i did not do the second one
#help-10
rough stump
rain storm
Its B I think
conjugate of z
half silo
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
frigid cliff
someone help me
obtuse pebbleBOT
someone help me
frigid cliff
<@&286206848099549185>
i dont understand what this is asking me to do
all it says is" Evaluate the function graphically."
ebon shard
i dont understand what this is asking me to do
the function is discontinuous here
ebon shard
what does that mean
Take the limits $lim_{x\to3}^+$ and $lim_{x\to3}^-$
In this case, both limits exist and are equal to 3, therefore the limit as $x\to3 = -3$
The issue here is that while the limit exists, the function itself doesn't exist at that point due to it having a point discontinuity
warm shaleBOT
The Cat Collective
frigid cliff
I did do that but it says its wrong
ebon shard
I did do that but it says its wrong
Is the question multiple choice? If so, what are the options?
or is it open ended
frigid cliff
its like this
ebon shard
ohhh
frigid cliff
i answered -3
ebon shard
wait
frigid cliff
but it says "incorrect"
ebon shard
the point was moved
frigid cliff
ok
ebon shard
it's not just discontinuous
frigid cliff
huh
ebon shard
it's defined piecewise such that at 3 it's equal to positive 3
so try +3 instead of -3
frigid cliff
wintry swift
Take the limits $lim_{x\to3}^+$ and $lim_{x\to3}^-$
In this case, both limits e...
it does not matter if the function is continous or not. f(3 is clearly definied from the graph.
ebon shard
it does not matter if the function is continous or not. f(3 is clearly definied ...
right
sorry
frigid cliff
it does not matter if the function is continous or not. f(3 is clearly definied ...
so is it positive 3
ebon shard
We glossed over the point being moved
ebon shard
so is it positive 3
yes
ebon shard
how do you get thatt i just want the thought proces
If the point is deleted and not just moved, then you check the limits and the best you can say is that the limit approaches a value
if the point is moved as in this case, then AFAIK at that x value, the function takes on the y coordinate of the moved point
frigid cliff
umm i cant understand that write in like the equation
frigid cliff
its ok how about this
ebon shard
its ok how about this
the limits are different, the function DNE at x=5
ebon shard
for this one the function just doesn't exist at x=5
because the left and right hand side limits don't approach the same value
empty maple
$(-infty, -4)\cup(5, \infty)$
ebon shard
yea
ebon shard
so on a graph
put it in desmos
frigid cliff
ok
empty maple
:/
ebon shard
ok
for desmos put in these 2 equations to get your graph
y = {x < -4: x}
y = {x > 5: 1}
frigid cliff
ok thanks!!
ebon shard
you're welcome
frigid cliff
okkk
frigid cliff
ok thank you!
ebon shard
you're welcome
frigid cliff
would this be correct?
ebon shard
I think so
frigid cliff
it says its wrong
ebon shard
damn
frigid cliff
how do i do this
i have one last try
wait i changed it to these arrows
do they mean infinite
ebon shard
I have never seen that type of arrow before
but you're welcome to try
because I don't know
ebon shard
heck yea!
frigid cliff
thank you so much!!!
ebon shard
good job /genuine
frigid cliff
but how does it work
ebon shard
thank you so much!!!
You're very welcome
ebon shard
but how does it work
wdym?
ebon shard
i need to show work for class as well
For the piecewise functions, you just have to graph them
frigid cliff
ohh ok!!
ebon shard
would you like me to explain the limits for this too?
frigid cliff
yes plz
ebon shard
okay
at x=-4, there is a left hand side (negative) limit
but no right hand side limit bc it becomes discontinuous
so the limit doesn't exist there
frigid cliff
ohh
ebon shard
at x=5 the limit doesn't exist either bc the here the right hand side (positive) limit exists but the left hand side limit doesn't
the limit doesn't exist between -4 < x < 5 either by definition of the function
frigid cliff
ohhh ty!!
ebon shard
the function exists by definition for x < -4 and x > 5
now we know that -4 and 5 are open circles due to the hard inequality
frigid cliff
yes
ebon shard
and the function isn't defined between them
therefore the limit not only doesn't exist at the points but between either
does that make sense?
ebon shard
:3
frigid cliff
wait but shouldnt it be 5 < x < -4
ebon shard
wait but shouldnt it be 5 < x < -4
negatives first
ebon shard
and if it were like that, we'd be saying that 5 is less than x is less than -4
and that doesn't make sense
5 isn't less than -4
ebon shard
howd you guys figure out that (5, 1) point is one
by the definition of the piecewise function
frigid cliff
ohh okk
ebon shard
its ok how about this
the first part of the definition can be read as "f of x equals x for all x values less than -4" and the second is "f of x equals 1 for all x greater than 5"
I expanded it like that so you can see what's going on
frigid cliff
ohh i kind of get it
ebon shard
so the function here can be defined in "chunks"
frigid cliff
not 100% though
ebon shard
like lemme make a new piecewise function
frigid cliff
ok!!
frigid cliff
um -1<x<1?
ebon shard
nope
ebon shard
so
here
y = 1 OR -1 only
it equals -1 for all x less than 0
and 1 for all x greater than 0
ebon shard
ebon shard
I'd recommend getting some practice with piecewise functions before doing any limit problems with them
ebon shard
and because its y = 1 it goes horizontal
yep
frigid cliff
I'd recommend getting some practice with piecewise functions before doing any li...
okokk
ebon shard
okokk
open desmos and play around with them, or google for some problem sheets online and try them
frigid cliff
wait can i add you
ebon shard
and/or if you have a precalculus book, they should have a chapter for this type of stuff
yea sure
obtuse pebbleBOT
@frigid cliff Has your question been resolved?
obtuse pebbleBOT
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brazen smelt
Is there a way to transform these equations to the uv domain?
x-2y = 0
x-2y = -4
x+y = 4
x+y = 1
obtuse pebbleBOT
Is there a way to transform these equations to the uv domain?
x-2y = 0
x-2y = -...
glossy yacht
Find their intersections
brazen smelt
sure
glossy yacht
do you just randomly dictate which is u and v?
Okay, yes, as long as you do not care about the sign of the area
Also don’t forget your Jacobean
I mention that because some funny stuff happens when you take the absolute of the determinant sometimes
Don’t ask me what I don’t remember
obtuse pebbleBOT
onyx flower
given a sequence a_n such that lim a_n = infinity, prove:
obtuse pebbleBOT
given a sequence a_n such that lim a_n = infinity, prove:
onyx flower
so in class we proved this with the constraint that a_n is a sequence of whole numbers
but i dont understand what that changes
why wouldnt the same proof work here as well?
gilded needle
what was the proof for that case?
onyx flower
they proved it with the definition
gilded needle
well the key difference between the case where a_n are whole numbers vs the general case is that in the former, you have a subsequence of (1 + 1/n)^n, and in the latter you do not
onyx flower
whats a subsequence?
dark stirrup
whats a subsequence?
A sequence of another sequence with some elements possibly removed
twilit meteor
shudnt a_n be necessarily a diverging series?
dark stirrup
shudnt a_n be necessarily a diverging series?
It is
gilded needle
a_n is a sequence, not a series
onyx flower
a_n is a sequence, not a series
whats the difference?
gilded needle
whats the difference?
series = sequence of partial sums
nobody's summing anything here
onyx flower
ah
so the proof from class is:
so why doesnt this proof work without the constraint?
obtuse pebbleBOT
@onyx flower Has your question been resolved?
onyx flower
<@&286206848099549185>
onyx flower
.close
obtuse pebbleBOT
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Available help channel!
short yew
obtuse pebbleBOT
static spade
-x - x/8 shouldn’t be -3x/2 in the second integral
obtuse pebbleBOT
@short yew Has your question been resolved?
ashen schooner
Can anyone verify what I'm doing wrong if anything?
obtuse pebbleBOT
Can anyone verify what I'm doing wrong if anything?
ashen schooner
First thing i do is distribute the -3 to each thing on the top
and the 2 to everything in the denominator
how do i get the - value positive? for a
well for all
Because everything on the top is now negative
how would i put it all over 1 if that makes sense
ivory owl
Instead of thinking of a fraction over a fraction, think of the numerator times the reciprocal of the denominator
Let me make an image in a quick mnute
*minute
ashen schooner
like for what i circled why do we not just -8 from 3/2
ashen schooner
you can do that-
ashen schooner
Yeah I know that part
But on the thing it multiplied everything by 2
Is it because it doesn't have a common denominator? (The exponents)
ivory owl
Ohh, I think I know what you mean
ashen schooner
Yeah
ivory owl
So you're doing (3/2) - 8
ivory owl
Yeah
And then you multiple 8/1 by 2/2
And because 2/2 = 1, and anything times 1 is itself it ends up working
ashen schooner
so if you move a negativen number to the denominator it's not negative anymore right
ashen schooner
I think I will because it seems generally easier
obtuse pebbleBOT
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.
ashen schooner
breathes..
ashen schooner
As a helper, please do not give out answers that could be copied as a homework s...
this is a practice problem dpmo
ashen schooner
I have all the answers i just wanna se the process like this for example
I'm confued how
ohh-
i didn't see the *5 😭
okay i think that's it thanks so much
.close
obtuse pebbleBOT
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Available help channel!
brazen smelt
obtuse pebbleBOT
brazen smelt
"S1 is the line segment
u = v, 0 ≤ u ≤ 5, so y = v = u and x = u2 = y2."
How is this obtained from transformations??
nvm, just bounds.
.close
obtuse pebbleBOT
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Available help channel!
onyx olive
could anone verify the answer to question 5d is 72 degrees?
obtuse pebbleBOT
could anone verify the answer to question 5d is 72 degrees?
@onyx olive Has your question been resolved?
mellow glacier
could anone verify the answer to question 5d is 72 degrees?
i think angle would be taken from positive x axis.
so 180-72
obtuse pebbleBOT
@onyx olive Has your question been resolved?
tardy roost
obtuse pebbleBOT
heavy shore
do you know the properties of the modulus function?
tardy roost
Nope
heavy shore
|-x| = x so you have to take both the values
tardy roost
Ah
heavy shore
for x<3/2; 2x-3 - (6-x) = 13 because the modulus function will make it positive
3/2 ≤ x<6; -2x+3 - (6-x) = 13
x≥6; -2x+3 - (x-6) = 13
solve for all three equations
@tardy roost you get it now?
rugged socket
Hello teachers
midnight parcel
question about the equation, I am pretty sure that this is a -cot with a period of 180, but I am lost on how to get to the equation
obtuse pebbleBOT
question about the equation, I am pretty sure that this is a -cot with a period ...
@midnight parcel Has your question been resolved?
midnight parcel
<@&286206848099549185>
tepid sapphire
problem is that this is not really a $-\cot(x)$ curve since $-\cot (0) = -\frac{\cos 0^{\circ}}{\sin 0^{\circ}} = -\frac{1}{0}$, which is clearly not defined, but in the graph we see $f(0)=1.5$.
warm shaleBOT
Crystopher
tepid sapphire
because the image you gave, which represents the graph of a function (the one you need to find) says so.
midnight parcel
how is it f(0)=1.5?
is that a trig function?
tepid sapphire
midnight parcel
that's the asymptote right?
tepid sapphire
no, the graph clearly states there is continuity, to prove that that point is in fact a removable discontinuity of $-\cot (x)$ you would need to prove that $\lim_{x\to 0} -\cot x = 1.5$. But if you tried to do that you would realize that the limit $\lim_{x\to 0} -\cot x$ does in fact not exist.
warm shaleBOT
Crystopher
midnight parcel
we haven't gotten to limits yet
this is review from precalc i think
so its supposed to be just trig functions
so i am very confused now
tepid sapphire
yes, it is a trig function, what I am saying is that $-\cot (x)$ is not that function.
warm shaleBOT
Crystopher
midnight parcel
yea, i know that, i just wrote -cot(x) as a base because I thought it was a -cot function
since it looks similar
tepid sapphire
try tan(x) instead
midnight parcel
yea, i know that, i just wrote -cot(x) as a base because I thought it was a -cot...
equation something like this
this one does not really fit to me as my graph thing does not start at 0.0
tepid sapphire
yes, but what if you shift it 1.5 units upwards?
tepid sapphire
equation something like this
or are you just required to answer as in one of these?
midnight parcel
sort of
tepid sapphire
well, even so, you can have C=1.5 there
tepid sapphire
I'm speaking of C as in the image you sent
tepid sapphire
yes, but you can also have tan(x)+1.5.
tepid sapphire
what does the graph say?
midnight parcel
to me, 180
tepid sapphire
yes, it is clearly 180
so, let's say your function is $f(x)=A\tan (B(x-\alpha))+C$
warm shaleBOT
Crystopher
midnight parcel
mhm
tepid sapphire
We begin noting that $B=1$, since the period is $180^{\circ}$
warm shaleBOT
Crystopher
midnight parcel
and that's always the case when its 180?
tepid sapphire
yes, in a tan or cot function, form $f(x)=\tan(Bx)$ the period is $\frac{180^{\circ}}{B}$, for sin and cos, namely as in $g(x)=\sin (Bx)$ it would be $\frac{360^{\circ}}{B}$
warm shaleBOT
Crystopher
tepid sapphire
next we look at $\alpha$
warm shaleBOT
Crystopher
midnight parcel
mhm
tepid sapphire
what does it represent?
midnight parcel
period? I am not certain though
tepid sapphire
no, that would be B, which we said has value 1 here, $\alpha$ represents a horizontal shift of the graph.
warm shaleBOT
Crystopher
tepid sapphire
so the graph of $\tan (x-\alpha)$ is the same as the graph of $\tan (x)$ but shifted $\alpha$ units to the right
warm shaleBOT
Crystopher
tepid sapphire
Yes, to have a better intuition you can try the next setup in desmos or geogebra to see how it behaves when $\alpha$ changes
warm shaleBOT
Crystopher
midnight parcel
so then the answer should be just tan (x+1.5)
tepid sapphire
no, first the 1.5 is outside, since it is vertical shift, $\alpha = 0$ since we see in the graph of the searched function that it is not in fact shifted horizontally, so far we have $f(x)=A\tan (x) + 1.5 $
Lastly we decide on A
warm shaleBOT
Crystopher
midnight parcel
midnight parcel
Lastly we decide on A
could you explain how to find A?
I thought it was the amplitude
warm shaleBOT
Crystopher
midnight parcel
right
tepid sapphire
and we have $f(x) = A\tan (x) + 1.5$
warm shaleBOT
Crystopher
midnight parcel
what does that do tho?
tepid sapphire
then $f(45^{\circ}) = A\tan (45^{\circ}) + 1.5$\
since $f(45^{\circ})=f(45^{\circ})$ then\ $A\tan (45^{\circ}) + 1.5 = 4.5$, then you just solve for $A$.
warm shaleBOT
Crystopher
tepid sapphire
you mean this? $A\tan (45^{\circ}) + 1.5 = 4.5$
warm shaleBOT
Crystopher
midnight parcel
yea
tepid sapphire
well, our goal is to find out what A is here, so it would come in handy to have an equation from which you can solve A, this equation was found by knowing that the point $(45,1.5)$ is on the graph. Since it is on the graph, then it must be true that $f(45^{\circ})=A\tan (45^{\circ})+1.5 = 4.5$, so we need to find the A value that makes $A\tan (45^{\circ})+1.5 = 4.5$ true.
warm shaleBOT
Crystopher
midnight parcel
then after we find A, do we use it in the equation? And that equation defines this whole function?
tepid sapphire
yes, if we chose an arbitrary A, it would be likely that $f(45^{\circ}) \neq 4.5$, which would mean that our function $f(x)$ would not represent the graph.
warm shaleBOT
Crystopher
midnight parcel
okay but I am still very lost
tepid sapphire
Well, you can see it as we needing to determine A so that our function goes through that point (45,4.5). It is like we are trying to fit our function to the graph it is supposed to represent.
midnight parcel
yes that makes sense
midnight parcel
**Crystopher**
but this thing does not for some reason
tepid sapphire
Alright, our function $f(x) = A\tan(x) + 1.5$, passes through the point $(45, A\tan(45^{\circ})+1.5)$, and we want it to pass through $(45,4.5)$, so both coordinates must be the same, so for x-coordinate $45=45$, obviously true, for y-coordinate we need $A\tan(45^{\circ})+1.5 = 4.5$ so that $f(45^{\circ}) = 4.5$, which is what we want to achieve.
warm shaleBOT
Crystopher
midnight parcel
okay so we have to calculate A
sorry its 1 am for me and my brain's dead
can we just plug that in to the calculator?
tepid sapphire
no need, should be easy enough to do by hand. Do you know what $\tan(45^{\circ})$ is?
warm shaleBOT
Crystopher
midnight parcel
1
tepid sapphire
yes, so it becomes $A+1.5=4.5$, which gives $A=3$
warm shaleBOT
Crystopher
midnight parcel
so 3tan(x)+1.5
tepid sapphire
yes, as a side note you could also do this with a -cot curve since $-\cot (x-90^{\circ})=\tan (x)$, so alternatively $f(x)=-3\cot (x-90^{\circ})+1.5$, but you see that it is simpler to just state it as a tan function instead.
warm shaleBOT
Crystopher
midnight parcel
that makes sense but what if a different point was given for this graph, this equation should work anyways
tepid sapphire
yes, you input whatever point is given and solve for A. In theory you could have 4 of these points to find all of $A,B,C,\alpha$ by forming an equation system and solving for all variables, but then you need to be sure that any pair of points do not give equivalent equations.
warm shaleBOT
Crystopher
midnight parcel
can i ask you a seperate question
tepid sapphire
go ahead
midnight parcel
what topic is this?
i know its still trig
or like if you know anything about this
tepid sapphire
trig + quadratics, an easy way of doing it is by replacing $u=\sin (t)$, then it becomes $2u^2-u-1=0$ which you should be able to solve, then, when you find the solutions you substitute back $\sin (t) = u$ and solve for $t$ in each root.
warm shaleBOT
Crystopher
midnight parcel
i wish my teacher would teach, all i was given was this https://openstax.org/books/calculus-volume-1/pages/1-3-trigonometric-functions
midnight parcel
trig + quadratics, an easy way of doing it is by replacing $u=\sin (t)$, then it...
i'll try doing that
tepid sapphire
this example is similar in essence
it does not substitute directly but the core idea is the same so to speak
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
woeful fox
obtuse pebbleBOT
timid silo
How do I find that theta is less than π/2
obtuse pebbleBOT
How do I find that theta is less than π/2
@timid silo Has your question been resolved?
timid silo
tender tusk
this might be a weird way of doing it but if theta = arcsec(x), then at x=1 theta = 0 and as x approaches infinity arcsec(x) approaches pi/2 as thus theta will approach pi/2
@timid silo
obtuse pebbleBOT
timid silo
.reopen
obtuse pebbleBOT
✅
timid silo
this might be a weird way of doing it but if theta = arcsec(x), then at x=1 thet...
Oh yeah, thanks I got it a little bit
Isn't there an easy way? I don't know much about trignometry limits
tender tusk
do u know what arcsec looks like?
timid silo
Yes
timid silo
Yes
tender tusk
u should be able to determine the range
timid silo
Range of sec inverse is (0,π)
timid silo
Oh sorry I got confused with cot
tender tusk
remember that x>1 so we are just looking at that line on the right
and notice the range for that line
tender tusk
correct
tender tusk
all good
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
hot rune
helppp!!!
hot rune
a
May i have help with how they got the N value
The question is quite big and is in multiple steps ^ those r the sol
like where did they get n = 12 from? what information tells us this?
thank youu!!
delicate kettle
I got d^3/8 g^5/8 d^1/4 g^1/4
Then I did d^3/8 x d ^1/4
And
G^5/8 x g^1/4
Oh shit sorry wrong channel
obtuse pebbleBOT
@hot rune Has your question been resolved?
full minnow
could someone please tell me why $\frac{a^4-(a+1)^2}{a^2-a-1}$ is equal to $a^2 + a + 1$?
obtuse pebbleBOT
could someone please tell me why $\frac{a^4-(a+1)^2}{a^2-a-1}$ is equal to $a^2 ...
warm shaleBOT
菜月
kind hawk
fossil crag
yep
full minnow
I mean
fossil crag
so what's a^4 - (a+1)^2
full minnow
a^4-(a^2 + 2a +1)?
fossil crag
what's x^2 - y^2?
I didn't ask you this question for nothing
fossil crag
no...
you told me that x^2 - y^2 is (x+y)(x-y)
which is correct
so
what is a^4 - (a+1)^2
big hint... a^4 = (a^2)^2
warm shaleBOT
菜月
fossil crag
yeah
full minnow
oof, the latex fail
fossil crag
well no, isn't it obvious already
fossil crag
the numerator is $(a^2+a+1)(a^2-a-1)$
warm shaleBOT
rafilou2003
full minnow
Thank you very much
fossil crag
so what happens when you divide by a^2 - a - 1
full minnow
you end up with a²+a+1
fossil crag
yes
full minnow
Tysm
But how would I do it in the future
like similar questions?
do I just need to recognise the pattern
fossil crag
yes
always learn your factorisation formulas
if you see some thing that looks like ...^2 + 2...*... + ...^2
start thinking about (a+b)^2
tepid bobcat
<@&268886789983436800>
obtuse pebbleBOT
hot hazel
yet another crypto scam link to add to the pile
obtuse pebbleBOT
Available help channel!
timid silo
obtuse pebbleBOT
timid silo
I was trying to proceed using Vieta's Formula
I do not know where to move after this
tulip prawn
$12^3=1728$
warm shaleBOT
nameless individual
obtuse pebbleBOT
@timid silo Has your question been resolved?
slate monolith
lemme try the q
timid silo
Do not we need to confirm that it have complex roots ?
obtuse pebbleBOT
Available help channel!
robust cargo
where does the - in the second row come from?
obtuse pebbleBOT
where does the - in the second row come from?
robust cargo
im only confused by the left side of the equation
warm canopy
partial fractions
robust cargo
ok thank you, is it some weird stuff? cause I seriously dont understand why that works on first look
weary reef
It's partial fractions. I assumed you knew that already.
https://www.mathsisfun.com/algebra/partial-fractions.html
obtuse pebbleBOT
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Available help channel!
vapid yacht
obtuse pebbleBOT
vapid yacht
Does anyone know hot to solve 3) a)?
tulip prawn
$2x+5$ is algebraic and hence continuous, just plug in
warm shaleBOT
nameless individual
obtuse pebbleBOT
@vapid yacht Has your question been resolved?
slate agate
when you use stokes theorem, the area integral is supposed to be the area contained by the path right??
obtuse pebbleBOT
when you use stokes theorem, the area integral is supposed to be the area contai...
slate agate
but if you have a vector field that is contained ina smaller area than that contained by the line/path, do you integrate only with respect to that area with the area integral?
shut lagoon
The vector field has to be defined and be C^1 in (and a bit around) the surface, so it wouldn't happen
worn yoke
if the vector field doesn't extend all the way to the path then it doesn't make sense to integrate
slate agate
im trying to figure out why my prof has integrated the surface with respect to a different boundary than he integrated the path
shut lagoon
Can you show?
slate agate
yeah im bringing it up to i can bring context
shut lagoon
Okok
slate agate
so the premise is theres a current carrying wire and I have to find the H field
anyways, the crux of the solution is applying stoke's theorem
So I understood the boundaries for the interior
but then for the outside, it seems like he integrates only to a
on the left side, the path is integrated to p
thats the only way the mathematics works here
if its not clear what im saying ill come back later with a more fleshed out solution
shut lagoon
If I'm not mistaken, the local field depends on the enclosed current, so once you go beyond the wire's radius, you're not adding more current, so you just stop at a
obtuse pebbleBOT
@slate agate Has your question been resolved?
slate agate
If I'm not mistaken, the local field depends on the enclosed current, so once yo...
i see, thak you
.close
runic rivet
can someone help with this?
obtuse pebbleBOT
can someone help with this?
errant lark
!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 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
1
errant lark
Do you know how to read a pie chart, firstly?
runic rivet
yes, i think the answer is 12 is that correct?
errant lark
Uh. No.
Oh wait.
Yes. I did not see the 35-40 sector. Lol.
So, yes, 12 is correct.
obtuse pebbleBOT
obtuse pebbleBOT
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unkempt lotus
obtuse pebbleBOT
unkempt lotus
Is this the final solution? Or do I need to do something else?
shut lagoon
Well the limit is supposed to be a number, there should be no x's remaining in there.
obtuse pebbleBOT
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chilly nest
I was wondering if anyone could help me with a question about surds?
obtuse pebbleBOT
I was wondering if anyone could help me with a question about surds?
chilly nest
I am quite confused on how to simplify it 😭
marsh geyser
Work on sqrt(8)
marsh geyser
Factor 8 in prime numbers
chilly nest
But once I have done that, I need to simplify the equation as a single surd
Thats the part I am confused about
reef tendon
asqrt(b)=sqrt(a^2b)
marsh geyser
a*sqrt(b)=sqrt(a^2*b)
This is not true always
chilly nest
I got √2^2*2
reef tendon
This is not true always
For example
marsh geyser
Ok, you can rewrite 2^2 * 2 the inside
reef tendon
2sqrt(3)=sqrt(12)
marsh geyser
And find the square root of one of then
marsh geyser
For example
It has to be >0
marsh geyser
Not because of imaginary numbers, but because you create false solutions
If you have -2sqrt(3) you cant say thats sqrt(-2)^2 * 3) = sqrt(12)
marsh geyser
I got √2^2*2
Now what can u do from here?
chilly nest
I don't know
marsh geyser
What if u write like
Sqrt(2^2) * sqrt(2) ?
Do you know this property for positive numbers?
marsh geyser
No
reef tendon
wait wut
marsh geyser
Sqrt(2^2) * sqrt(2) ?
This is a product not a sum
chilly nest
Ok
marsh geyser
Lets go back to here
reef tendon
sqrt(8) does not equal sqrt(4)+sqrt(2)
so change the plus sign
to multiply
what do you have?
marsh geyser
Sqrt(8) = sqrt(2^2) * sqrt(2)
marsh geyser
That looks good
gritty grotto
is there anyway to find the c value without a calculator?
obtuse pebbleBOT
is there anyway to find the c value without a calculator?
gritty grotto
the answers use solve N for the c value
i cant seem to use it
is there any other way to do this manually e.g first integrate (but the Cs cancel out)
brittle swan
is there anyway to find the c value without a calculator?
I think finding the indefinite integral is the most direct way
obtuse pebbleBOT
@gritty grotto Has your question been resolved?
molten torrent
Is this a viable way to derive the volume of a sphere?:
$r^3-2\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\sqrt{r^2-x^2-y^2}dydx$
obtuse pebbleBOT
Is this a viable way to derive the volume of a sphere?:
$r^3-2\int_{-r}^{r}\int...
warm shaleBOT
Josh ♚
molten torrent
If so I'm not sure how to actually go about solving the integral
keen verge
Is this a viable way to derive the volume of a sphere?:
$r^3-2\int_{-r}^{r}\int...
Ah, I don’t think so
keen verge
Is this a viable way to derive the volume of a sphere?:
$r^3-2\int_{-r}^{r}\int...
I’m not 100% sure, but using gauss div theorem, i don’t think the divergence of this works out to be a constant mass (F) if that makes sense.
keen verge
<@648060265459482624> is there a way to use double integrals to calculate it?
Yes gauss divergence theorem
keen verge
No idea what that is 👍
It’s fairly straight forward if you’re comfortable with parameterisation and double integrals
keen verge
Can I explain my logic?
Idk, give it a go.
molten torrent
I'm somewhat new to double integrals tbh
I know my integral is wrong but I was thinking
Consider a sphere of radius r and centre (0,r,0)
Half of it's volume will be the volume of a cylinder centred on (0,0,0) with height r, minus the volume between the sphere and the xy plane above the circle x²+y²=r², no?
keen verge
Half of it's volume will be the volume of a cylinder centred on (0,0,0) with hei...
*minus the volume between…, but yeah
keen verge
Ah, okay I see your logic now
molten torrent
But the integral was scuffed lol
keen verge
Um, I suppose so, but it would make significantly more sense to use cylindrical coordinates or spherical coordinates to describe the region of intergration as opposed to Cartesian coords. It would be much easier to solve. Even then you’d have to describe the region in 3 dimensions and when solving the triple integral directly you’d get at least one of those variables of integration would be fairly trivial to solve pretty much giving you a double integral anyways
molten torrent
So uhhh
What 😭
Bro I'm an A level student, in the grand scheme at things I'm at the braindead infant skill level idk half of what you're talking about
keen verge
So ig long story short if you want to derive the volume of a sphere using integrals look at changing how you describe the region
Using sphereical coordinates or cylindrical coords most likely
keen verge
Yeah?
molten torrent
Much more basic
I just did the double integral thingy for a cylinders volume successfully, with the following integral
$\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}h;dxdy$
warm shaleBOT
Josh ♚
molten torrent
Where h is the cylinder height (dxdy should say dydx)
keen verge
I think you’ve got your x and y mixed up
molten torrent
Yes I have
It's late lol I can't write
But anyways so let's say the circular face was an eclipse
*ellipse
How would the r and -r bounds change?
I know how the other bounds would change
But would they just stay the same if r is the furthest point on the ellipse form the origin?
keen verge
Hmm, well, assuming you’ve got the other one right, it would just be a matter of integrating over the x variable since you’ve restrained your y variable to region anyways
molten torrent
Okay let's assume an ellipse centred at (0,0,0)
The ellipse will be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
warm shaleBOT
Josh ♚
molten torrent
$\int_{?}^{?}\int_{-\sqrt{b^2(1-\frac{x^2}{a^2})}}^{\sqrt{b^2(1-\frac{x^2}{a^2})}}h;dydx$
warm shaleBOT
Josh ♚
molten torrent
What would the question marks be @keen verge?
keen verge
Well, it’s should be -a and a I’m pretty sure
molten torrent
Yeah I thought that at first
Also that integral looks cursed in latex
Anyways ty for the help @keen verge
keen verge
All good.
obtuse pebbleBOT
@molten torrent Has your question been resolved?
daring steppe
how does this pass the rules for a vector field? thanks!
obtuse pebbleBOT
how does this pass the rules for a vector field? thanks!
daring steppe
like distibutivity of scalar multiplication over scalar addition
(1+1)*1= 1x1+1x1 -> 2 !=1
or is it the case that 1+1 = 1 so (1+1)*1= 1 = 1x1+1x1
given that the set is only 1 would r always have to be 1?
daring steppe
yeah
rigid plaza
given that the set is only 1 would r always have to be 1?
The only application is this yeah
With the current set
daring steppe
gotcha
but if the set was s={x} where x is in R^1
it would not be a vector space
if the rule was
1+x=1
because (1+1)2 =4 = 2x1+2x1 !=1
rigid plaza
The only application is this yeah
Wait actually nvm, I think that one they are trying to show scalar multiplication with it
daring steppe
2 just being a random choice from R^1
rigid plaza
Not a product of two elements which need not be defined
So r can be anything in the field
daring steppe
So r can be anything in the field
i dont know what a field is yet
rigid plaza
So any real number
rigid plaza
Yes
daring steppe
thats so funky
rigid plaza
It is basically turning every operation on 1 into identity
daring steppe
right
daring steppe
but if the set was s={x} where x is in R^1
was this the correct thought
rigid plaza
We don't have the multiplication rule defined for when it is not acting on 1
So the question doesn't really make sense
rigid plaza
Then you don't have closure
obtuse pebbleBOT
rigid plaza
Np
obtuse pebbleBOT
Available help channel!
opal vine
The problem says it’s either 16.5, 6, or 7.5 and I am unsure of what it is as I got 10.67. What is x?
obtuse pebbleBOT
The problem says it’s either 16.5, 6, or 7.5 and I am unsure of what it is as I ...
tulip prawn
!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 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
please show your steps @opal vine
in particular, I think one of the options are correct indeed.
tulip prawn
What is the rationale for $LN+NP=MN+NO$?
warm shaleBOT
Your LaTeX Helper
opal vine
I did it because it was on the same side so the line would continue on and total that amount but idrk
tulip prawn
Have you been taught that $LN/NP=MN/NO$?
warm shaleBOT
Your LaTeX Helper
opal vine
I do an online thing and they only provide videos and they didn’t give a video explaining that so I wasn’t taught that shape
tulip prawn
Given that $LN/NP=MN/NO$, then, can you solve the problem?
warm shaleBOT
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obtuse pebbleBOT
obtuse pebbleBOT
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open spoke
can someone explain what this dfa accepts and why?
obtuse pebbleBOT
can someone explain what this dfa accepts and why?
graceful rapids
can someone explain what this dfa accepts and why?
"accepts" meaning any state sequence that starts at the arrow and ends in a double circle?
open spoke
"accepts" meaning any ~~state~~ sequence that starts at the arrow and ends in a ...
yeah
graceful rapids
see if you can think of some example sequences maybe?
open spoke
see if you can think of some example sequences maybe?
00, 000, 001
graceful rapids
sure, that's the middle part
something, followed by any number of zeros, followed by something
graceful rapids
sure, that's the middle part
something, followed by any number of zeros, followe...
wait mb that's not really a good way of phrasing it
reading off the chart directly i would say
any number of 1s (any number, including none), then 0, then 0, then anything
tulip prawn
If that answers your question, please type .close to allow other people to use the help channels.
open spoke
**If that answers your question, please type `.close` to allow other people to u...
i have another question, should i close then post it in a seperate one?
tulip prawn
i have another question, should i close then post it in a seperate one?
sure
open spoke
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
alpine bay
claim
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
tough copper
before I move onto the second part I want to check my progress on this question, fair warning it is super messy
obtuse pebbleBOT
before I move onto the second part I want to check my progress on this question,...
tough copper
obtuse pebbleBOT
@tough copper Has your question been resolved?
tough copper
<@&286206848099549185>
obtuse pebbleBOT
@tough copper Has your question been resolved?
night crown
How would I do this? I don't want the answer, just need how to do it explained, thanks?
obtuse pebbleBOT
How would I do this? I don't want the answer, just need how to do it explained, ...
night crown
and could someone explain the terminology too
night crown
yeah
molten stirrup
rational here based on the context is natural number
timid silo
Yes!
molten stirrup
equivilent is equak
silk hollow
Hello
timid silo
How would I do this? I don't want the answer, just need how to do it explained, ...
- 3/5 = 9/15
Hope this helps!
obtuse pebbleBOT
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.
molten stirrup
he clearly said explained
night crown
19. 3/5 = 9/15
okay so it is basicly just find whatever number makes 5 into 15 and apply it to 3
timid silo
he clearly said explained
So, if you notice the 5 is just tripled (5 • 3) to = 15, you can repeat this same pattern and multiply the 3 on top by the same process, meaning the 3 would equal 9! 🙂
timid silo
How would I do this? I don't want the answer, just need how to do it explained, ...
20 is 44/60. Hope this helps!
night crown
20 is 44/60. Hope this helps!
I was just about to ask, lol
again, I do not want an answer, just explain what the work is and how to do it please, thanks
obtuse pebbleBOT
@night crown Has your question been resolved?
haughty coral
~help
haughty coral
help
calc 1
please
i've been doing this homework for the last uh...2 hours and 30 minutes
and like i understand the concept its just i keep somehow getting the answer wrong
this is the work i have
but when i plug in it ends up being like
7k then 700k
which i know is wrong
please
please
please
plz
plz
<@&286206848099549185> its been 15 minutes yo i need help 🔥
shut lagoon
I think you're plugging things in wrong then, because your formula is right.
shut lagoon
Yes
shut lagoon
But it doesn't yield 7k
shut lagoon
Are you doing (-3.01)^2
shut lagoon
🙂
haughty coral
im embarrassed and sad and frustrated
shut lagoon
Your disappointment is immeasurable
fallen inlet
Ik the shapes don’t look the same but aren’t they vertical angles? Bc their tips are touching like that idk
obtuse pebbleBOT
Ik the shapes don’t look the same but aren’t they vertical angles? Bc their tips...
fallen inlet
And why do the vertical angles work here if the shapes also don’t look the same …
heavy depot
zoom in on the angle itself
the shapes they connect to dont matter
its vertical 'angles' not vertical shapes
fallen inlet
The angles look about the same to me
heavy depot
theyre vertical
they are the same
theyre vertical bc you can mirror the whole angle system to make the two congruent
fallen inlet
Ik the shapes don’t look the same but aren’t they vertical angles? Bc their tips...
So this is vertical?
But there’s still not enough info?
heavy depot
theres never really enough information to tell if an angle is vertical
you kinda just have to guess
i mildly hate how its just assumed
the reason vertical angles work is because they are formed by two lines intersecting each other and creating angles
as an example for the first image they should really tell you that "Point E such that it lies on segments CF and BG" but its just assumed most of the time 🤷♂️
timid silo
can $-[1-x]$ be written as $-1-[-x]?$ (if $[\cdot]$ denotes greatest integer function)
obtuse pebbleBOT
can $-[1-x]$ be written as $-1-[-x]?$ (if $[\cdot]$ denotes greatest integer fun...
warm shaleBOT
pun pun
ashen dove
Mb
timid silo
np
ashen dove
can $-[1-x]$ be written as $-1-[-x]?$ (if $[\cdot]$ denotes greatest integer fun...
Yes
Assume n ≤ x < n+1 where n is an integer
U can easily prove it
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
outer nexus
hey can any one help me tomorrow at 8:30 with some Hw it’s an online class pls i would do it but im out of time and my class finishes tm.
obtuse pebbleBOT
hey can any one help me tomorrow at 8:30 with some Hw it’s an online class pls i...
@outer nexus Has your question been resolved?
limber nebula
please help idk what to do
