Alright thanks
#help-17
dull geode
wispy mirage
vocal sleetBOT
wispy mirage
I don‘t get it how do we know what x should be
vast shale
^
let's start with a) ax - 2a + b = 5 - 3x
i suck at explaining but i did it in paint and u get for a) that a is -3 and b is -1
just match the coefficients and constant terms on both sides
and ummm its all the same
b) ax−a+bx+3b=3x+1
(im just removing the parenthesis btw I say it directly)
so
same as before
coefficient of x is that a+b = 3?
constant terms is -a +3b = 1
vocal sleetBOT
i suck at explaining but i did it in paint and u get for a) that a is -3 and b i...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
vocal sleetBOT
@wispy mirage Has your question been resolved?
tired matrix
a drink has 1L of pepsi. if the pepsi drains at a rate of .5 mL/s, and the ice in the drink fills it with water at .1 mL/s for every centimeter of surface area (1.5 cm), when the drink has 4 identical ice cubes, what is the function of the ratio of pepsi to water?
vocal sleetBOT
a drink has 1L of pepsi. if the pepsi drains at a rate of .5 mL/s, and the ice i...
wispy mirage
r u here buddy <@864784192768049162>
Yes sry im in bio class rn
wispy mirage
uh i assume ur solving for a and b
Yes but we should we let x be
elder scaffold
well u should try to cancel out one of the variables
vocal sleetBOT
Yes but we should we let x be
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
@tired matrix Has your question been resolved?
tired matrix
question still stands
a drink has 1L of pepsi. if the pepsi drains at a rate of .5 mL/s, and the ice in the drink fills it with water at .1 mL/s for every centimeter of surface area (1.5 cm), when the drink has 4 identical ice cubes, what is the function of the ratio of pepsi to water?
vocal sleetBOT
@tired matrix Has your question been resolved?
vocal sleetBOT
@tired matrix Has your question been resolved?
vocal sleetBOT
quick tree
arctic mantle
quick tree
.close
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static egret
how do i prove this?
vocal sleetBOT
@static egret Has your question been resolved?
vocal sleetBOT
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
@static egret Has your question been resolved?
digital skiff
how do i prove this?
What have you tried/thought about doing?
static egret
What have you tried/thought about doing?
I tried using the fact that M • adj(M) =0 but don’t know how to proceed from there
near zephyr
hi
fiery jetty
hi guys
fiery jetty
i have an exercise, find W which belongs to complex number, and w squared is = z. given : z = 1+i
why cant we say that w = square root of 1+i
is it because we cant square root imaginary number
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mild trench
ok so
mild trench
how can i write a sentence that hits the point of the markschemes final box
pls anyone
ive been asking for 30 mins
mild trench
that's a lot of words with vague instructions
gimme a sec
mild trench
what even is the original question
flat whale
all you need after the third box is
$(x+13/4)^2 + 87 / 16 > 0$ and $(x+13/4)^2 + 87 / 16 = x^2 + (6 + 1/2)x + (18 - 2)$
twin meteorBOT
riemann
flat whale
implies $x^2 + (6 + 1/2)x + (18 - 2) > 0$
twin meteorBOT
riemann
flat whale
Then re-arrange to get the starting inequality
mild trench
all you need after the third box is
omg im so sorry i wrote my question wrong
mild trench
i meant box 4
flat whale
you really don't need that much in the 4th box
mild trench
what will be my concluding statment that hits the ms point
flat whale
you can completely ignore it
mild trench
you can completely ignore it
i kinda need it
flat whale
and replace it with the 3 inequality/equations i wrote
mild trench
i need like a worded statement
flat whale
vocal sleetBOT
@mild trench Has your question been resolved?
mild trench
anyone?
golden iron
looks ok to me
sharp rain
Go on....
sharp rain
Finding the differential
Putting it equal to zero
Getting x
Putting it back in original equation
To get its minimum value
mild trench
ty
vocal sleetBOT
@mild trench Has your question been resolved?
onyx sage
I have a question about fixed points
vocal sleetBOT
I have a question about fixed points
onyx sage
suppose we have a continuous function from [a,b] to [a,b]
we know by IVT that there is at least one fixed point p in (a, b) such that f(p)=p
suppose now that the function is differentiable
do we know that the fixed point is unique if f'(x) is not equal to 1
or do we need |f'(x)| < 1
hushed pewter
do we know that the fixed point is unique if f'(x) is not equal to 1
Do you mean $f'(x)\ne 1$ for all $x\in(a, b)$?
twin meteorBOT
SWR
hushed pewter
or do we need |f'(x)| < 1
btw consider what MVT would tell you
onyx sage
Do you mean $f'(x)\ne 1$ for all $x\in(a, b)$?
yes
hushed pewter
suppose we have a continuous function from [a,b] to [a,b]
And you mean f(a)=a and f(b)=b? Or is [a,b] simply the codomain? Or is [a,b] actually the image of f?
onyx sage
And you mean f(a)=a and f(b)=b? Or is [a,b] simply the codomain? Or is [a,b] act...
no of course not f(a) = a f(b)=b, the domains and codomains
and yes by mean value theorem and the second condition we can prove that
im interested to know if
you can prove with the first hypothesis
only assuming $f'(x)\ne 1$ for all $x\in(a, b)$
twin meteorBOT
Erijoni
onyx sage
<@&286206848099549185>
hushed pewter
no of course not f(a) = a f(b)=b, the domains and codomains
what?
worn stone
no of course not f(a) = a f(b)=b, the domains and codomains
I don't think you mean this
Unless the function is just y = x
worn stone
**Erijoni**
But then this condition fails
hushed pewter
@onyx sage are there any other conditions on $f$ other than that it is continuous, its domain is $[a, b]$ and its codomain is $[a, b]$?
twin meteorBOT
SWR
hushed pewter
Unless the function is just y = x
y=x^r on domain [0, 1] has image [0, 1] for all r>0
worn stone
y=x^r on domain [0, 1] has image [0, 1] for all r>0
No he said f(x) = x for all of the domain
That would fail
hushed pewter
No he said f(x) = x for all of the domain
where did he say that?
worn stone
no of course not f(a) = a f(b)=b, the domains and codomains
That's what this sounded like to me
Oh
I apologise I need to put my glasses on
worn stone
do we know that the fixed point is unique if f'(x) is not equal to 1
If f'(x) never equals zero on the co domain then yes f(x) is strictly unique
For a bounded continuous function
onyx sage
<@485780127364677632> are there any other conditions on $f$ other than that it i...
differentiable on (a,b)
worn stone
1 wouldn't tell us anything
onyx sage
@worn stone
worn stone
Yes
onyx sage
do you understand the problem now
worn stone
It's not that I didn't understand it I just literally didn't have my glasses on lol
But yes I understand it
hushed pewter
differentiable on (a,b)
right
hushed pewter
Okay, and f(a) could be any value in [a, b], correct? it is not required that f(a)=a or f(a)=b? (same question with f(b)
hushed pewter
It's not that I didn't understand it I just literally didn't have my glasses on ...
@onyx sage ping me if needed
onyx sage
Okay, and f(a) could be any value in [a, b], correct? it is not required that f(...
no no its not
worn stone
lol
f'(x) never equaling 1 doesn't tell us anything because say f'(x) (-1,1) then it could turn and the point could be non unique
Again just a could
Where as if f'(x) >0 for [a,b] or f'(x) < 0 for [a,b] then it could never turn on itself then the IVT point is unique
@onyx sage
onyx sage
just to clarify
i prove by ivt that if the codomain is also [a,b]
and i proved by mvt that if |f'(x)|<1 for all x in (a,b) then the fixed point is unique
worn stone
and i proved by mvt that if |f'(x)|<1 for all x in (a,b) then the fixed point is...
Can I see this proof?
onyx sage
yes
worn stone
f'(x) never equaling 1 doesn't tell us anything because say f'(x) (-1,1) then it...
Nvm this point is completely irrelevant
Ignore that
onyx sage
worn stone
It is true then I apologise
For some reason my brain had convinced itself we were talking about injectiveity
worn stone
Yes this makes sense
Basically all you need to do is sub in g'(\epsilon) into the line |g'(\epsilon)|*|p-q|
And you can see it is true
hushed pewter
do we know that the fixed point is unique if f'(x) is not equal to 1
@onyx sage
oh you said not
onyx sage
Basically all you need to do is sub in g'(\epsilon) into the line |g'(\epsilon)|...
exactlyy
thank you guys i think i got it
.close
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vocal sleetBOT
hushed pewter
@onyx sage this was the counterexample I meant
f'=1 at x=1/2 yet the fixed point is not unique.
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restive arch
How do I do this?
vocal sleetBOT
How do I do this?
silent basin
what do you know?
restive arch
None
silent basin
so variables are like numbers, but every variable has the same value once chosen
so we want to combine them in order to make the equation simpler
make sense?
restive arch
Yea
silent basin
so something like -2b and 6b although we don't know the value of b, they both share b as a "like term"
restive arch
How do I know that 3cb and 6b aren't like terms then? Or are they
silent basin
How do I know that 3cb and 6b aren't like terms then? Or are they
we don't know if cb=b so we can't combine them
assuming we add them, we could do something like 3cb+6b=3b(2c+1) but 2c is not a constant
restive arch
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wise pelican
vocal sleetBOT
wise pelican
help
i just need someone to explain it to me
i got the answer
just
im so confused
on how i get there
oak magnet
You should probably have that the range is [-4,5) ?
Well what the x value of the minimum of that function ?
In other words, the vertex
-b/2a
Ok
So 0/2
x = 0
0 is in the domain, wunderbar
f(0) = -4
So you have the least
And now, you get the upperbpund by f(sup[domain])
wise pelican
im confused right
wise pelican
it says the derivitive or something
oak magnet
Oh
oak magnet
Wait can you ss the sol and show me ?
oak magnet
Yeah so actually when you have a quadratic, there is only one critical point
wise pelican
whats a critical point
oak magnet
Which is the point (-b/2a , f(-b/2a))
oak magnet
whats a critical point
Somewhere where the derivative equal 0
But we have easier result for quadratic
Than using derivative
wise pelican
whats a derivative 😭
oak magnet
Thats why im saying to not use this
oak magnet
Which is the point (-b/2a , f(-b/2a))
And use this instead
wise pelican
i need to understand why though
i was watching this video
4nb
4b
he explains it
oak magnet
Im not enough wake up to make a proper demonstration but you can find one online
wise pelican
xD
oak magnet
Anyway
oak magnet
Yes
wise pelican
howd we go from that
oak magnet
From whar
wise pelican
Which is the point (-b/2a , f(-b/2a))
.
wraith python
What math class are you taking?
vocal sleetBOT
@wise pelican Has your question been resolved?
silent basin
not the exact same problem but the same flavor
vocal sleetBOT
not the exact same problem but the same flavor
silent basin
@tidal dock
silent basin
show that f_n(x) converges to 0 pointwise but not uniformly
tidal dock
okay i am not very familiar with uniform convergence but i see how this is true
indeed the intervals [0, 1/n] and [1/n, 2/n] both get arbitrarily small (smaller than epsilon) as n grows
silent basin
if you pick a single point, it's 0, but over all points, it's not continious
silent basin
indeed the intervals [0, 1/n] and [1/n, 2/n] both get arbitrarily small (smaller...
but is that a good enough argument to show continuity
tidal dock
okay i am not very familiar with uniform convergence but i see how this is true
oh shit
i meant uniform convergence
not continuity
actually $f_n$ is nonzero on $(0, 2/n)$, so for every $\epsilon > 0$ we can find $N$ s.t. $\forall n \in \mathbb{N}$: $n > N \implies f_n(\epsilon) = 0$
twin meteorBOT
artemetra
tidal dock
specifically set $\epsilon = \frac{2}{n} \implies N = \frac{2}{\epsilon}$
twin meteorBOT
artemetra
tidal dock
so $\lim_{n\to\infty} f_n(\epsilon) = 0$ for all $\epsilon > 0$ and $\lim_{n\to\infty} f_n(0) = 0$
twin meteorBOT
artemetra
tidal dock
but is that a good enough argument to show continuity
i think this is enough to claim pointwise convergence
playing with the graph to see why there cannot be uniform convergence
@silent basin i hope this was at least somewhat helpful lol
dense eagle
not the exact same problem but the same flavor
yeah i mean if you think about what the function actually is
it's basically just a spike of 1 at 1/n
we know it converges pointwise to 0 so we just have to show it doesn't uniformly converge to 0
what's the max value of our f_n?
silent basin
yeah i mean if you think about what the function actually is
good sorry I was doing somethign
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muted arrow
vocal sleetBOT
muted arrow
I’m stuck on how to find my Mew and fit in the inequality
vocal sleetBOT
@muted arrow Has your question been resolved?
muted arrow
<@&286206848099549185>
vocal sleetBOT
@muted arrow Has your question been resolved?
knotty kestrel
how would i do this with a graph of f(x)
vocal sleetBOT
how would i do this with a graph of f(x)
knotty kestrel
i cant plug in to solve bcuz its not F(x)
pale perch
so like for a) i could do 2 * 2 * 1/2?
not quite right
knotty kestrel
the height would be 3?
pale perch
its the area bounded by the function and the x axis
youre bounding it with y=1
so it should be a trapezium area formula
or a rectangle + triangle if you want
knotty kestrel
so i use the formula for area of a trapezoid?
pale perch
itll work
pale perch
remember integrals do signed area
pale perch
signed area means the parts under the x axis will be negative
eg from 5 to 8 would be negative area
vocal sleetBOT
jagged cargo
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main totem
not sure how to approach this problem; my first step is to parameterize C, but sometimes I see my professors parameterize functions in terms of x=cost and y=sint but also x=t and y=2t^2 for example. Im not sure which to do, and in general I have no intuition for when to do either method
vocal sleetBOT
not sure how to approach this problem; my first step is to parameterize C, but s...
flint idol
well the first parametrization you mentioned describes a unit circle
the second one you mentioned is the curve C
do you understand where that comes from?
main totem
do you understand where that comes from?
well yes?
flint idol
alright do you know how to go from there?
parametrizing the curve is usually the hardest part for these problems
main totem
so since the curve isnt some circle or anything
I think the parameterization of it is [t,2t^2, 0] ?
flint idol
well this is a 2d vector line integral so you can drop the third component
main totem
yeah
flint idol
just consider the x and y
main totem
so next, I should input points A & B into my parameterization right?
To get the bounds of my integral
flint idol
if you find that better then sure
main totem
well tbh i dont even need to do that breh
flint idol
usually i like to rewrite it so that i have everything in terms of t
main totem
erm
flint idol
usually for these integrals we wanna rewrite it
$\int_{C}\vec{F}\cdot d\vec{r}=\int_{t_1}^{t_2}\vec{F}(r(t))\cdot r'(t)dt$
twin meteorBOT
y0shi
main totem
well for r(t), my x component is t, my y component is 2t^2
main totem
so I imagine I have to input r(t) into F
flint idol
mhm yep
main totem
giving me (2t^2)^2i + (t)^2j
flint idol
yep
main totem
dotted with i+4tj
flint idol
mhm
main totem
i result with an integral bounded from 0-1
the function I will be integrating is 4t^4+4t^3
main totem
i get 9/5
flint idol
,w int_{0}^{1} 4t^4+4t^3 dt
twin meteorBOT
flint idol
yeah thats good
flint idol
ywyw
flint idol
mhm yep
if it has a radius of 1
essentially youre varying the angle is what youre doing there
main totem
essentially youre varying the angle is what youre doing there
I have another question
there is a change only occuring in the j axis
so does that mean my bounds are 0 to 2pi
flint idol
mhm yeah
well its more so they gave you a parametrization
and if you plug in 0 and 2 pi
it gives you those points
main totem
yeah but for the previous question the bounds were in the i axis instead
flint idol
well it was just the fact that our parametrization before and here involved a t
in the x direction before
and the y direction now
so that it looks like its because of that
i couldve chose that to be 2t instead and change the parametrization up
but doesnt necessarily mean that the t ranges from 0 to 2pi
its just something to keep in mind in case the parametrizations arent that nice
main totem
so whats a general way for determing bounds in line integrals
flint idol
its just to see where we want our endpoints
and find what t value we would need to range from to get those points
main totem
so lets say the line we were trying to integrate over was in 3 dimensions
so like A = (1,2,3) & B = (5,6,7)
How would I figure out the bounds here?
flint idol
it would depend on our parametrization
just consider a simple line from (0, 0, 0) to (1, 0, 0)
i can do a parametrization of x=t, y=0, z=0
thatll give us the bounds from 0 to 1
main totem
I mean I can understand it in terms of 1 and 2 dimensions
flint idol
but i can also do a parametrization of x=2t, y=0, z=0
the bounds would be different
ranging from 0 to 1/2
main totem
well lets use the question here for reference
lets say my 2 points were A = (1, 2pi, 3) and B = (2, 3pi, 4)
what would my bounds here be
flint idol
okay so like that would depend on your parametrization
i dont believe the curve they give you there covers those points
you need a specific parametrization that covers those points
and you need to find the specific "t" values you have to range through to go between those points
thats essentially what those bounds would be
vocal sleetBOT
@main totem Has your question been resolved?
misty citrus
main totem
heres my working out for this question
i integrated what I got for my dot product via wolfram
flint idol
yep looks good
main totem
but its wrong apparently
flint idol
yw!
main totem
that tripped me up really badly lol
flint idol
just do more practice
itll come more naturally eventually
usually problems want the exact value unless they said to round it so
vocal sleetBOT
@main totem Has your question been resolved?
near finch
I just want to double check this for my quiz tommrow. So for this one I would input -3 , f(-3). But if it was (x-3) I would input postive 3
vocal sleetBOT
I just want to double check this for my quiz tommrow. So for this one I would in...
near finch
outer warren
yes
near finch
Ty!
vocal sleetBOT
@near finch Has your question been resolved?
vocal sleetBOT
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grizzled frost
vocal sleetBOT
grizzled frost
i dont know where to start or what to do
bronze osprey
i dont know where to start or what to do
can you at least write down the null and alternative hypotheses?
"significantly different" is where you need to read
bronze osprey
yeah that's correct
then for the test-statistic, you need $t = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$
twin meteorBOT
south, just south
grizzled frost
2.86
bronze osprey
,calc (23-21)/(3.5/sqrt(25))
twin meteorBOT
Result:
2.8571428571429
yeah fair enough
then you need to go look up a t-distribution table
degrees of freedom is 25 - 1 = 24
two-tailed probability is 0.05
for example this one
grizzled frost
this is still for part B right?
bronze osprey
no we've moved on to part c now
bronze osprey
yep
then this is our confidence interval (z should be t for this question)
calculate both ends of this interval and see if 21 is inside
grizzled frost
ok let try that
bronze osprey
oh shit the question wants you to use the p-value instead
grizzled frost
oh yea
bronze osprey
ah you need your graphing display calculator or I can do it in R
bronze osprey
then this is our confidence interval (z should be t for this question)
this is part d then
bronze osprey
anyways
bronze osprey
yes that's correct for part d
bronze osprey
2.064
the other way is by comparing critical t-values
2.857 > 2.064 so reject the null
I just found that way to be easier personally
but unfortunately the question wants you to find the p-value
bronze osprey
anyways
't-distribution probability' then use the t-value that you calculated, not the one from the table, and df = 24
grizzled frost
how do you find the p value without using a online calc
bronze osprey
how do you find the p value without using a online calc
impossible
well technically, the p-value is some integral
but to evaluate the integral you need technology
grizzled frost
so would part C be 0.009 < 0.025 so we reject null hypothesis
bronze osprey
so would part C be 0.009 < 0.025 so we reject null hypothesis
no it's 0.009 < 0.05
grizzled frost
i see
how would part D go about?
would it be something like 2.857 > 2.064 so reject null hypothesis
?
bronze osprey
would it be something like 2.857 > 2.064 so reject null hypothesis
no part d is just like you found the critical values from the table are 2.064 and -2.064
so you label your t-distribution sketch with that
bronze osprey
you're done!
you might also want to label t = 2.86 with a vertical line but that's optional
vocal sleetBOT
bronze osprey
we did do quite a lot for only 4 parts
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vast shale
how many even numbers less than 500 can made formed using the digits 1,2,3,4,5
each digit can only be used once
vocal sleetBOT
how many even numbers less than 500 can made formed using the digits 1,2,3,4,5
e...
vast shale
Q)how many even numbers less than 500 can made formed using the digits 1,2,3,4,5
each digit can only be used once
restrictions:
must be a even number, so ends it 2 or 4
must be less than 500 so must begin with : 1,2,3,4
is a 3 digit number
4P1 3P1 2P1
as we can have 4,2 in start and end i decided to do it sperately
1,2,3 3P1 4
3P1 3P1 1P1
= 9
1,4,3 3P1 2
3P1 3P1 1P1
=9
9+9 = 18
answer is 28 ?!
can anyone help me and tell me why my thinking was wrong?
shut canyon
uhh
so you can have either 2 or 4 in one digit place
wait
i solved im also getting 18
as the answer
im pretty positive that 18 is correct
vocal sleetBOT
@vast shale Has your question been resolved?
vast shale
im pretty positive that 18 is correct
okk, ima just see if anyone also gets the same answer
thanks for the help
vocal sleetBOT
@vast shale Has your question been resolved?
cosmic cloud
@vast shale I'm getting 28 too
take cases
first case : - three digit
second case : - two digit
one digit number can be either 2 or 4 so two possibilities
vocal sleetBOT
vast shale
<@456226577798135808> I'm getting 28 too
yes i got it thank you
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olive jewel
How would I do 2 ii)?
vocal sleetBOT
How would I do 2 ii)?
olive jewel
What I've done so far
olive jewel
Yeah
olive jewel
the integration onem
Sorry integration one
shut canyon
oh okay thats a relief i hate complex no.
olive jewel
I think it's kind of to do with complex but it's more putting it in the form of the previous question then integrating
But I feel like I've done it wrong though
shut canyon
you did a calculation mistake
in the end
check from tan²x/sec⁶x step
@olive jewel
so it should be sin²xcos⁴x not sin⁴xcos⁴x
vocal sleetBOT
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neon rain
I need help
vocal sleetBOT
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vast shale
vast shale
<@&286206848099549185>
vocal sleetBOT
@vast shale Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
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vast shale
How can I eliminate t from the 2 equations? Whats the procedure?
bronze osprey
How can I eliminate t from the 2 equations? Whats the procedure?
do you have the original question?
cause $x_A (t)$, $\phi(t)$ and so on could be literally any function of $t$
twin meteorBOT
south, just south
bronze osprey
it's not possible then to eliminate t, unless you have more information
vocal sleetBOT
vocal sleetBOT
Available help channel!
barren eagle
guyz does anyone here know how to place these? or like how to do it?
i know how to balance but this activity is a little confusing for me bc of how its placed
barren eagle
this is stoichiometry
bronze osprey
guyz does anyone here know how to place these? or like how to do it?
i know ho...
indeed a creative way to teach stoich
okay so if you look at the ratio of two things at a time, you have 1 pound chicken = 1 casserole
so 4 pounds chicken = 4 casseroles
should be smooth sailing from here
3 tbsp vinegar -> 1 onion
15 tbsp vinegar -> ?? onions
barren eagle
OHHH thats literally just itt?
bronze osprey
yes
barren eagle
OKAYY, thank youuu
bronze osprey
OKAYY, thank youuu
no worries!
cooking is a great way to tech chem fr
and I love adobo
barren eagle
r u filipinooo?
bronze osprey
r u filipinooo?
no
barren eagle
cooking is a great way to tech chem fr
rttt
steep crater
@barren eagle W math question
barren eagle
reall
lucid meteor
guyz does anyone here know how to place these? or like how to do it?
i know ho...
What did I just even witness bro 
Never thought stoichiometry can be taught that way
bronze osprey
yes PH education system gets a lot of flak
good to know there are teachers like that who really care about their students
minor wren
can I ask chemistry questions here?
vocal sleetBOT
can I ask chemistry questions here?
viral copper
minor wren
west bone
how can i help you
bronze osprey
luckily it's stoich
vocal sleetBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
I don't know where to begin.
minor wren
I used gpt to get the answer but I dont understand it
a) What mass of the sulfide of mercury was produced in the second experiment?
To answer this, we need to determine the limiting reagent in the second experiment. The limiting reagent is the reactant that is completely consumed first, thus determining the maximum amount of product that can be formed.
From the first experiment, we know that 1 g of mercury reacts with excess sulfur to produce 1.16 g of mercury sulfide. This means that 1 g of mercury reacts to form 1.16 g of product.
In the second experiment, we have 1.50 g of mercury and 1 g of sulfur. Using the ratio from the first experiment, we can calculate the theoretical yield of mercury sulfide from each reactant:
Mercury: 1.50 g Hg * (1.16 g HgS / 1 g Hg) = 1.74 g HgS
Sulfur: 1 g S * (1.16 g HgS / 1 g S) = 1.16 g HgS
Since sulfur produces less mercury sulfide (1.16 g) than mercury (1.74 g), sulfur is the limiting reagent. Therefore, the maximum amount of mercury sulfide that can be formed in the second experiment is 1.16 g.
b) What mass of which element (mercury or sulfur) remained unreacted in the second experiment?
We have already determined that sulfur is the limiting reagent, so it will be completely consumed. To find the remaining mass of mercury, we subtract the amount of mercury that reacted from the initial amount:
Remaining mercury = Initial mercury - Mercury reacted
= 1.50 g - (1.16 g HgS * (1 g Hg / 1.16 g HgS))
= 1.50 g - 1 g
= 0.50 g
Therefore, 0.50 g of mercury remained unreacted in the second experiment.
here is the answer gpt got
west bone
It asks for the mass of mercury sulfide produced in a reaction where 1.50 g of mercury and 1 g of sulfur are mixed.
bronze osprey
I used gpt to get the answer but I dont understand it
!nogpt
vocal sleetBOT
I used gpt to get the answer but I dont understand it
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
minor wren
ok my bad
minor wren
yes
west bone
so put the number in
minor wren
add the masses?
west bone
Moles of Hg = Mass of Hg / Molar mass of Hg = 1.50 g / 200.59 g/mol ≈ 0.00748 mol
for me i calculate like this
minor wren
we didnt take mols yet so I dont think we are supposed to use it
west bone
well \
sorry i dont know how to calculate without mole
mole is actually basic for chemistry
vocal sleetBOT
@minor wren Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
static brook
if E is parallel to the surface , flux is gonna be 0 , yeah?
vocal sleetBOT
if E is parallel to the surface , flux is gonna be 0 , yeah?
static brook
cuz vector area and surface are perpendicular to each other , so theta should be 90 and cos 90 is 0
now am i wrong? cuz our teacher said it should be cos0 which gives 1
vocal sleetBOT
@static brook Has your question been resolved?
golden iron
been a while since i've done this, but wikipedia agrees with you
θ is the angle between the electric field lines and the normal (perpendicular) to A.
maybe they're using a different convention 
vocal sleetBOT
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Available help channel!
crisp ginkgo
What is this haha I understand nothing
vocal sleetBOT
What is this haha I understand nothing
crisp ginkgo
Do I find the area of the rectangle
crisp ginkgo
Nope
What is that haha
I got a bit further
I don’t get what’s happening tho I’m just following the instructions lol
vast shale
Have you done calculus before tho?
Cause this question may require knowledge of calulus
crisp ginkgo
Yes I’m doing calc 1
It’s for that class
I solved it but I understand nothing lol
rancid otter
I solved it but I understand nothing lol
Okay Ill walk you through
So do you understand why the first step is xy?
or part I should say
crisp ginkgo
Yea cuz that’s how you calculate the area right?
Area of a rectangle is length times width
rancid otter
Yes correct
crisp ginkgo
Part 4 is where I’m lost
rancid otter
Now do you understand how we get 1480=2y+x
Okay so have a look at that diagram
there's fencing in the shape of a rectangle and one side is already covered so theres no fencing there
Then the question tells you
The perimeter will use 1480 meters of fencing
crisp ginkgo
Ohh so there are 2 y’s and 1 x
rancid otter
Exactly
rancid otter
Now we have, A=xy and 1480=2y+x, and we're probably going to have find the maximum area possible
So we need to use A=xy and someone simplify it down to 1 variable
doesn't matter which one we use
crisp ginkgo
So we can solve for y?
rancid otter
You can, you will get the same answer at the end
But you should solve for x as they've asked you to do it like that
crisp ginkgo
Solving for x you get that equation
What does it do tho, why do you need to isolate a variable
rancid otter
So we can plug it into A=xy
crisp ginkgo
Like practically what does it mean when we do that
rancid otter
To remove one of the variables
rancid otter
Like practically what does it mean when we do that
So think of it like this, we have 2 things that can vary, and thats kind of complicated to think of. So instead we found a relationship between these 2 things that vary and replace one of them and get it in terms of the other to simplify the whole thing
crisp ginkgo
Like I take one part of the fence and equal it to the other two parts?
rancid otter
You can imagine it as x not existing anymore and just being replaced by 1480-2y on the diagram
crisp ginkgo
It doesn’t make sense tbh, x is the smaller side right?
How can it equal to two big sides
crisp ginkgo
Ohh oops, so x is the two bigger sides minus the entire fence?
Still doesn’t make sense
rancid otter
Okay look
We have one side that is x
and 2 sides that are y
one side we forget about as its already covered
crisp ginkgo
Yes by the water
rancid otter
we also know that the total fencing is 1480
crisp ginkgo
But if 1480 is the entire fence right, that minus 2y would give us x ohh
I understood as I wrote it lol
rancid otter
yes now do the next bit
crisp ginkgo
Yes why do you take the derivative?
What does that do
Oh wait there’s a step before that
crisp ginkgo
To find the area you multiply the length of x times y
And that gives you what?
Like practically
Cuz the answer is -2y^2+1480y
But again not sure what it means
crisp ginkgo
So we do that so we can find only what the length is equal to or only what the width is equal to
rancid otter
So we do that so we can find only what the length is equal to or only what the w...
We do that to write A in terms of only one variable and nothing else
So then we can differentiate it
And if you set the first derivative equal to 0
(for quadratics) you get the minimum/maximum value
crisp ginkgo
Ohhh critical points
Are maximum and minimum
How do you know if you got the maximum or the minimum then
rancid otter
if you have y=ax^2+bx+c, can you remember the rule to know if the curves extends upwards or downwards?
crisp ginkgo
The first term being positive or negative?
rancid otter
Yes
crisp ginkgo
So it’s negative so it’s a local maximum right?
rancid otter
So in this case the first term is?
rancid otter
So it’s negative so it’s a local maximum right?
Yes, it's also global in this case
So using that we found the value of y that maximizes A
crisp ginkgo
Ohh I see, so we found the maximum value of the area
rancid otter
No
crisp ginkgo
Oh
rancid otter
We found the value of y that maximizes A
we still need one more step to find the actual Area
crisp ginkgo
Ohh the biggest value that we can plug to y
Now we need the biggest value that we can plug to x?
rancid otter
Ohh the biggest value that we can plug to y
No
crisp ginkgo
🥲
rancid otter
We found the value of y which gives the greatest possible value of A, all other values result in smaller areas
crisp ginkgo
Ohh okay I understand, cuz other values would result in a smaller area
Please🤣
This is correct right🤣
rancid otter
Ohh okay I understand, cuz other values would result in a smaller area
Yes
Now what
crisp ginkgo
Okay good haha
rancid otter
We re not done yet
crisp ginkgo
Now we need the value of x that would give the biggest area
rancid otter
No
crisp ginkgo
What🥲
rancid otter
Cuz the answer is -2y^2+1480y
Remember you had this
you dont need x anymore
just use this to calculate Area
rancid otter
Oh yes then you find that
crisp ginkgo
Why do we plug the y value to the x equation and not to the derivative?
That also makes no sense to me
I thought we are supposed to use the derivative to find maximum
rancid otter
The derivative is a one time use equation for us in this problem, we find the value of y that maximizes the area then we use y to find the other stuff as they're linked together with the equations we had gotten before
crisp ginkgo
But we could also take the derivative with respect to x to find the maximum of x?
rancid otter
You could but you would need to get A in terms of x then differentiate it then set that equal to 0 and calculate it etc. etc.
its pointless
crisp ginkgo
Ohh I see okay as long as it’s possible then it’s logical to me
But also wouldn’t x just be twice y?
Like always
Cuz it’s the longer side
And rectangle works like that I think
The longer side is twice the shorter side
rancid otter
The longer side is twice the shorter side
This is not true
crisp ginkgo
Oof
rancid otter
I can draw a rectangle with any ratio of long to short side length
crisp ginkgo
Ohh yea, so it’s just luckily works here
rancid otter
Ohh yea, so it’s just luckily works here
If you don't mind me asking, what's your major?
rancid otter
Computer science
May god help you, you have a long journey of maths left to complete
crisp ginkgo
Oh god🤣😭
rancid otter
You should probably start doing alot of maths because everything will get very difficult very fast if you dont have a very solid foundation
watch 3b1bs essense of calculus on yt its really good for gaining intuition
crisp ginkgo
The professor is currently useless unfortunately
But I’m still getting an A haha
Paying 1200$ for this class btw😁
Thank you for the help! You made it make sense in my mind
vocal sleetBOT
Available help channel!
foggy sphinx
can i ask my question here ?
vocal sleetBOT
can i ask my question here ?
sinful sail
Wat
fervent rain
yes
foggy sphinx
im solving calculus
i've been using symbolab and chatgpt as a guider/correcter
chatgpt is making mistakes with Parametric equation
i dont know how to add two equations in symbolab
anyone knows or dealed with this before ?
flat tiger
!nogpt
vocal sleetBOT
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
flat whale
!done
vocal sleetBOT
If you are done with this channel, please mark your problem as solved by typing .close
foggy sphinx
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
foggy quiver
i can do the math for the contour integrals but i don’t understand holomorphic planes and functions
vocal sleetBOT
i can do the math for the contour integrals but i don’t understand holomorphic p...
bitter pilot
this is me doing contour integrals
cool stuff
foggy quiver
i can do the math for the contour integrals but i don’t understand holomorphic p...
it’s just a very abstract topic for me and it’s kinda confusing
to be fair i’m only a senior in high school so this is super difficult but it’s really fun
vocal sleetBOT
@foggy quiver Has your question been resolved?
foggy quiver
<@&286206848099549185>
my main question is just what is the importance of holomorphic planes and functions
flat whale
my main question is just what is the importance of holomorphic planes and functi...
Holomorphic functions just means the complex function has a Taylor series
https://en.m.wikipedia.org/wiki/Holomorphic_function
Not all functions have Taylor series
foggy quiver
Holomorphic functions just means the complex function has a Taylor series
<https...
how does a holomorphic function come into play when using contour integrals
and why is defining the holomorphic function important as well
flat whale
Derivatives are important in residue theorem
bitter pilot
Cauchy's integral theorem
foggy quiver
i was also curious about how solving a contour integral with the cauchy integral formula is different then the residue theorem
flat whale
What does "defining the holomorphic function" even mean
foggy quiver
Cauchy's integral theorem
i know this
bitter pilot
i know this
I am sorry
flat whale
i was also curious about how solving a contour integral with the cauchy integral...
What
Did you compare them
foggy quiver
like what’s the difference
flat whale
What does your book say
foggy quiver
i don’t have a book
i’m teaching myself
i just started learning about contour integrals today
flat whale
I recommend reading a complex analysis book then to teach yourself
foggy quiver
sweet
foggy quiver
I recommend reading a complex analysis book then to teach yourself
i know a lot of stuff but my knowledge is limited to what i’ve learned so far
omg wait
ok
how would that applications differ
when do you use the cauchy and when do you use the residue
vocal sleetBOT
@foggy quiver Has your question been resolved?
hardy jewel
Three random numbers
𝑋
X,
𝑌
Y, and
𝑍
Z are independently and uniformly picked between 0 and 1. What is the probability that
𝑋
+
𝑌
X+Y and
𝑋
+
𝑍
X+Z are both less than 1? Express your answer as a reduced fraction.
hardy jewel
spiral turtle
@hardy jewel have you ever heard of the term "convolution?"
spiral turtle
(n.b. this integration is a convolution.)
hardy jewel
<@309755154033541120> have you ever heard of the term "convolution?"
no i have not
i tried just doing it geometrically but it’s not an easy shape to find the volume of
i know i guess i could use intervention to find the volume , but was wondering if there was an easier way
vocal sleetBOT
@hardy jewel Has your question been resolved?
half imp
did you see my messages
onyx sage
i have an integral which i can't solve
vocal sleetBOT
i have an integral which i can't solve
onyx sage
$\int \frac{dx}{(x^3-1)\sqrt{(x-1)(x-2)}}$
twin meteorBOT
Erijoni
heavy yoke
,w integrate 1/((x^3-1)*sqrt((x-1)(x-2)))
twin meteorBOT
onyx sage
i had the idea of taking the subs $x-1=t$, $\frac{1}{\sqrt{(x-1)(x-2)}=t$
heavy yoke
where did you get this problem from?
heavy yoke
was that the entire problem? was it just the integral or was it something involving that integral?
onyx sage
my course is divided into two parts, problem solving and theory
this was from the first part
the problem was just solve the integral
which i think is crazy
if we let $t=x-1$, we have $\int \frac{dt}{t(t^2+t+1)\sqrt{t^2-t}}$
twin meteorBOT
Erijoni
grim lotus
u = 1/t?
onyx sage
what do you think about u = 1/sqrt{t^2-t}
onyx sage
neither do i for 1/t idk
grim lotus
i mean at least with that you can write t in terms of u easily
onyx sage
i agree
dt = -1/u^2 du
$\int \frac{\frac{-1}{u^2} du}{\frac{1+u+u^2}{u^3}\sqrt{\frac{1-u^2}{u^2}}}$
twin meteorBOT
Erijoni
onyx sage
$-\text{sgn}(u) \int \frac{u^2du}{(u^2+u+1)\sqrt{1-u^2}}$
twin meteorBOT
Erijoni
onyx sage
it seems very hard now since we have a quadratic and if we complete the square we have (u+\frac{1}{2})^2 which is not the same as the term inside which is only u^2
so i dont think we can proceed using this method
any other idea?
twin meteorBOT
Erijoni
onyx sage
<@&286206848099549185>
$\int \frac{dt}{t(t^2+t+1)\sqrt{t^2-t}} = \int \frac{dt \cdot \left (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}}\right )}{t(t^2+t+1)\sqrt{t^2-t} (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}})}=-2\int \frac{(t-1)d(\frac{t}{\sqrt{t^2-t}})}{t(t^2+t+1)}$
$\int \frac{dt}{t(t^2+t+1)\sqrt{t^2-t}} = \int \frac{dt \cdot \left (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}}\right )}{t(t^2+t+1)\sqrt{t^2-t} (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}})}=-2\int \frac{(t-1)d(\frac{t}{\sqrt{t^2-t}})}{t(t^2+t+1)}$
twin meteorBOT
Erijoni
onyx sage
If $u = \frac{t}{\sqrt{t^2-1}}$, then we have $t = \frac{1}{\sqrt{1-u^2}} $, and $\int \frac{dt}{t(t^2+t+1)\sqrt{t^2-t}} = \int \frac{dt \cdot \left (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}}\right )}{t(t^2+t+1)\sqrt{t^2-t} (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}})}=-2\int \frac{(t-1)d(\frac{t}{\sqrt{t^2-t}})}{t(t^2+t+1)}= -2\int \frac{\frac{1-\sqrt{1-u^2}}{\sqrt{1-u^2}} du}{\frac{1}{\sqrt{1-u^2}} \left ( \frac{1}{1-u^2} + \frac{1}{\sqrt{1-u^2}} + 1 \right )}$
onyx sage
im still trying to figure it out
vast shale
carry on
twin meteorBOT
Erijoni
onyx sage
If we let $t=x-1$, we have $\int \frac{dx}{(x^3-1)\sqrt{(x-1)(x-2)}}=\int \frac{dt}{t(t^2+3t+3)\sqrt{t^2-t}}$, then if $u = \frac{t}{\sqrt{t^2-1}}$, then we have $t = \frac{1}{\sqrt{1-u^2}} $, and $\int \frac{dt}{t(t^2+3t+3)\sqrt{t^2-t}} = \int \frac{dt \cdot \left (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}}\right )}{t(t^2+3t+3)\sqrt{t^2-t} (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}})}=-2\int \frac{(t-1)d(\frac{t}{\sqrt{t^2-t}})}{t(t^2+3t+3)}= -2\int \frac{\frac{1-\sqrt{1-u^2}}{\sqrt{1-u^2}} du}{\frac{1}{\sqrt{1-u^2}} \left ( \frac{1}{1-u^2} + \frac{3}{\sqrt{1-u^2}} + 3 \right )} = -2 \int \frac{(1-u^2)(1-\sqrt{1-u^2})du}{-u^2+\sqrt{1-u^2}+2}$
twin meteorBOT
Erijoni
onyx sage
If we let $t=x-1$, we have $\int \frac{dx}{(x^3-1)\sqrt{(x-1)(x-2)}}=\int \frac{dt}{t(t^2+3t+3)\sqrt{t^2-t}}$, then if $u = \frac{t}{\sqrt{t^2-1}}$, then we have $t = \frac{1}{\sqrt{1-u^2}} $, and $\int \frac{dt}{t(t^2+3t+3)\sqrt{t^2-t}} = \int \frac{dt \cdot \left (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}}\right )}{t(t^2+3t+3)\sqrt{t^2-t} (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}})}=-2\int \frac{(t-1)d(\frac{t}{\sqrt{t^2-t}})}{t(t^2+3t+3)}= -2\int \frac{\frac{1-\sqrt{1-u^2}}{\sqrt{1-u^2}} du}{\frac{1}{\sqrt{1-u^2}} \left ( \frac{1}{1-u^2} + \frac{3}{\sqrt{1-u^2}} + 3 \right )} =-\frac{2}{3} \int \frac{(1-u^2)(1-\sqrt{1-u^2})}{1-u^2+\sqrt{1-u^2}}du$
twin meteorBOT
Erijoni
vocal sleetBOT
@onyx sage Has your question been resolved?
onyx sage
If we let $t=x-1$, we have $\int \frac{dx}{(x^3-1)\sqrt{(x-1)(x-2)}}=\int \frac{dt}{t(t^2+3t+3)\sqrt{t^2-t}}$, then if $u = \frac{t}{\sqrt{t^2-1}}$, then we have $t = \frac{1}{\sqrt{1-u^2}} $, and $\int \frac{dt}{t(t^2+3t+3)\sqrt{t^2-t}} = \int \frac{dt \cdot \left (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}}\right )}{t(t^2+3t+3)\sqrt{t^2-t} (\frac{-t}{2(t^2-t)\sqrt{t^{2}-t}})}=-2\int \frac{(t-1)d(\frac{t}{\sqrt{t^2-t}})}{t(t^2+3t+3)}= -2\int \frac{\frac{1-\sqrt{1-u^2}}{\sqrt{1-u^2}} du}{\frac{1}{\sqrt{1-u^2}} \left ( \frac{1}{1-u^2} + \frac{3}{\sqrt{1-u^2}} + 3 \right )} =-\frac{2}{3} \int \frac{(1-u^2)(1-\sqrt{1-u^2})}{1-u^2+\sqrt{1-u^2}}du$
twin meteorBOT
Erijoni
onyx sage
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
crude scaffold
Use u-substitution to evaluate:
Indefinite Integral: (5-4x)/[sqrt(2x-1)] dx
crude scaffold
My answer is:
1/2 (2x-1)^1/2 - 2/3(2x-1)^3/2 + C
But the correct answer is:
3(2x-1)^1/2 - 2/3(2x-1)^3/2 + C
Why?
crude scaffold
Can you answer the question
silk osprey
$\int \frac{5-4x}{\sqrt{2x-1}} dx$
crude scaffold
How did you solve the problem?
Started with
U = 2x-1
du = 2dx
dx = 1/2 du
twin meteorBOT
knief
crude scaffold
and x = (U+1)/2
crude scaffold
and x = (U+1)/2
then i plugged things in and got
silk osprey
so 5-4x = 5-2u-2 = 3-2u
crude scaffold
$\int \frac{5-4u-4}{4u^1/2} du$
twin meteorBOT
ΛƧ☆ЯΛ
twin meteorBOT
ΛƧ☆ЯΛ
crude scaffold
wth
silk osprey
^{}
silk osprey
use curly braces
crude scaffold
$\int \frac{5-4u-4}{4u^{1/2}} du$
twin meteorBOT
ΛƧ☆ЯΛ
silk osprey
so 5-4x = 5-2u-2 = 3-2u
^
silk osprey
u = 2x - 1 -> 2x = u + 1 -> 4x = 2u + 2
5-4x = 3-2u
$\int \frac{3-2u}{2\sqrt{u}} du$
twin meteorBOT
knief
twin meteorBOT
knief
silk osprey
and you can change the second fraction to u^1/2
and you should be able to integrate this from here
silk osprey
you’re welcome