It means that $\sum_{n=0}^N x_n$ converges to finite value as N to infinity.
#help-10
slate zephyr
warm shaleBOT
M8732
slate zephyr
In mathematics details really matter to avoid confusion.
slate zephyr
I think I will explain some more things about series in general.
First, there is a thing called comparison test.
slate zephyr
All it really means is that if you have a second series y_n that you know to be SMALLER and y_n IS DIVERGENT, then your bigger x_n's are also gonna be divergent as a series.
digital remnant
yup
slate zephyr
Why? Well quite simply $\sum_{n=0}^N x_n > \sum_{n=0}^N y_n \to \infty$
warm shaleBOT
M8732
slate zephyr
and by rules of limes this means the first is also gonna pproach infinity.
warm shaleBOT
demiryolu mühendisi
warm shaleBOT
M8732
digital remnant
yeah
i should divide by n^a?
but then its wrong
$\frac{ln(n)}{n^a} < \frac{1}{n^{a-1}}$
warm shaleBOT
demiryolu mühendisi
digital remnant
a - 1 can be < 1
slate zephyr
there is also a second part to the convergence test. It says if a series is POSITIVE and termwise LESS than another one, then we know for sure it is convergentz.
digital remnant
i dont see it
slate zephyr
$\ln(x) < x$, you agree?
Okay, I realize this will only do it for a > 2.
slate zephyr
a - 1 can be < 1
yeah, it won't work for convergent a between 1 and 2.
slate zephyr
What do you mean with riemann?
digital remnant
n^b with b > 1
digital remnant
**demiryolu mühendisi**
like this
because we do the serie and we say whats on the right converges hence whats on the left converges also
slate zephyr
yeah
I would say it is just geometric series. We know that geometric series converge.
digital remnant
yeah
slate zephyr
for this
slate zephyr
I want to spell out a warning again.
You can combine comparison test with integrals, BUT it only works if the function is monotone.
if your function is oscilatory, then you pick up the intermediate values in the integral and god knows what will happen
slate zephyr
I recommend looking into the different "tests"/"criteria" and closely study their proofs and preconditions.
slate zephyr
**M8732**
The good news is that none of that will be much more complicated then this argument lol.
Anyway,. let's try to solve the problem at hand.
digital remnant
how about i do $ln(n) > 1$ at a given moment. SO we have $\frac{ln(n)}{n^a} < \frac{1}{n^a}$
for a > 1 still
and we conclude with riemann
slate zephyr
💀
warm shaleBOT
demiryolu mühendisi
slate zephyr
I am afraid you used the inequality the wrong way.
digital remnant
oh yes
slate zephyr
you get the last line in the other direction
digital remnant
doesnt make sense lmao
slate zephyr
We need ln(n) < something
convergence requires limiting size, so we need less then something.
digital remnant
i have idea
lemme write it
i still have a problem for a < 2
this is so terrible really
slate zephyr
We need a stronger estimate.
slate zephyr
**demiryolu mühendisi**
What does this really mean?
digital remnant
it means $\frac{ln(n)}{n^a} \rightarrow 0$ when $x \rightarrow +\infty$
warm shaleBOT
demiryolu mühendisi
slate zephyr
yes!
The issue is that this just says it goes to zero
nobodyx is summing yet.
However
your statement is still not good enough.
It's true though also for a between 0 and 1 I think.
Do you think you can use that? Where is the statement from?
slate zephyr
,wolf ln(n)/n^0.1, n -> oo
warm shaleBOT
slate zephyr
yes
digital remnant
slate zephyr
🇫🇷 😬
slate zephyr
I will trust you, because piecing it together with basically nonexistant french is going to be difficult.
slate zephyr
this is so sad
aw come on, from our conversation here it doesn't seem so bad.
digital remnant
yeah
slate zephyr
Let me try think of a onvincing argument why it is also true for a between 0 and 1.
digital remnant
but showing that ln(n) = o(n^a) isnt really required
we never had to prove it in course
digital remnant
yes
warm shaleBOT
M8732
slate zephyr
yay
with this we get an easy proof, I think, let's see.
I claim ln(n) < n^a + A for some A for any a > 0.
if we just apply exp, which is monotone and preserves inequalities, it is equivalent to
n < exp(A) exp(n^a)
and that is true because if you use power series for second exp, you can consider the term with sufficiently high power k, such that k > 1/a.
exp(x) > C x^k for some C, hence exp(n^a) > C n, and A = ln(C) solves the original claim.
slate zephyr
I claim ln(n) < n^a + A for some A for any a > 0.
and then we can use this to proof an o thing if we want I think.
But it also directly solves our divergence problem, so forget about the o thing.
slate zephyr
I claim ln(n) < n^a + A for some A for any a > 0.
Does this make sense? @digital remnant
digital remnant
it does
slate zephyr
great
now we have ln less than something
and n^a + A is gonna be eventually smaller than just n if a is small.
digital remnant
this has no solution ig
i had one
but it makes more cases
thanks for the help though
definetly learnt new things
ill ask more people tomorrow
slate zephyr
this has no solution ig
no?
obtuse pebbleBOT
@digital remnant Has your question been resolved?
open meteor
obtuse pebbleBOT
grizzled shore
$\frac{1}{2\sqrt{Gm}}\int_d^0\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{d}}}, dr$
warm shaleBOT
Frosst
grizzled shore
is that right?
$\frac{1}{2\sqrt{Gm}}\int_d^0\sqrt{\frac{rd}{d-r}}, dr$
perhaps like this?
warm shaleBOT
Frosst
open meteor
yes
open meteor
perhaps like this?
where do go from here though
grizzled shore
im guessing this is physics?
open meteor
yes
grizzled shore
are there any restrictions on the constants
open meteor
two particles of mass m are distance d apart, how long for them to meet
grizzled shore
d has to be negative here dont they
open meteor
particle is moving from distance d apart to distance 0 apart
grizzled shore
you'll get a negative answer
grizzled shore
you should've sent this as your first picture
open meteor
sry my fault
grizzled shore
not just 1 single part of a larger picture
grizzled shore
Hmm
Perhaps let r = d - x or d + x
Something of that sort
Then maybe they divided as well
I think they were just rescaling the limits from d to 0 to 0 to 1
obtuse pebbleBOT
@open meteor Has your question been resolved?
slate zephyr
I think they were just rescaling the limits from d to 0 to 0 to 1
yes
just substitution to get boundaries
x=r/d
obtuse pebbleBOT
@open meteor Has your question been resolved?
sinful seal
obtuse pebbleBOT
harsh remnant
obtuse pebbleBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
Exponetial?
sinful seal
y?
sinful seal
Idk is it 3?
sinful seal
I dont know then
sinful seal
1?
sinful seal
ok im dumb sorry. so its just one?
sinful seal
Doesnt have one for that question?
How do i find the x-int and the equation of asymtotes?
Ok so i know to find the x-intercept y=0 but i cant figure out how to rearrange when the power is x
<@&286206848099549185>
obtuse pebbleBOT
@sinful seal Has your question been resolved?
sinful seal
.close
fiery knot
obtuse pebbleBOT
obtuse pebbleBOT
@fiery knot Has your question been resolved?
fiery knot
<@&286206848099549185>
obtuse pebbleBOT
@fiery knot Has your question been resolved?
thick knoll
obtuse pebbleBOT
polar fossil
just make another
obtuse pebbleBOT
Available help channel!
tribal marten
help me
obtuse pebbleBOT
@tribal marten Has your question been resolved?
timid silo
So the question is asking to change the parameter equation into cartiesian equation.
The answer key says that the equation is y = sqrt(x^2-1) but I got y = xsin(t)?
obtuse pebbleBOT
So the question is asking to change the parameter equation into cartiesian equat...
tardy epoch
So the question is asking to change the parameter equation into cartiesian equat...
Well you didn't finish?
timid silo
oh to get rid of t
right
ok nvm i redid it again
i used 1 + tan^2(t) = sec^2(t) trig identity instead
alr thx
.close
scenic scaffold
I am still confused on the concept of differentiability? How would you differentiate this function?
obtuse pebbleBOT
I am still confused on the concept of differentiability? How would you different...
scenic scaffold
3x^2 + x + 5
trail musk
d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x)
d/dx(c*f(x)) = c * d/dx f(x)
This is called linearity. It's the first rule.
scenic scaffold
6x + 1
scenic scaffold
6x + 1
This is the derivative of the 3x^2 + x + 5. How would you proceed from here?
scenic scaffold
d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x)
d/dx(c*f(x)) = c \* d/dx f(x)
This ...
Sorry, I am a little confused still.
scenic scaffold
You get 6x + 1 using the power rule. How do you differentiate this however?
scenic scaffold
So when it says to differentiate, is it just asking us to find the derivative?
hasty viper
yea
scenic scaffold
I see, so what if it is asking if it is differentiable at a certain point?
hasty viper
you check the graph and see if theres a hole in it
or a discontuinity
if there is it isnt unless its a removeable discontuinity
based on the graph tbh
scenic scaffold
And this is the graph of the derivative correct?
hasty viper
no the original function
outer furnace
obtuse pebbleBOT
outer furnace
slate zephyr
!status
obtuse pebbleBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
erm this is more like basic differential eqn but shouldnt the (P^-1)(P)(Dt) be at the top of the exponential
instead of (P^-1)(P) be below
outer furnace
u = exp(P^-1DPt) = P^-1 exp(Dt) P
u = exp(P^-1DPt) = P^-1 exp(Dt) P
how can you bring down the P^-1 and P
slate zephyr
exp is defined via a power series
$\exp(P^-1DP) = \sum_{n=0}^{\infty} \frac{(P^{-1}DP)^n}{n!}$
warm shaleBOT
M8732
outer furnace
can I think of it this way: P^-1*P = 1. e^1 = 1
slate zephyr
can I think of it this way: P^-1*P = 1. e^1 = 1
I don't see how it is relevant.
outer furnace
All the middle P cancel
Ig it somewhat represents this?
slate zephyr
$\exp(P^{-1}DPt) = \sum_{n=0}^{\infty} \frac{(P^{-1}DPt)^n}{n!} = \sum_{n=0}^{\infty} \frac{P^{-1}D^nPt^n}{n!} = P \sum_{n=0}^{\infty} \frac{D^nt^n}{n!}P^{-1} = P^{-1}\exp(Dt)P$
warm shaleBOT
M8732
slate zephyr
oh yeah
outer furnace
then couldnt u remove it entirely
slate zephyr
can I think of it this way: P^-1*P = 1. e^1 = 1
Sorry, if that is what you meant here, then it is relevant.
outer furnace
ah okay i think i somewhat understood it
slate zephyr
I thought e as exponential function 
vapid fox
given the image, determine f(-2)
how do i solve this question?
obtuse pebbleBOT
given the image, determine f(-2)
how do i solve this question?
high lily
simple substitution
vapid fox
do i substitute x with -2?
vapid fox
ok il try that thansk
the answer should be -0.5, when i substiutded the x with -2 it gave me 7.5
scenic echo
not really sure tbh maybe give a domain for x as there are some values that x cannot be?
glossy grove
Does it have something to do with area of rectangle
There is only 1 equation linking the length to 80
and the area of trapzeium equation is derived from it.
scenic echo
how did you find ad in terms of x
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
brisk arrow
Given there's six different item and four identical boxes, then how many way of placements there are?
obtuse pebbleBOT
Given there's six different item and four identical boxes, then how many way of ...
brisk arrow
so basically i just listed all the possiblities out
then count them out indivisually
however, this is not a good idea IMO
is there any better ways?
near helm
你要看谁会看中文😭
near helm
Given there's six different item and four identical boxes, then how many way of ...
我看到都有点头晕😵💫
brisk arrow
Given there's six different item and four identical boxes, then how many way of ...
didn't I just given out the question in English at the very beginning
near helm
didn't I just given out the question in English at the very beginning
“Didnt I just give out the question in English at the very beginning?”
brisk arrow
thank you for correcting me
near helm
👍
Correct my Chinese too lol
I am in america and try my best to type pinyin
But 我也是中国人
我数学不好🥲
near helm
it is a shame to be inadequat at math as a chinese
American standards, I am okay
Chinese standards, 肯定很低😣
美国就是这样，没办法
妈妈叫我上多点数学课，我不喜欢数学🙁
I am better in biology
obtuse pebbleBOT
@brisk arrow Has your question been resolved?
obtuse pebbleBOT
@brisk arrow Has your question been resolved?
agile oriole
Given there's six different item and four identical boxes, then how many way of ...
have you done (2)
brisk arrow
No, but I think I know how to do that one
agile oriole
What if you take your answer to (2) and divide it by the number of times you overcounted
brisk arrow
What if you take your answer to (2) and divide it by the number of times you ove...
well, that's a way to do it. However, it is a little bit complicated to actually apply it into the question
.close
delicate kettle
obtuse pebbleBOT
delicate kettle
LCD is 14
so it becomes 35p - 35q - 2q - p - 2p + 2q
yes?
<@&286206848099549185>
fickle turret
so it becomes 35p - 35q - 2q - p - 2p + 2q
Why didn't you multiply the left numerator by 2?
delicate kettle
Why didn't you multiply the left numerator by 2?
wdym
fickle turret
And don't forget the brackets. Because -(2q-p) = -2q + p
fickle turret
There is a - before the fraction. So it is basically -(2q-p)
delicate kettle
no, -2(p+q) ≠ -2p + q.
does that mean i am correct?
delicate kettle
do you know what the symbol ≠ means?
no
delicate kettle
-2(p+q = -2p+-2p
royal basin
generally, if you see somebody using a symbol you don't know, you SHOULD ask them what it means.
or any other notation.
delicate kettle
oh ok mb
delicate kettle
two typos here.
hm?
oh ok
-2(p+q)
u give -2 to both p and q
so it becomes -2p - 2q
royal basin
but the -2(p+q) was over 7 not over 14.
royal basin
$\frac{35(p-q) - (2q - p) - 4(p+q)}{14}$ is how i would have written the first step.
warm shaleBOT
Ann
delicate kettle
so its just -4p- 4q
royal basin
also don't try to kill 2 birds with 1 stone
usually that ends poorly unless you have superhuman dexterity.
don't try to expand brackets AND combine fractions into one all at once.
delicate kettle
alr lemme slow down
-4(p+q) = -4p + -4q then it becomes -4p - 4q
and since - (2q-p), -2q + p
and, 35(p-q) = 35p - 35q
all together,
35p - 35q -2q + p -4p - 4q
or no?
royal basin
yes correct now
delicate kettle
ok
now
we group
35p - 35q -2q + p -4p - 4q
35p + p -4p
- 35q - 4q -2q
35p+p = 36p -4p = 32p
35q-4q = 31q = 2q = 29q?
32p and 29q?
royal basin
35p+p = 36p -4p = 32p
do not write like this
= is not a "next step" symbol
also the first q term is not 35q, it is -35q
delicate kettle
oh mb
-35q - 4q is -31q- 2q = -29q
so 32p and 29q
how do i know if i should use + or minus
royal basin
well... not stripping -29q of its minus sign would've been a good start.
also don't treat negative numbers as positive numbers' evil twins
you had a bunch of p terms and a bunch of q terms added together in a big sum
oh also -35q - 4q is not -31q.
it's -39q.
and adding -2q to that makes it into -41q.
you sorted out the p terms and got 32p, and you sorted out the q terms and should have gotten -41q,
for a total value of 32p + (-41q)
or 32p - 41q
royal basin
yes?
delicate kettle
oh alr
royal basin
also yes 2n might be the gcd of those first two terms but that does you no good
since there is also that pesky constant 12
which cannot be ignored
delicate kettle
so am i correct that its 2n(^3+8)
royal basin
no
delicate kettle
since there is also that pesky constant 12
then, ill do 2n^3 + 12
royal basin
2n(^3+8) is even nonsensical
delicate kettle
teacher said group fractions
royal basin
teacher said group fractions
there are no fractions here
delicate kettle
2^3/ by 2n
royal basin
n^3 isn't the product of n and a dangling superscript 3.
n^3 means n * n * n
you know this, right?
delicate kettle
yes
royal basin
anyway,
factoring sth out of 2 terms and leaving the third one behind is useless.
for factoring by grouping you need at least four terms to group them two by two.
in some way.
but you can't do this here.
delicate kettle
yes
royal basin
the most you can factor out of all three terms is a 2, leading to 2(n^3 + 8n + 6).
and now this cubic polynomial probably requires some more advanced tech to factorize.
i am not sure if it even does factorize quickly.
it's a bit surprising to me that they'd give you this
it's quite a jump in difficulty from the prev question
royal basin
do you not even do algebra yet
delicate kettle
do you not even do algebra yet
im currently doing algebra right now.
royal basin
and the question with p's and q's was a jump above what they used to give you?
delicate kettle
and the question with p's and q's was a jump above what they used to give you?
yep
royal basin
now i'm curious what they gave you before.
can you share some earlier problems?
also i just checked, n^3 + 6n + 8 is unfactorable. i can't think of an elementary explanation, but it does not have any clean factorization.
delicate kettle
also i just checked, n^3 + 6n + 8 is unfactorable. i can't think of an elementar...
according to ans,
its correct.
delicate kettle
now i'm curious what they gave you before.
here
this is a bit more advanced
then what they usually give
but yea
i also cant really understand this.
royal basin
wait what
?????
you're confusing me
this doesn't sound like it answers my question
royal basin
and you deem this LESS complex than the stuff you opened the channel with?
royal basin
i don't mean length i mean difficulty...
you had some trouble collecting like terms and adding fractions
royal basin
wym "can't get both"
delicate kettle
wym "can't get both"
idk how to answer either
royal basin
"well i don't get either one"
delicate kettle
huh
royal basin
clearer wording.
delicate kettle
oh alr
royal basin
anyway that's 2 topics for you to review.
royal basin
do what?
royal basin
that one's a dead ringer for factorization by grouping.
for example you could group the terms as (4p^3 + 8p^2) + (3p+6)
delicate kettle
yup
royal basin
then factor out as much as you can from each pair of terms
royal basin
no
that p on the very left was pulled out of a hat
and you didn't actually factor out as much as possible from the pair of terms
royal basin
no now that's even wronger
it feels as if you still treat p^2 as meaning p multiplied by a small dangling 2.
AND the first term is also screwed up.
delicate kettle
huh
royal basin
p^2 is NOT the product of p and ^2. do you understand?
royal basin
do you understand?
royal basin
yes or no
royal basin
no, you can factor out more than 4p.
but you can factor out just 4p and see what happens.
if you do it correctly you will see what else can be factored out.
delicate kettle
my teacher says
u can factor out
4p^2
but i dont get it
since ^3 and ^2 is different
royal basin
well then let's write out those exponents for what they are
instead of $4p^3 + 8p^2$ write $4ppp + 8pp$
warm shaleBOT
Ann
royal basin
then is it clearer that you can factor out not just 4p but 4pp?
delicate kettle
yup
delicate kettle
alright
oh
i see
u can just remove the 2
and keep the one
so does that mean, if its actually 4p^2, does that mean that, 4p^2(p^1 + 2)
delicate kettle
oh, im sorry. but, since there is also 3p+6, it becomes 3(p+2)
yes?
shouldn't they both be the same
though
royal basin
yes, 3p+6 factors as 3(p+2).
royal basin
shouldn't they both be the same
who are "they"?
royal basin
why would the two pairs factor out to the exact same thing
they started out different
they will remain different
why would two different things have the same factorization
royal basin
i severely doubt that.
i severely doubt that they'd give you two pairs of the exact same terms for a factorization problem.
anyway, to bring together everything we've done thus far, we now have: $$4p^3 + 8p^2 + 3p + 6 = 4p^2(p+2) + 3(p+2)$$
warm shaleBOT
Ann
royal basin
no it is not the final, we are not done yet.
delicate kettle
no it is not the final, we are not done yet.
oh alr
royal basin
yes, i see your example.
v^2 (5v-2) and 5(5v-2) aren't identical.
nor should you expect them to be.
but they do have a COMMON FACTOR.
i don't know why you could confuse "these things have something in common" for "these things are ONE AND THE SAME"
in our case we also have a common factor, (p+2), that you now need to factor out.
delicate kettle
in _our_ case we also have a common factor, (p+2), that you now need to factor o...
oh ok
that part
i do not know
please guide me
royal basin
you should drill the distributive law into your head.
because that law is all over the place when it comes to factorization
a(b+c) = ab + ac
you should be able to recall this law if you got woken up at 3 in the morning and placed against a wall.
4p^2(p+2) + 3(p+2) = (4p^2 + 3)(p+2)
that's all i can say
delicate kettle
you should drill the distributive law into your head.
alr
royal basin
thats already the final answer?
yes, because i cannot really break this step down any further.
you either know the distributive law or you don't.
delicate kettle
oh alriught, thank you so much
i will learn it right now.
so basically
u just put
the ones inside the ()
inside a box
then
u add the ones outside the box?
so its 4p^2 and 3
then u have to add them?
delicate kettle
yes, because i cannot really break this step down any further.
or somethings u gotta subtract them
?
royal basin
so basically
u just put
the ones inside the ()
inside a box
then
u add the ones outside the box?
i have no idea how to even respond to this.
i guess if it helps you this wording might be okay, but i really cannot make much sense of it at all.
delicate kettle
ok
so since 4p^2(p+2) + 3(p+2) = (4p^2 + 3)(p+2)
they both got
p+2
u put it inside the parenthesis
so (P+2)
delicate kettle
i _guess_ if it helps you this wording might be okay, but i really cannot make m...
then,
4p^2
3
do u always add them
or u sometimes subtract
royal basin
whichever + or - sign was between them before, it stays the same after ...
u put it inside the parenthesis
so (P+2)
not really... the (p+2) is OUTSIDE the parentheses.
obtuse pebbleBOT
@delicate kettle Has your question been resolved?
timid silo
obtuse pebbleBOT
timid silo
Can someone explain me where did the pi/2 happened there ?
Let me translate for you the excersise : it says convert the transient function into to polar coordinates.
thin star
idk man it's all Greek to me
timid silo
Welcome 2 math
xD
Find the output y(t) system with the function transient H(jω) when x(t)
And below it says the transient function into to polar coordinates is given by H(jω)
obtuse pebbleBOT
@timid silo Has your question been resolved?
stuck herald
obtuse pebbleBOT
agile oriole
there are 8C4 ways to make the choice
out of those, you could subtract the ways that choose none of the friends, and then subtract the ways that choose two of the friends
agile oriole
i tried 2/8 x 6/7 x 5/6 x 4/5 = 1/7 but its wrong
have you considered that there has to be exactly one of the friends chosen?
Not more
stuck herald
have you considered that there has to be exactly one of the friends chosen?
ya
i think so i did
stuck herald
ok
royal basin
i am getting an answer exactly 4 times bigger
and i think i now know why
2/8 x 6/7 x 5/6 x 4/5
this assumes that (a) the students in the group are ordered (i.e. it matters who is first, who is second etc.) and (b) the girl that is picked comes first specifically.
stuck herald
> 2/8 x 6/7 x 5/6 x 4/5
this assumes that (a) the students in the group are _ord...
so what you did to get 4 times
royal basin
i did it with combinatorics
$\frac{\binom{2}{1} \cdot \binom{6}{3}}{\binom{8}{4}}$ plus some tricks for cancellation.
stuck herald
i did it with combinatorics
oh so without resctriction in denminator?
warm shaleBOT
Ann
royal basin
it can if you account for the 4 possible positions of the girl on the team
stuck herald
it _can_ if you account for the 4 possible positions of the girl on the team
so times 4 i should be doing?
royal basin
yes
warm trench
<@&286206848099549185>
Can someone help me with question 8 and 9..
Question 8.the book has given only the lowest. temperature. And they have told me to write the difference between the highest and the lowest of these temperature. How do i get the highest temperature
Q9) i want to know if i should do anything with the tide time or just substract the highest high tide and the lowest low tide and do nothing wifh fhe tide time
royal basin
Question 8.the book has given only the lowest. temperature. And they have told me to write the difference between the highest and the lowest of these temperature. How do i get the highest temperature
you just want to take the lowest and highest of the numbers written
-89.2 and -23 resp
Q9) i want to know if i should do anything with the tide time or just substract the highest high tide and the lowest low tide and do nothing wifh fhe tide time
you don't need to worry about the timestamps, only which tides are high and which are low
royal basin
.close
jagged summit
Help with question B please!
obtuse pebbleBOT
Help with question B please!
royal basin
!status
obtuse pebbleBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
- I don't know where to begin
that
im unsure how to correctly utilise the form f(x)=(x-a)Q(x)+r
which is what im pretty sure the solution is kinda
or the working
upper shuttle
notice that
1-x^2 = (1-x)(1+x)
upper shuttle
so
P(x) = Q(x) (1-x)(1+x) + (4-x)
what remainder do we get when we divide that by 1+x?
upper shuttle
well, notice Q(x) (1-x)(1+x) is divisible by (1+x). so there is no remainder from Q(x) (1-x)(1+x). there is only remainder from the (4-x) part.
jagged summit
oh truu
upper shuttle
can you do it from here?
jagged summit
yeah thanks
upper shuttle
no problem.
frigid zephyr
need help figuring out why this would not be 3
obtuse pebbleBOT
need help figuring out why this would not be 3
frigid zephyr
royal basin
why would it be 3
frigid zephyr
Earlier question has a similar problem, i have no idea why this second one is 0
agile oriole
it is x approaching -infty, not y
frigid zephyr
the lim x-> inf
frigid zephyr
ok let me explain my thought process
so i learned that lim x->-inf comes from the left side and vice versa
I'm primarily confused because in my mind "if im coming from the left it "stops" at 3" and same thing for the right side
i'm applying this incorrect but im not sure what to specfically look for
royal basin
you are very incorrect
x -> -∞ means x goes off TOWARDS the left.
off to infinity.
frigid zephyr
x -> -∞ means x goes off TOWARDS the left.
so more like this?
royal basin
yes except don't tie it to the graph
frigid zephyr
yes except don't tie it to the graph
wait so then isnt the answer just this?
royal basin
no
you're confusing x with f(x)
i was a little alarmed at the positioning of your red arrows. i was mistaken to do so.
when you go off to the left ON THE GRAPH, what does y approach
frigid zephyr
i was a little alarmed at the positioning of your red arrows. i was mistaken to ...
no problem haha
frigid zephyr
when you go off to the left ON THE GRAPH, what does y approach
Ohhhh wait so -1
royal basin
-1 yes
frigid zephyr
and then other side would also be -1 right?
royal basin
guess so yes
frigid zephyr
guess so yes
does this sound right to you? [x-> -inf = -1 and x-> +inf = -1] || for positive infinity i'm going off to the right so it would share -1 as the limit so it looks like the aformentioned red line going horizontally across
agile oriole
does this sound right to you? [x-> -inf = -1 and x-> +inf = -1] || for positi...
looks right
frigid zephyr
ok awesome! thank you for your patience and help! rly appreciate it from u both. many thanks much love
.close
crimson zodiac
Absolute value of f inverse x = 1 + f inverse x
obtuse pebbleBOT
Absolute value of f inverse x = 1 + f inverse x
crimson zodiac
How do i do this
agile oriole
Absolute value of f inverse x = 1 + f inverse x
could you write it in symbols, or take a picture of the original
royal basin
$|f^{-1}(x)| = 1 + f^{-1}(x)$
warm shaleBOT
Ann
crimson zodiac
Yes
royal basin
can you solve the equation $|a| = 1+a$
warm shaleBOT
Ann
crimson zodiac
hmm
-0.5
Fits the criteria
I get it
So i just equate f inverse x to -0.5
Thanks
.close
woeful fossil
how would i find loga 18?
obtuse pebbleBOT
how would i find loga 18?
agile oriole
,rotate
warm shaleBOT
agile oriole
how would i find loga 18?
you can factorize 18 into primes
fossil crag
18 = ... * ...
vast willow
$4\log_{a}{3}=\log_{a}{3^4}$
warm shaleBOT
calculus is fun
vast willow
and 3^4=81 not 18 so wrong
vast willow
you can factorize 18 into primes
try this
boreal egret
Try taking log base a 18 as log base a (2x3^2)
woeful fossil
rightt
boreal egret
u can take out the square root of 3 as a power
then u get,
log base a 2 + 2 log base a 3
i think u can continue without any issue after this :)
woeful fossil
yep got it
boreal egret
:)
echo gazelle
integral cos^3xdx
obtuse pebbleBOT
integral cos^3xdx
fossil crag
!status
obtuse pebbleBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
ok so i broke cos^3xdx into
| cos^2xcosx
then made cos^2x = 1-sin^2x so
|(1-sin^(2)x)cosxdx
then idrk how to proceed i tried a couple ways but i know im doing it wrong
frank monolith
sinx = u
vast willow
ok so i broke cos^3xdx into
| cos^2xcosx
then made cos^2x = 1-sin^2x so
|(1-si...
why did you even do this if you didnt think of a substitution
i mean what motivated you to do this in the first place
vast willow
didnt you ask your prof why you should do that
you should know the reason behind doing something you shouldnt just walk with what your prof says without knowing the "why"
echo gazelle
he doesn't really explain his teaching or answer many questions he just reads the notes from the board so im looking back at the notes but im having trouble understanding
vast willow
"why" is the question that enables you to solve similar problems to the one you are solving
vast willow
he doesn't really explain his teaching or answer many questions he just reads th...
did you try asking him
echo gazelle
i didn't
vast willow
ok ask your prof about reasons to understand the idea deeply
you can do that and you can avoid it but its better to ask
echo gazelle
true
all the trig integral stuff is so confusing to me i need to get a tutor bc im struggling to understand like everything he says
ok i understand how to do the problem now
bronze basalt
Hola, I need help with this one
obtuse pebbleBOT
Hola, I need help with this one
bronze basalt
Setting it up that is
I can do the algebra but setting up the equation I’m having difficulty
I need to find PTU and QTP
Angles PTU and QTP*
I did find PTU 5x-32=3x which simplified is 48
48+Y (Y=the supplementary of 48) which is Y=132 132+48=180
132=PTU
Then QTP to find that, I take 48 (which 5x-32=48) and the 89 interior corresponding angles QTM=89 89+48=137 then we take 180-137=43. QTP=43
.close
rocky pebble
Need help with this modular arithmethic problem: Let n and m be positive integers and let d = gcd(n,m). Prove that the equation pair x = a mod(n), x = b mod(m) has integer solutions x if and only if a = b mod(d), and prove that the solutions are unique mod(lcm(n,m)).
obtuse pebbleBOT
Need help with this modular arithmethic problem: Let n and m be positive integer...
rocky pebble
I have proven the first part but I have trouble proving that the solutions are unique mod(lcm(n,m))
My intuition says that I should use the chinese remainder theorem somehow
obtuse pebbleBOT
@rocky pebble Has your question been resolved?
rocky pebble
<@&286206848099549185>
obtuse pebbleBOT
@rocky pebble Has your question been resolved?
obtuse pebbleBOT
@rocky pebble Has your question been resolved?
obtuse pebbleBOT
@rocky pebble Has your question been resolved?
echo dirge
I need help, but i am unsure what my specific problem and i think me trying to explain it will make it confusing.
obtuse pebbleBOT
I need help, but i am unsure what my specific problem and i think me trying to e...
echo dirge
Im trying to send an image of a screenshot, but its going a but slow
I need help with how to calculate the 4.5x
echo dirge
sorry. Im fragmenting my messages to much
b The videogames has a calculation for points. Its adds two scores with eachother, then uses the number in the too right.
It combines my score(16700) with the enemy score(2728). then multiplies it by the 4.5 in the top right corner
Im asking how can i multiply the 4.5 by the i would get from combining the 16700 and the 2728?
my english isnt the greatest so this was the best way i could kinda explain it
orchid wind
Im asking how can i multiply the 4.5 by the i would get from combining the 16700...
Bc I don’t understand this
sorry
echo dirge
my primary language is english, but im not terribly good at english. its a bit complicated
echo dirge
the way the total score is calculated is both fighter scores are first added together
the two scores being added are 16700 and 2728
and then the number is multiplied by 4.5x
echo dirge
How do i multiple that number by 4.5
echo dirge
Im looking for any way of solving the problem
echo dirge
i tried that. The total number the game gave was around 53000
I used a calculator at first
and got 87000
but the game says the total number was around 53000
so i am assuming there was a different way of multiplying the number
orchid wind
Then your score is around 11777
orchid wind
If the game mutiples your score by 4.5
And your end score was 53000
Then you can 53000 divided by 4.5
To find your original score
echo dirge
then i think theres more to the equation if thats the case
if 11777 would have to be the original score.
i think i have all i need.
Unless you think i may be incorrect
Im assuming there is division somewhere in the calculation that i wasnt told about
im thinking since 11777 would have to be the total score before the 4.5 is applied, that there must be division somewhere in the equation since 11777 is lower than the scores being added together
My math isnt really the best either. so i appologize if this seems like a basic topic. I do thank you for helping me with the problem.
.close
timid silo
answer is the same but the 3rd term is 2x^3. cant find my mistake
tardy epoch
closed your other channel
bold cairn
Rewrite the a^2 - 2ab + b^2 part again
the work on the right side is correct
but I think you added the equations incorrectly
timid silo
it’s not
timid silo
the work on the right side is correct
it doesnt add up right
bold cairn
You forgot to subtract in some places
bold cairn
it doesnt add up right
It does add up correctly on the right side. You distributed correctly
bold cairn
it doesnt add up right
Where does it not?
I'm confused why the right side of the page is not correct
versed eagle
How do I do this properly? I’ve been flying by the seat of my pants
obtuse pebbleBOT
How do I do this properly? I’ve been flying by the seat of my pants
versed eagle
The x intercept is 1 btw
(x-1)
For reference the answer should look something like this
nocturne minnow
,rotate
warm shaleBOT
versed eagle
How do I do this properly? I’ve been flying by the seat of my pants
This the problem btw