ah okay thanks
#help-49
last slate
last slate
we know this because theres nonzero vertical acceleration
i guess that makes sense ill write it down
.close
midnight plankBOT
last slate
just try to model the throw in ur head
yeah I've just done that now. I think i get it. If in theory i could throw something completely horizontally, it starts off wiht no initial vertical velocity because its going straight but then moves down
so just before it hits the ground it has to have a velocity becasue gravity puls it down
Thas correct right?
last slate
wdym by starts off with no initial velocity?
no i mean no initial vertical velocity because its going in a literal straight line
dull yoke
the moment the ball leaves ur fingers, it has initial vertical velocity when u throw it straight up
last slate
the moment the ball leaves ur fingers, it has initial vertical velocity when u t...
But this isnt being thrown up?
Its being thrown sideways
dull yoke
im confused, are we referring to an example where the ball is thrown straight up? or perfectly horizontally?
dull yoke
yeah I've just done that now. I think i get it. If in theory i could throw somet...
yea i was just confused cuz u said "completely vertically" here
last slate
yea i was just confused cuz u said "completely vertically" here
oh shit im an idiot
i meant horizontally
ive changed it now it should make sense
dull yoke
hmm lemme think about this actually
last slate
im using a pen in my hand to think about the example as opposed to a pebble or a ball
and im assuming i am able to throw it completely horizontally straight
dull yoke
cuz as soon as u release the ball, its under the effects of earths gravity, which is acceleration in the vertical component
hmmm
last slate
But the moment i release it, its completely horizontal no? to start off
Im going off the fact that u = 0 in the answer given by my teacher
dull yoke
yea i think so. the moment u release it, its completely horizontal. but i think its just for that moment
that infinitesimal moment
last slate
yeah i think so too. I dont know if thats how it actually is but if there is a small vertical initial velocity, I dont think I would be asked to find that at this stage of Maths. Im only doing Alevel maths after all.
dull yoke
yea, as soon as it leaves ur fingers it may be horizontal infinitesimally, but after that, due to earths gravity, it starts to drop
dull yoke
yeah i think so too. I dont know if thats how it actually is but if there is a s...
thats british right
last slate
Yeah ill just take that assumption foward when doing these types of questions
last slate
thats british right
Yeah
dull yoke
is that high school?
last slate
thats the second year after highschool I think. We have secondary school which starts from age 11 to age 16. Then sixth form where we do Alevels from age 16-18
umbral rampart
What about the second one?
dull yoke
ah ok gotcha
umbral rampart
mb didnt know more ppl here
hearty rune
i sensed some british in here
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static lion
answer = pi/2
how to get this?
midnight plankBOT
answer = pi/2
how to get this?
final monolith
it's the same as x->infinity of arctan(x)
which is like "what's the angle of a triangle with a really tall opposite side"
static lion
arctan aprroach pi/2 but 1/x goes to infinity
static lion
which is like "what's the angle of a triangle with a really tall opposite side"
i dont get what u meant
final monolith
like this limit is similar to arctan(10000)
static lion
but isnt that with a spevific number instead of infinity?
subtle blaze
Let u = 1/x
as x -> 0⁺ , u -> + inf
$\lim_{x\to0^+} \tan^{-1}(1/x) = \lim_{u\to+\infty}\tan^{-1}(u)$
grand pondBOT
Frosst
static lion
why we need to sub u?
grand pondBOT
twilit field
the temperature of a body at t=0 is 160F and it decreases as $\frac{dT}{dt}=-k\left(T-80\right)$ , where K is a positive constant, if T(15)=120F , find T(45)
midnight plankBOT
the temperature of a body at t=0 is 160F and it decreases as $\frac{dT}{dt}=-k\l...
grand pondBOT
ƒ(Why am. I here)=I don't know
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
subbing in the original conditions gives me $C=-ln(80)$
grand pondBOT
ƒ(Why am. I here)=I don't know
main current
One of those beautiful cases in ODE where subbing in too early can screw you up
You'll see this commonly done as:
ln(T - 80) = -kt + C
T - 80 = Ae^(-kt)
For some constant A
twilit field
thought of that, but all the answers are integers, and I can't use a calculator
thanks a lot though
so that gives A=80
substituting the second condition
$40=80e^{-k(15)}$
grand pondBOT
ƒ(Why am. I here)=I don't know
twilit field
now how do I find k without a calculator
hearty rune
logs
twilit field
no log table either
hearty rune
you want the decimal form?
twilit field
I'm not sure
the answers are integers
so idts
maybe some info hasn't been printed in the question
hearty rune
ln(1/2)=-15k
15k=ln(2)
you could probably approximate with taylor series ig
idk what else you could do
umbral rampart
Can't understand M & L part
I keep getting same amount between them
midnight plankBOT
Can't understand M & L part
I keep getting same amount between them
sly patrol
its bascially saying that there are more A members than L
there are 20% more A members than L
do u understand?
M-40%
B-20%
A-30%
L-10%
im pretty sure
midnight plankBOT
@umbral rampart Has your question been resolved?
midnight plankBOT
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tawdry river
How would i solve this?
midnight plankBOT
How would i solve this?
novel herald
How would i solve this?
at a cursory glance, you could differentiate both sides and simplify
you could also use product rule to expand both sides and simplify
er integrate by parts*
uh
at a closer look
the RHS is equivalent to $\int_0 ^ x -(t^2 + 1)f(t) + 1dt$
grand pondBOT
M8 of 48
novel herald
set the two integrands equal to each other and solve for f(t)
tawdry river
wheres the +1 come from?
novel herald
differentiate x and put it back into the integral
tawdry river
isnt x outside the integral?
novel herald
$\int_0 ^ x -(t^2 + 1)f(t) + 1dt = \int -(t^2 + 1)f(t)dt + \int 1dt = \int -(t^2 + 1)f(t)dt + x$
grand pondBOT
M8 of 48
midnight plankBOT
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simple osprey
is there a way to do this if it wasnt a multiple choice problem
midnight plankBOT
is there a way to do this if it wasnt a multiple choice problem
simple osprey
or do i have to go thru each option and differentiate it
junior flower
there are tricks to do it but it's pretty intuitive. write y'' = 4y. so you want a function that, when differentiated twice, is the original function times 4
simple osprey
so e^2x
and if it was y''=-4y sin2x or cos2x would work
ok i think i can do it without the mcq options
.close
junior flower
in general, solutions to differential equations are hard to come up with
midnight plankBOT
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unkempt quail
Hello, I'm trying to convert a double integral from rectangular to polar coordinates. What would the bounds of the integral be if they are -inf to inf
midnight plankBOT
Hello, I'm trying to convert a double integral from rectangular to polar coordin...
still rose
there are two ways of doing this
either you use the equations, which is hellish
or you think of it logically
-inf to inf covers the whole xy plane yes?
in polar, that means you cover all the degrees (so 0 to 2π)
and you cover every point between the origin (r=0) and infinity (r=∞)
unkempt quail
ah so 0 to 2pi? and 0 to inf
that makes sense
thanks! doing stuff with guassian integrals is mind melting
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tiny gate
I’m trying to get to the blue underlined form but I’m so far off?
midnight plankBOT
I’m trying to get to the blue underlined form but I’m so far off?
soft token
For one, $\left(\frac{1}{r} - 1\right) \neq \frac{1 - r}{r^2}$
grand pondBOT
Enemagneto
tiny gate
Ohhh that should be 1-r/r?
soft token
(1-r)/r
soft token
It's not exactly wrong as long as you have raised LHS to the power of r as well.
Although, remember that you need a Q in RHS so perhaps try looking for the term Q in RHS and then replace that with Q. Then, it should become simpler.
tiny gate
Hmm ok
In the final form K is kind of isolated and im really not seeing how that can happen
soft token
If you can't see far enough ahead, then just try doing it step by step and then think further.
tiny gate
Yeah no the more I do the worse this gets
tiny gate
Doesn’t look useful
Especially if my LHS is to the power of smth that just seems unnecessary
tiny gate
I’m trying to get to the blue underlined form but I’m so far off?
But don’t see how else u get the 1-r power in the final form
tiny gate
Is this even right or worth doing
Only reason I did that
soft token
Okay, I am having trouble understanding what you have done. Where is the expression in top left coming from?
tiny gate
Is this even right or worth doing
Tried subbing in Q to the top
soft token
Either I can't understand what you did, or you did something wrong. In any case, let me just tell a bit.
After this line
tiny gate
Lmaooo I’m probably wrong
soft token
Since $Q = A[ \delta \cdot k^r + (1-\delta) \cdot L^r]^{\frac{1}{r}}$
grand pondBOT
Enemagneto
tiny gate
It’s not even the question itself it’s just the form my professor likes his answer is always long
soft token
$\frac{Q}{A} = [ \delta \cdot k^r + (1-\delta) \cdot L^r]^{\frac{1}{r}}$
grand pondBOT
Enemagneto
tiny gate
**Enemagneto**
Yhhh
soft token
$[ \delta \cdot k^r + (1-\delta) \cdot L^r]^{\frac{1-r}{r}} = \left([ \delta \cdot k^r + (1-\delta) \cdot L^r]^{\frac{1}{r}}\right)^{1-r}$
grand pondBOT
Enemagneto
median quest
help me find l5
soft token
$[ \delta \cdot k^r + (1-\delta) \cdot L^r]^{\frac{1-r}{r}} = \left(\frac{Q}{A}\right)^{1-r}$
grand pondBOT
Enemagneto
soft token
Now, try replacing that in original and try.
tiny gate
The tragedy is that the final form wants K instead of A wtf
I’ll make a note anyway and see what happens lmaoo
soft token
So, you now have $\left[\delta A k^{r-1} \left(\frac{Q}{A}\right)^{1-r}\right]$
grand pondBOT
Enemagneto
soft token
Which is same as $\left[\frac{\delta A}{k^{1-r}} \cdot \left(\frac{Q}{A}\right)^{1-r}\right]$
grand pondBOT
Enemagneto
soft token
$\left[\delta A \cdot \left(\frac{Q}{A \cdot k}\right)^{1-r}\right]$
grand pondBOT
Enemagneto
tiny gate
**Enemagneto**
Would’ve not got that
tiny gate
Just take A out of bracket and you will have what you need.
Yeah okay
soft token
Would’ve not got that
Well, do you get it now or not? It's important that you understand.
So that thing there is just this.
$$a^b = \frac{1}{a^{-b}}$$
grand pondBOT
Enemagneto
tiny gate
No I get that tbf
It’s just it’s not something I’d see to do
Like I don’t have that intuition
soft token
Well, practice helps. You'll slowly get better.
Also, look at the goal and try to achieve that. For example, you needed to get k inside the bracket and so think about that and so on.
tiny gate
Yeah I’m hoping there’s more questions like this in the textbook because there aren’t many qs per exercise
Idk why our professor made it like that because don’t have other resources I know of
Thank you a lot though cause I don’t think I’d have ever seen that and even I did itd be too late
nova yoke
because a series can be convergent without being absolutely convergent
and this is such a series
well, if you find that it's absolutely convergent then it's also convergent
so it's only when you find that it's not absolutely convergent that you have to do more work
midnight plankBOT
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Available help channel!
tawny belfry
midnight plankBOT
tawny belfry
yeah get the same denominator but i need help with 5(x-3) x 6
grand pondBOT
faiyrose
tawny belfry
so it would be 30x - 3 over 12?
grand pondBOT
faiyrose
midnight plankBOT
@tawny belfry Has your question been resolved?
coral bobcat
Hi how would i do 5?
midnight plankBOT
Hi how would i do 5?
coral bobcat
So 2=ab^1
last slate
yes
coral bobcat
And 8=ab^2
last slate
i mean no ab isn't the same as ab^2 so they don't cancel
coral bobcat
So would you square the first equation?
last slate
try and solve for b here
but anyways this
you are trying to do some elimination thing but honestly substitution is often much more feasible in those situations
,align
ab &= 2 \
b &= {???}
grand pondBOT
coral bobcat
Wait so b=2/a
coral bobcat
And then it would be 8=a(2/a)
last slate
uh no you messed up ur algebra
last slate
ye
coral bobcat
a=1/2
coral bobcat
B=4?
coral bobcat
:DD
last slate
by the way alternatively
you could have divided the equations
<@&268886789983436800> this person has been trolling (they deleted a few messages previously)
coral bobcat
you could have divided the equations
What would i divide by?
coral bobcat
Yeah
torpid shadow
<@&268886789983436800> this person has been trolling (they deleted a few message...
no bro i already got banned for 1 month not again pls😭
coral bobcat
8=ab^2/2=ab^1?
last slate
uh one sec
last slate
8=ab^2/2=ab^1?
\begin{tabular}{rl}
& $ab= 2$\
$\divsymbol$ &$ab^2 =8$\
\hline
&$\51b =\514$\
\end{tabular}
something like this
grand pondBOT
last slate
so you can solve for b and get b = 4
midnight plankBOT
last slate
nww
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Available help channel!
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stable dawn
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chrome ibex
Hello guys
chrome ibex
one question
if i have a Poisson Distribution
that describes the amount of calls the some call center recieves in one minute
After receiving a telephone call at the center, what is the probability of needing to wait more than 10 seconds before receiving another call?
latent elk
Can u post the whole problem?
chrome ibex
the problem is written in portuguese but i can translate it wait a sec
The number of calls arriving per minute at a company's telephone exchange is a random variable with a Poisson distribution.
(a) Determine the average number of calls arriving per minute, knowing that the probability of 9 calls arriving in a minute is equal to the probability of 10 calls arriving in a time interval of the same duration.
(b) What is the probability that 16 calls will arrive at the exchange in a 2-minute period?
(c) What is the probability of at least 540 calls arriving in an hour?
(d) After receiving a phone call at the exchange, what is the probability of having to wait more than 10 seconds before receiving another call?
my question is in the d)
lethal path
The number of calls arriving per minute at a company's telephone exchange is a r...
The Poisson process is memoryless
That is to say, P(X > m + n | X > m) = P(X > n) if X follows a Poisson distribution
chrome ibex
!close
lethal path
!close
The bot is down, be patient with it
.close
See .close is the proper way to type it but it doesn't work
chrome ibex
Thank you
but can you help me with d)
i know i have to use an exponencial distribution
lethal path
i know i have to use an exponencial distribution
Yeah, so the exponential distribution has the same lambda as the Poisson distribution
If your lambda is calls per minute, then your t is just 10, 10 minutes
chrome ibex
Yeah, so the exponential distribution has the same lambda as the Poisson distrib...
So yo call T a random variable "Time between calls" right ?
lethal path
So yo call T a random variable "Time between calls" right ?
Yes
chrome ibex
but why does it have the same lambda as the poisson distribution ?
lethal path
but why does it have the same lambda as the poisson distribution ?
Cause of the memoryless property I mentioned earlier
Each event of a Poisson process is independent from all others
chrome ibex
i am not that good at english i dont know what that means let me research sorry
lethal path
So how long you wait for the next call doesn't depend on what has happened before
chrome ibex
but since its a different "time lengh" like minutes and seconds shouldnt it change ?
lethal path
but since its a different "time lengh" like minutes and seconds shouldnt it chan...
Yes, lambda depends on the time length you're using
So 1 call a minute is 60 calls an hour
So that means your t has to be in minutes if your lambda is 1 call a minute
chrome ibex
ohhh ok
one question
can i use the "proporcionality of Poisson"
i dont know how to call it
like 1 m lambda = 10 so in 1 s lamba = 0.6
lethal path
like 1 m lambda = 10 so in 1 s lamba = 0.6
Yes
chrome ibex
and now i do T>10 = 1 - T<= 10
lethal path
Yeah
lethal path
No worries
midnight plankBOT
@chrome ibex Has your question been resolved?
lethal path
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midnight plankBOT
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main sandal
I found that the series converges but I am not sure how to approximate it correct to .001. I think I have to use the alternating series estimation theorem but I do not know the steps to do it.
midnight plankBOT
I found that the series converges but I am not sure how to approximate it correc...
lethal path
I found that the series converges but I am not sure how to approximate it correc...
So the maximum error in an alternating series is just (the absolute value of) the next term
That's what the theorem is
main sandal
not sure I understand. I've been confused about series from the start of learning about them.
lethal path
not sure I understand. I've been confused about series from the start of learnin...
Well, so if you take the absolute value, then you can 'get rid' of the (-1)^(k + 1)
(-1)^(k + 1) can only be either -1 or 1
lethal path
So we can just substitute k + 1 into k
$\frac{(k + 1)^2}{e^{k + 1}} < 0.001$
lethal path
to find out whether it converges
It does converge btw yes
Well you have to show it converges using the ratio test
For example, 1 + 1/2 + 1/3 + ... + 1/n
The sequence goes to 0 as 1/n goes to 0 as n goes to infinity
But then the summation diverges: it goes to infinity
main sandal
right, and then to find the error I just plug in k+1 and solve?
grand pondBOT
south
main sandal
and that applies mostly all the time? Where I would just plug in k+1 or n+1 for k or n?
lethal path
and that applies mostly all the time? Where I would just plug in k+1 or n+1 for ...
Yeah, so the kth term is just when you substitute k = k
The (k + 1)th term is when you substitute k + 1 = k
Well, the kth and (k + 1)th terms of the sequence and not the summation
The kth term of the summation would be the sum of the sequence from k = 1 to k = k
(I should really use another variable, I know)
main sandal
no it's fine I think I get it. Can I solve and double-check my answer with you?
main sandal
and do I set it to less than .001 or .0001. cause for some reason I thought you set it to the next decimal
lethal path
and do I set it to less than .001 or .0001. cause for some reason I thought you ...
It's just 0.001
main sandal
ok
ok I know this is the simple math part of the problem but I am struggling with it
lethal path
ok I know this is the simple math part of the problem but I am struggling with i...
Yeah so you can just use Desmos or your calculator
It's not solvable using algebra
main sandal
so it would be 10.853?
lethal path
so it would be 10.853?
Yeah
So k + 1 should be 11 to have an error smaller than 0.001
main sandal
ok I see. and if i was asked to show my work would I just put desmos on the paper?
lethal path
Yeah
midnight plankBOT
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edgy onyx
how do i go on about solving this:
Evaluate the arclength of y=(x^3)/6)+1/2x as 0≤x≤3
midnight plankBOT
how do i go on about solving this:
Evaluate the arclength of y=(x^3)/6)+1/2x as ...
tribal tartan
do you remember the arc length formula?
edgy onyx
integral from a to b sqrt 1 + [f(x)]^2
tribal tartan
well it’s f’ not just f
edgy onyx
yeah
tribal tartan
so we’d have to take the derivative of this function
edgy onyx
quick question
tribal tartan
yeah?
edgy onyx
i have seen another equation similar to the one i just typed but instead of the f of x part it has dy/dx
tribal tartan
yeah
tribal tartan
yep
tribal tartan
is that f’(x) squared?
edgy onyx
no im doing that
grand pondBOT
y0shi
tribal tartan
the derivative should just be using the power rule which should give us $f’(x)=\frac{x^2}{2}+\frac{1}{2}$
grand pondBOT
y0shi
tribal tartan
that isn’t right
edgy onyx
what
tribal tartan
**y0shi**
derivative should be this
not x^4
it wouldn’t make sense if the degree went up when we differentiated
the degree goes down during differentiation
edgy onyx
umm
tribal tartan
are you mixing differentiation with anti derivatives?
edgy onyx
i used differentiation rule
grand pondBOT
y0shi
edgy onyx
d/dx(x^3)/6+d/dx(1/2)
tribal tartan
so using the power rule
tribal tartan
(x^4)-1/2x^2
up here
edgy onyx
1/6*3x^2-2/2x^2
you simplify this
edgy onyx
cancel out 6 and 3
tribal tartan
howd you get the second term to be that?
edgy onyx
wait
tribal tartan
it’s $\frac{d}{dx}\frac{x}{2}$
grand pondBOT
y0shi
tribal tartan
which would just be 1/2
edgy onyx
you end up with 1/2*x^2-1/2x^2
tribal tartan
and also that what you had up there cannot be higher than a degree of 2
so it can’t possibly give us an x^4 term
regardless of what we do to manipulate it
edgy onyx
now we write all numerators above the least common denominator 2x^2
tribal tartan
you end up with 1/2*x^2-1/2x^2
the second term isn’t 1/2(x^2)
it’s $\frac{d}{dx}\frac{x}{2}$
grand pondBOT
y0shi
tribal tartan
what does this evaluate to?
edgy onyx
im not sure
edgy onyx
where dont you agree with my solution?
tribal tartan
well since the derivative of x/2
is not 2/2x^2
or whatever you had up there
you did the derivative of the first term correctly
but not the second term
edgy onyx
use the differentiation rule
grand pondBOT
y0shi
tribal tartan
because of the power rule
edgy onyx
d/dx(1/f)=-(d/dx)(f)/f^2
tribal tartan
the denominator here is a constant
not a variable
we don’t use the quotient rule
edgy onyx
so you say that the derivative of 1/2x is just 2?
edgy onyx
how
tribal tartan
$\frac{d}{dx}\frac{x^1}{2}=\frac{1x^0}{2}=\frac{1}{2}$
grand pondBOT
y0shi
edgy onyx
isn't x a varaible?
edgy onyx
but you said it is not
tribal tartan
i never said that
edgy onyx
the denominator here is a constant
.
edgy onyx
isnt it 2x
tribal tartan
huh
edgy onyx
well it is 1/2x
tribal tartan
where’d you get $2x$ from $\frac{x}{2}$
grand pondBOT
y0shi
tribal tartan
they are completely different
grand pondBOT
y0shi
edgy onyx
yes
tribal tartan
oh bruh
that’s why we need to put parenthesis
since 1/2x is very vague
can be interpreted as (1/2)x or 1/(2x)
edgy onyx
wait did you think it was 1/2*x
tribal tartan
yes
edgy onyx
my bad
tribal tartan
because most of the time people do that since they are lazy
edgy onyx
idk latex
tribal tartan
well then $\frac{x^2}{2}-\frac{1}{2x^2}$ should be right
grand pondBOT
y0shi
edgy onyx
not finished
tribal tartan
as the derivative
so we need to square it now
or just plug it into the arc length formula and then simplify
tribal tartan
hm?
edgy onyx
where did you get x^2/2
edgy onyx
yes but now you simplify it
tribal tartan
it’s already simplified
edgy onyx
no
edgy onyx
(x^4)-1/2x^2
tribal tartan
i would recommend just plugging this into the arc length formula
edgy onyx
now you square it
edgy onyx
(x^8)-(2x^4)+1/4x^4
then you get this
tribal tartan
at least put the necessary parenthesis
tribal tartan
(x^4)-1/2x^2
this can be viewed as $x^4-\frac{1}{2x^2}$ and will most likely be viewed that way by everyone
edgy onyx
at least put the necessary parenthesis
((x^4)-1)/(2x^2) better?
grand pondBOT
y0shi
tribal tartan
then you get this
same with this one
edgy onyx
((x^8)-(2x^4)+1)/(4x^4)
tribal tartan
okay now that makes sense
so now we just plug that into the arc length formula
and we’ll get $\int_{0}^{3}\sqrt{1+\frac{x^8-2x^4+1}{4x^4}}dx$
grand pondBOT
y0shi
edgy onyx
can we cancel out -2x^4 with 4x^4
tribal tartan
yeah
not really cancel out
but they combine together
as like terms
so we should have $\int_{0}^{3}\sqrt{\frac{x^8+2x^4+1}{4x^4}}dx$
grand pondBOT
y0shi
tribal tartan
good so far?
tribal tartan
well 1 is the same as $\frac{4x^4}{4x^4}$
grand pondBOT
y0shi
tribal tartan
so since they share the same denominator, i can combine the fractions to this:
$\frac{4x^4+x^8-2x^4+1}{4x^4}$
grand pondBOT
y0shi
edgy onyx
yeah i get you are trying to create 1 with (4x^4)/(4x^4) but if we create 1 with that where does our original 4x^4 go
edgy onyx
oh alright
tribal tartan
alright so you understand how we got that integral right?
edgy onyx
yes
tribal tartan
alright so any ideas on how we can simplify this
we have to remove the square root
so we can integrate
edgy onyx
how do i do that
edgy onyx
yea
tribal tartan
where a is x^4
this
edgy onyx
why x^4?
tribal tartan
well since $(a+b)^2=a^2+2ab+b^2$
grand pondBOT
y0shi
tribal tartan
if our first term is x^8
edgy onyx
yes
tribal tartan
a has to be x^4
tribal tartan
alright so let’s convert it to that form
and then you will see that the square root just cancels out with that square
edgy onyx
((x^4)+1)^2
tribal tartan
yeah so $\sqrt{\frac{(x^4+1)^2}{4x^4}}$
grand pondBOT
y0shi
tribal tartan
now we can cancel out the square root in the numerator
and the denominator
to get $\frac{x^4+1}{2x^2}$
grand pondBOT
y0shi
tribal tartan
well just split the fractions really
tribal tartan
yw!
edgy onyx
@tribal tartan
help
when substituting with 0
i got undefined
did we do smth wrong or is it actually supposed to be like that
edgy onyx
yes
tribal tartan
cuz the original function is undefined at 0
so the arc length would just go up to infinity
we didn’t even need to do the problem then if it’s on that interval
cuz we can just tell from the start
that the function approaches infinity when x approaches 0 from the right
tribal tartan
it wouldn’t change anything
tribal tartan
the original function was unbounded at the start regardless
the arc length would be infinity on that interval
heady stump
midnight plankBOT
heady stump
What exactly did they do to get rid of the 1?
I'm guessing it has something to do with the raising all the n's by 1
Does raising everything (n) by 1 remove 1 from the series?
tribal tartan
yeah so notice how the first term of the series is 1
if we just got rid of that and kept the original content of the series the same
we get what we have up there
or we could’ve just changed the starting point to n=1
heady stump
alright makes sense thank you but I gotta keep starting point at n = 0
heady stump
stupid how these textbooks answers expect us to understand everything they do
.close
midnight plankBOT
tribal tartan
haha yeah they skip lots of steps
midnight plankBOT
Available help channel!
bleak pier
midnight plankBOT
zenith stone
,rotate
grand pondBOT
zenith stone
in the third line
when u expand (3x+y)^3
why subtract it with -3xy?
(3x+y)(3x+y)^2 = (3x+y)^3
adding the subtracting term will invalidated the equality
zenith stone
It is (3x)^3 + y^3
oh the LHS is (3x)^3 + y^3?
bleak pier
I'm assuming you're trying to use the cubic expansion (under)?
YES
still nacelle
ok so here a = 3x, b = y
but a^2 + b^2 is (3x)^2 + y^2, not (3x+y)^2
you did (3x+y)^2 instead of (3x)^2 + y^2
still nacelle
yh i was gonna say, it felt like you forgot your bracket in the last line XD
bleak pier
And one more thing
still nacelle
And one more thing
the +-?
still nacelle
oh yh you wrote 49 last time so I didn't even see you wrote 37 this time
keen seal
for the D - dv
E - exponentials
T - trig
A - algebraic
I - inverse
L - logarithms
midnight plankBOT
for the D - dv
E - exponentials
T - trig
A - algebraic
I - inverse
L - logarithm...
runic hamlet
why should it be a logarithm?
keen seal
why should it be a logarithm?
i kinda assumed since its base of the natural logarithm
its log
wait but if its e alone without any exponent
what section does it
belong to
acc thats just a constant right
okay tysm!
.close
midnight plankBOT
midnight plankBOT
Available help channel!
twin ridge
This is a proof I am trying to solve. I am a little stuck at the moment though.
twin ridge
This is where I'm at so far.
hard shard
this seems like you can use law of cosines to get the job done
twin ridge
this seems like you can use law of cosines to get the job done
I'm not sure if I'm allowed to use that tbh. I also haven't taken trig.
midnight plankBOT
@twin ridge Has your question been resolved?
umbral rampart
midnight plankBOT
umbral rampart
This is my work, not sure if I'm doing something wrong
tacit blade
This is my work, not sure if I'm doing something wrong
Wouldn’t bi be a partal circle that ends at the boundaries of the field?
umbral rampart
Wouldn’t bi be a partal circle that ends at the boundaries of the field?
this u mean?
tacit blade
Ye
umbral rampart
alr
tacit blade
there's still no intersection though
Try checking the scaling again ig
umbral rampart
Try checking the scaling again ig
how
midnight plankBOT
@umbral rampart Has your question been resolved?
umbral rampart
<@&286206848099549185>
tacit blade
<@&286206848099549185>
One meter is 100 centimeters, you need a tenth of a meter. Mesure out 10 cm and draw a circle
umbral rampart
One meter is 100 centimeters, you need a tenth of a meter. Mesure out 10 cm and ...
im confused
midnight plankBOT
@umbral rampart Has your question been resolved?
tacit blade
That looks right
umbral rampart
.close
midnight plankBOT
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Available help channel!
sand badger
Hey guys
sand badger
Got a math test tomorrow in algebra 2
latent elk
Got a math test tomorrow in algebra 2
Good luck!
sand badger
Need a bit of help with logarithims and would apprecaite it if someone can walk me through some of the problems
sand badger
Good luck!
Thanks mna
My grade is not too hot but I want to improve it sure you guys understand
latent elk
Sure
latent elk
💀
sand badger
what's so skull emoji
latent elk
Just ask
sand badger
A problem I'm confused on open to anyone is switching a natural log to a exponential form.
For a regular log for example I did log3 29 = x + 2 would be 4^x+2 = 9 which I think is right
but for a natural log
ln(7x) = 13
Really not sure where to start
mossy spindle
Ln is basically log to the base ' e'
sand badger
So changing to exponential form
Beacuse on log3 29 = x + 2
im lowkey forgetting how to even do that
sand badger
here let me send a picture of my paper
This is kind of what I'm working with here
I do need to figure out expanding and condensing next
sand badger
Yeah of course but
mossy spindle
The ans to 4 is wrong btw
mossy spindle
what is it
x = 5
sand badger
Oh yeah idk why I put 4
good catch
Changing to epxonential form seems kind of easy like
If It's log2 42 = 5x + 4 It'd just be 2^5x+4 = 42
just have to move hte numbers around
but for the ln(7x) = 13 getting to exponential form I got no idea
latent elk
e^ln(7x) = e^3
.
mossy spindle
Ln is basically log to the base ' e'
.
sand badger
what are the es
latent elk
There a rule
sand badger
.
ooooooooooooooh
I'm picking up what you're putting down here
So It's pretty much
logE (7x) = 13?
mossy spindle
Yes
sand badger
e^13 = 7x?
latent elk
sand badger
hey blues do you play minecraft
latent elk
yes why
sand badger
yes
but iget off track easy
ooh I see
do you guys make maps for bedrock
I've talked to a few people who run studios
Very cool
sand badger
e^ln(7x) = e^3
Ok so
This is the exponential form
wait
but you still have the ln there
latent elk
Read this
sand badger
e^13 = 7x?
Is this not right though
Chatgpt is telling me It's right
It just needs to be in exponential form does not need to be solved
because it'd be loge (7x) = 13
and then you switch em around so it'd be
e^13 = 7x
latent elk
Is this not right though
Its right, the thing is u dont know how to get it
sand badger
Its right, the thing is u dont know how to get it
But if I have ln(7x) = 13 which it gives me
that's loge (7x) = 13
cause ln is jsut loge?
latent elk
cause ln is jsut loge?
yes
sand badger
yes
Gotcha
So moving to exponential form and backwards is just about moving the numbers around and natural logs is turning it into loge
latent elk
U chat too much, instead, write it down on paper then post it here
It will be more effective
sand badger
Ok gotcha
But moving onto expanding logairthims
I'll send a picture
The middle one
What's the kind of rule on expanding and condensing
Not sure if I am on the right track if It's log3x etc
latent elk
yep
sand badger
yep
Try to make out my shitty hand writing, but when it comes to condensing is it just backwards?
loga^5b^3
latent elk
- => *
-=> /
mossy spindle
@sand badger I would suggest that you watch a lecture on logarithms on YouTube
sand badger
sand badger
<@160138514766364672> I would suggest that you watch a lecture on logarithms on ...
Do you have any suggestions
mossy spindle
Do you have any suggestions
Idk i never have seen them but almost every video would be the same imo
midnight plankBOT
midnight plankBOT
Available help channel!
latent elk
where the log? @sand badger
midnight plankBOT
where the log? <@160138514766364672>
sand badger
where the log? <@160138514766364672>
ig in the front forgot to put lol
latent elk
it should be log(a^5/b^3)
midnight plankBOT
midnight plankBOT
Available help channel!
bleak pier
is this true?
midnight plankBOT
is this true?
obtuse basin
is this true?
For positive b and positive a besides 1, yes
latent elk
For positive b and positive a besides 1, yes
Yep 👍
I forgot the mention abt its range
But thats just basic knowledge
bleak pier
why positive?
stiff terrace
is this true?
yes log one side and it will equal the other
bleak pier
a^x here a should be positive?
stiff terrace
is it not positive?
bleak pier
logN log_a (a)= log_a(N)?
obtuse basin
why positive?
Why positive a or b?
obtuse basin
a^x here a should be positive?
For real numbers a and x, a^x, when defined, it always positive given a > 0
bleak pier
okay
obtuse basin
logN log_a (a)= log_a(N)?
Is the base of the first logarithm also a?
bleak pier
Is the base of the first logarithm also a?
yes
obtuse basin
Then yes
obtuse basin
Because log_a(a) = 1
stiff terrace
N should be in the brackets not a bc this rule
bleak pier
logN=n no0?
obtuse basin
they are mentioning it would be N?
No, they are not
bleak pier
i am stil not understandin how is this true
latent elk
I suggest watching some tutorial videos
latent elk
Bruh 😭
bleak pier
maybe there are many things which are simple for me compare to you
doesn't mean everything is comparable
latent elk
lemme think of an example
midnight plankBOT
latent elk
this term just simple return the value such that a to the power of that value = N
Right?
So if u take that value, then power a to itself, then it should return N
Simply because the log purpose is to do that
Like x + 5 = 0
then x = -5
then if u plug -5 + 5 then u get 0
Like log_3(9) = 2
Then 3 ^ 2 = 9
Then its the same as
3^(log_3(9)) = 9
See?
let 3 = a, 9 = N then:
a^(log_a(N)) = 9
Cause thats it job
@bleak pier
latent elk
.close
latent elk
The = will not happen
latent elk
Do what?
bleak pier
by multiply by negative sign
last slate
midnight plankBOT
rain wasp
whats your doubt?