfin berryBOT
@glass silo
cerulean oxide
ok
glass silo
Then you're rearranging $x = -\frac1y - \frac1C$ and all
elfin berryBOT
@glass silo
cerulean oxide
1 sec then
cerulean oxide
what
how is that not true
i jsut forgot to write C_int on the lefdt one
oh
u r on about my algebra
i just forget a -x
dont rly know if that does antything good for me tho 
@glass silo
glass silo
Multiplying by -1 should get you that intermediate line as $y= -\frac1{x - C_{int}}$ though, which leads you to the same $C = \frac1{C_{int}}$ but the final solution a bit different
elfin berryBOT
@glass silo
glass silo
<@788085606483361802>
messed up here a bit, should have been $y = \frac{C}{1 - Cx}$ rather
elfin berryBOT
@glass silo
cerulean oxide
messed up here a bit, should have been $y = \frac{C}{1 - Cx}$ rather
where do i insert 1/C_int
i thought the idea was to find the intergration constant
C_int=1/C
glass silo
C_int=1/C
Yea that's what I used to get the y = ...
You rearrange it for $C_{int} = \frac1C$ and then use that
elfin berryBOT
@glass silo
cerulean oxide
Yea that's what I used to get the y = ...
ok nice
so we get this
@glass silo u also got this
glass silo
Yep, same thing! 
glass silo
Well you want to pick some value such that the function is well defined and stuff
glass silo
<@788085606483361802>
The fact that you have it in this form means that, for example, you're defined at x=0
cerulean oxide
yeah, but what is this K thing exactly?
glass silo
The largest input value of x you want so that the function is well defined
cerulean oxide
largest?
glass silo
E.g. you could say, if C = 1, then your function is y = 1/(1 - x) and it is well defined, say, when x is between 0 and 0.5 inclusive
It also is well defined for say 0 to 0.9 for example, as long as x is not 1
cerulean oxide
it says "for the given C"
cerulean oxide
It also is well defined for say 0 to 0.9 for example, as long as x is not 1
i thought there would be a time where i got a value of C
glass silo
You won't find an explicit value of C here - the idea is that your K will depend on C
cerulean oxide
so when my C>0 then my K what
i am just not really understading how an equation without K can depend on K
cerulean oxide
You won't find an explicit value of C here - the idea is that your K will depend...
how
glass silo
What do you think the K would be? For example, what values do you think you can't choose for x?
cerulean oxide
if i dont know C idk?
cerulean oxide
What do you think the K would be? For example, what values do you think you can'...
i cna chose any values for x?
i can just adjust me C?
glass silo
What are you not allowed to divide by?
cerulean oxide
What are you not allowed to divide by?
but that isnt a value of K tho?
so my K>x*C
glass silo
but that isnt a value of K tho?
You pick K based on that
cerulean oxide
what have i even said
x*C=1 if x=C 😭
(1/C)*C = 1 yh
so K < 1/C?
@glass silo i have a new question, also in ODES
glass silo
cool cool 
cerulean oxide
<:catThink:1073406415575728200> cool cool <:roopopcorn:616252654175256579>
but
in the one before
why isnt x just called k
if we are inserting for x
glass silo
Because you want a range of values for x, from 0 to K - anything in between those should be valid to use
cerulean oxide
arh, ok
so K = 1/C will give us division by 0, which is not allowed
K > 1/C will give us a negative value, which is not on [0,K]
glass silo
K > 1/C would also include the choice K = 1/C, for which you'd be dividing by zero to and so similarly isn't allowed
cerulean oxide
I think i've solved the first one correct (a)
idk what i am even being asked in the second one
@glass silo
glass silo
Looks good to me
glass silo
idk what i am even being asked in the second one
Have you dealt with second order differential equations before?
cerulean oxide
Have you dealt with second order differential equations before?
no, only in euler/runge kutta method
but i doubt that helps here
glass silo
well presumably they want you to solve that differential equation under the initial conditions they gave, or?
cerulean oxide
yes yes
cerulean oxide
<:sadcat:635281037462601789> well presumably they want you to solve that differe...
Nonhomogeneous 2nd Order Differential Equations
is that what we are dealing with
glass silo
Nonhomogeneous 2nd Order Differential Equations
Well this one is homogenous and second order, but likely that would have been covered along the way!
glass silo
Yep it is 
cerulean oxide
Yep it is <:catThumbsUp:662911317090369546>
how do i solve it tho
glass silo
when it isnt equal to 0
Yep, when it's not equal to zero, it's nonhomogenous
And usually massive summary is that you "assume" the solution is something of the form $y = e^{\lambda x}$ and then use that to form a corresponding quadratic equation, noting that $y'' = \lambda^2 e^{\lambda x}$ and find possible values for $\lambda$ from there
elfin berryBOT
@glass silo
glass silo
Not sure I can condense it into something useful enough haha
cerulean oxide
Not sure I can condense it into something useful enough haha
so we guess that y = e^lamda*x
glass silo
Kinda like that, there should be some source that explains it somewhere that does it better
cerulean oxide
well
ay'' + by' + cy = 0
$$ ay'' + by' + cy = 0$$
$$y = e^{\lambda x} $$
$$ a\lambda^2 e^{\lambda x} + b\lambda e^{\lambda x} + ce^{\lambda x} = 0 $$
$$e^{\lambda x} (a\lambda^2 + b\lambda + c) = 0 $$
$$ a\lambda^2 + b\lambda + c = 0 $$
$$ y = e^{\lambda x} $$
elfin berryBOT
// mav
glass silo
Well you basically need to solve $\lambda^2 - 4 = 0$ and find the two values of $\lambda$ from there
elfin berryBOT
@glass silo
glass silo
glass silo
See the two real roots section from the above
glass silo
That's pretty much it, now use the initial conditions
put x=0, you should get D, and then when you differentiate that then put x=0, you should get 0
glass silo
Yea - the constants A and B will also depend on D
charred flicker
hello
cerulean oxide
Now i see that D = A + B
however made this must've thought they were well funny writing DAB
@glass silo
cerulean oxide
Now use that y'(0) = 0 - find y'(0) in terms of your constants <:roopopcorn:6162...
mh?
glass silo
so if i differentiate y
Yep, that's it
cerulean oxide
y = Aexp(2x) + Bexp(-2x);
this is what i differentiate?
so 2Aexp(2x) - 2Bexp(-2x)
glass silo
Yep
glass silo
that's zero
cerulean oxide
ok
how do i now find D
from this
do i reinsert into the old equation
but this doesnt give me anything new
it just says A=B
D=2A or 2B then
glass silo
So A and B are both D/2
cerulean oxide
so this is every D that satisfies the equation
D/2
or do i now
insert:
y = Aexp(2x) + Bexp(-2x);
A= d/2 and d/2 on B
glass silo
Yep, put A = B = D/2 in
cerulean oxide
glass silo
And that's your general solution for you!
cerulean oxide
that is D
glass silo
Well $D$ is supposed to be constant - leave your solution as $y = \frac{D}2 e^{2x} + \frac{D}2 e^{-2x}$
elfin berryBOT
@glass silo
cerulean oxide
oh, so no y
glass silo
That's what y is in terms of x and D
cerulean oxide
y = something x and D
glass silo
Yea what you found
marsh citrusBOT
Available help channel!
leaden berry
0 to 2000 integration e^{x/2}dx where {} fractional function
marsh citrusBOT
0 to 2000 integration e^{x/2}dx where {} fractional function
patent blaze
So you need $\int_{0}^{2000} e^{x/2} dx$?
elfin berryBOT
Jordi
patent blaze
Well if that's the case, try to find an antiderivative for $e^{x/2}$
elfin berryBOT
Jordi
steel sluice
Well if that's the case, try to find an antiderivative for $e^{x/2}$
isnt it just 2e^(x/2)
patent blaze
Indeed. There you have your answer
Now recall that $\int_a^b f(x) dx=F(b)-F(a)$, where $F'(x)=f(x), a,b\in\mathbb{R}$, and you have the solution
elfin berryBOT
Jordi
novel juniper
The fractional part function is just x-floor (x) , so the question is a bit more involved than that
steel sluice
Indeed. There you have your answer
can u pls solve this also https://discord.com/channels/268882317391429632/903486480985489478
patent blaze
The fractional part function is just x-floor (x) , so the question is a bit more...
What do you mean?
patent blaze
Oh, I assumed they meant with fractional part of x/2 that x/2 was a fractional exponent
If that is indeed the fractional part (x/2), the question is indeed a lot more complicated @steel sluice
novel juniper
0 to 2000 integration e^{x/2}dx where {} fractional function
Notice that the function is periodic with a period of 2
Or atleast I think it's periodic
patent blaze
Or atleast I think it's periodic
It is
steel sluice
If that is indeed the fractional part (x/2), the question is indeed a lot more c...
what does fractional part mean?
marsh citrusBOT
@leaden berry Has your question been resolved?
ebon depot
_C_R_M_B_
A_A_A
CARAMBA
A can’t be next to each other
What is the total amount?