I am going to go in a anohter window so if anyone comes to help feel free to @ me
#help-0
tidal gust
raven rover
Start with a diagram
raven rover
A diagram is a picture of what’s going on
tidal gust
Yeah I can visualize it and I have sketched one in geogebra
I have also thought about using discriminant
I think that seems promising
like something so that with what values of a the discriminant is negative=no tringles. With what values of a the discriminant is 0=one triange. With what values of a the discriminant is positive=two triangles.
<@&286206848099549185>
No progress yet
alpine sable
Still this problem?
tidal gust
Yeah
alpine sable
Use quadratic question formula to find in which case you get no triangles...
Let's call AC = c
tidal gust
Allright
alpine sable
You get
$c² - 8√3 c + 64 - a² = 0$
ocean sealBOT
bs292929
alpine sable
Thus for c to exist
$(8√3)² - 4(64-a²) \ge 0$
ocean sealBOT
bs292929
alpine sable
Here you would get your 3 conditions,
If less than 0 means triangle not possible
= 0 means only one possible value of c is possible
> 0 means 2 possible value of c is possible which means 2 triangles are possible
tidal gust
So I was in the right tracks all along.
alpine sable
Yup
tidal gust
Let me think about this for a second
alpine sable
Cool
tidal gust
Thank you so much. Now that I see it it is pretty obvious but you know that is how it always is with math haha
frank gorge
lone heartBOT
frank gorge
very confusedd how to get the equation
pine kettle
@frank gorge do you know what parent function it is
frank gorge
<@486171952600580116> do you know what parent function it is
yes i know for this one its y= √ x
but im not sure what to do after
alpine sable
Because we know that pink is of the form sqrt(kx), and that at 12 it has the value 6, we can solve for k by substituting sqrt(12k) = 6
frank gorge
Because we know that pink is of the form sqrt(kx), and that at 12 it has the val...
what os sqrt(kx)
alpine sable
square root of k * x
frank gorge
why do we want to solve for k thou
alpine sable
Because the equation of the transformed function is f(kx)
and we know f, so to find the transformed function only need to figure out k
frank gorge
so we divide 12 by 6?
alpine sable
To solve sqrt(kx) = 6 you should start by squaring both sides, so that you can isolate k
or in this case sqrt(12k) not sqrt(kx)
frank gorge
what does that do thou
alpine sable
It gets rid of the square root around k, so that we can solve for it
frank gorge
what do u do after u remove the square root?
alpine sable
Squaring both sides leaves you with 12k = 36, from here you can just divide both sides by 12 and get k
alpine sable
That means the transformed function is f(3x), or in this case sqrt(3x)
alpine sable
Its possible that they drew the graph somewhat off-proportion and expected you to get that result by looking at the difference between the two graphs?
But im not sure how they got that
Is there any more info other than that graph?
alpine sable
Do you know the formula for the green graph?
frank gorge
tis y=1/x?
alpine sable
Seems like it
Now what is something immediately evident about the pink graph, in comparison to the green one?
frank gorge
pink doesnt touch 2?
alpine sable
(I mean in relation to its sign)
frank gorge
ohh its reflected?
alpine sable
So what does that tell us about the form of its equation?
frank gorge
how do we find that
alpine sable
I would recommend looking at the point x = 1
frank gorge
what does that do?
frank gorge
wait wdym equal to -6
alpine sable
If you look at the pink graph at x = 1
alpine sable
(actually on closer expectation it appears to be equal to -5!)
So we would then have -1/k = -1/5
frank gorge
what does that mean
alpine sable
If -1/k = -1/5 then we must have k = 5
which would mean the pink graph’s equation is y = -1/5x
(though im really eyeballing the value here)
frank gorge
wait why -5
alpine sable
So the pink graph is y = -1/kx, and we know that when x =1 we have y = -5
so if we plug those in we get: -5 = -1/k
alpine sable
Then we can use the equation -5 = -1/k to solve for k
frank gorge
where did the x go?
alpine sable
we plugged in the value x = 1 because we know from the graph that when x = 1 we have y = -5
alpine sable
Starting with -5 = -1/k, we can multiply both sides by -k
that gets us 5k = 1
then we can divide both sides by 5 and get: k = 1/5
frank gorge
why did the k go to the other side?
alpine sable
Because we multiplied both sides by k, and k/k = 1
alpine sable
Well now that we know that the pink graph is y = -1/kx and that k = 1/5 we can just plug that in
versed hornet
help me: what is the formula for finding the area of a cylinder?
alpine sable
help me: what is the formula for finding the area of a cylinder?
use an unoccupied help channel next time, but its pi * r^2 * h
Wait
no thats volume
frank gorge
Well now that we know that the pink graph is y = -1/kx and that k = 1/5 we can j...
ohh i seee
alpine sable
its 2pirh + 2pi*r^2
frank gorge
what if u used a diff point that isnt 1,5?
modern moss
anyone need help?
alpine sable
You would get a different equation, but with values such that you get the same result for k
versed hornet
anyone need help?
yes
modern moss
with?
alpine sable
Please move to an unoccupied help channel
versed hornet
kk
frank gorge
You would get a different equation, but with values such that you get the same r...
ohh
ok i see thankss
.close
cosmic furnace
lone heartBOT
cosmic furnace
i only need help with part a
swift musk
figure out how to find the angles inside the triangle, then use the sine or cosine rule
lone heartBOT
@cosmic furnace Has your question been resolved?
lone heartBOT
lone heartBOT
Available help channel!
balmy schooner
Can anyone explain to me what this means? (so that i can complete these problems?)
merry depot
for the first one it means
for any $\epsilon >0$ you can define a $\delta > 0$ such that if $|x-4|<\delta$ then $|(2x-5)-3|<\epsilon$
ocean sealBOT
Zybikron
balmy schooner
what does the epsilon stand for? as well as the delta
apologies, my professor is not fluent in english and does not explain problems
i have fallen behind in comprehension
merry depot
epsilon can be any positive real number.
balmy schooner
okay! thank you
merry depot
delta is a number you get to define (usually based on epsilon) to obtain the inequality |f(x) - L| < epsilon
balmy schooner
Alright, so for any positive real number, the delta will be greater than zero for the given equation?
merry depot
so, delta isn't something like 2
you'll usually end up with something like, 'let delta = epsilon/4'
balmy schooner
so the equation has to equal a positive number or it doesn't exist?
merry depot
delta is also a positive number. But it's a number you get to define
unlike epsilon, which can be any positive number
balmy schooner
okay, how do you go about defining delta
merry depot
My old prof used to say "My enemy has chosen an epsilon and refuses to reveal it. We must find a delta to prove that this is continuous and thwart him!"
merry depot
okay, how do you go about defining delta
I typically just leave it as delta until I need it to be something.
balmy schooner
Okay.
So
I'm so sorry lmfao I'm slow
to prove the first problem then
i don't even know where to start
he's done examples in class but i don't think they match these
topaz flame
.
balmy schooner
and there isn't any process (or at least that's what it seems like)
quasi vector
I'm so sorry lmfao I'm slow
trust me, there are way slower people that show up in this server.
topaz flame
yep
merry depot
**Zybikron**
^^^^
balmy schooner
i remember from class that he tells us to solve (usually) or break the equation up if it equals 0 and somehow rework it
but i have no idea if i'm supposed to do that here or not
merry depot
You'll want to manipulate it so that you're always maintaining equality, or increasing the magnitude
balmy schooner
i've never had any limit problems that equal anything
right so i'm looking for equality in a solution
??
and in this case the limit as x --> 4 input in x solves directly, is that all i have to do?
merry depot
no, you need to prove it with the delta epsilon argument
balmy schooner
**Zybikron**
and this proves it?
merry depot
with some work, it can yes
balmy schooner
okay, so simplify ?
merry depot
sure, what do you get?
balmy schooner
|2x-2|?
merry depot
check that again
balmy schooner
|2x-8|
merry depot
can you factor anything out?
balmy schooner
|x-4|
merry depot
2|x-4| yeah
balmy schooner
oh right
balmy schooner
epsilon?
merry depot
not yet
balmy schooner
oh
merry depot
$|x-4|<\delta$, so what is $2|x-4|$ less than??\
Or how can you use $|x-4|<\delta$ to put an upper bound on $2|x-4|$?
ocean sealBOT
Zybikron
balmy schooner
I don't think I even have proper background comprehension for this lmfao
i'm a little overwhelmed
organic chem is too easy but this is muddling my brain
🤦♀️
merry depot
you're overthinking it
balmy schooner
okayokayokay
merry depot
what's the difference between |x-4| and 2|x-4|?
balmy schooner
2
merry depot
one of them is multiplied by 2
merry depot
ok
balmy schooner
2 delta?
merry depot
so $|x-4| < \delta$, get $2|x-4|$
ocean sealBOT
Zybikron
merry depot
2 delta?
So,
You have $|(2x-5)-3| = 2|x-4| < 2\delta$\
If you can pick a $\delta$ so that $2\delta < \epsilon$ then you're done\
Can you use $2\delta < \epsilon$ to find that delta?
ocean sealBOT
Zybikron
merry depot
or even $2\delta = \epsilon$
balmy schooner
ugh
ocean sealBOT
Zybikron
merry depot
solve for delta
balmy schooner
how do i know what epsilon is
merry depot
you don't. You jsut know it's a number
balmy schooner
any number greater than 0?
merry depot
yep
merry depot
**Zybikron**
we're trying to take this and get\
$|(2x-5)-3| = 2|x-4| < 2\delta <\cdots < \epsilon$
ocean sealBOT
Zybikron
merry depot
we can fill in the dots by choosing delta
balmy schooner
How do I choose delta?
merry depot
**Zybikron**
solve for delta
peak bough
As a side-note (I hope this helps), when thinking about the epsilon-delta definition of a limit, it can be very useful to think of |x - x₀| as the distance between x and x₀ and and likewise with |f(x) - L|.
In this way, the definition says that the limit as x -> x₀ for f(x) is L if for any given arbritray value ε (no matter how small) such that f(x) and L is at the most ε close to eachother (that is |f(x) - L| < ε) we always want us being able to find a δ such that if we just have that the distance |x - x₀| is smaller than this δ, we can always gurantee that the distance |f(x) - L| is smaller than ε.
Here it can be useful to think of it as a game like Zubikron said, where an enemy comes and demands you to find a value ε such that the distance |f(x) - L| is less than this value and you need to fight him with a value δ no matter his choice of ε.
Sorry for the long text 
Some people have different ways of learning different concepts. But thinking about the absolute value as a distance instead, was what really made the definition make sense for me. So I just wanted to share that in hope that it also will help you
balmy schooner
reading and trying to comprehend, but i appreciate it thank you <3
merry depot
Yeah, no one picks up delta-epsilon proofs right away. Mathematicians did them wrong (or at least in a very handwavy way) for a very long time before making them rigorous.
balmy schooner
It's been a few weeks now in class and I'm still not grasping haha
merry depot
they take practice
merry depot
https://tutorial.math.lamar.edu/classes/calcI/defnoflimit.aspx
Here's a site with several worked examples if that helps also
merry depot
but anyway, back to this.
We want to fill in $2\delta < \cdots < \epsilon$\
Some of the < can also be = in this case
ocean sealBOT
Zybikron
balmy schooner
well the original |(2x - 5) -3| < epsilon right
merry depot
how do i know which ones
Pick delta so it depends on epsilon (since your enemy can change it)
balmy schooner
let me rewrite what i've done so far to see if it helps
quick question too
How did we get these out of the limit equation?
merry depot
Basically. You want $2\delta < \epsilon$ or $2\delta = \epsilon$\
Since you can pick delta, just solve one of those for delta.
ocean sealBOT
Zybikron
balmy schooner
delta is a half an epsilon?
merry depot
delta is a half an epsilon?
yep
balmy schooner
oh okay
merry depot
Now. if you pick $\delta = \epsilon/2$ then\
$|(2x-5)-3| = 2|x-4| < 2\delta = 2(\epsilon/2) = \epsilon$
ocean sealBOT
Zybikron
merry depot
How did we get these out of the limit equation?
this comes from the delta-epsilon definition. $x_0=4$, $f(x) = 2x-5$ and $L=3$
ocean sealBOT
Zybikron
balmy schooner
okay!
i didn't register that 3 = L
progress 💪
okay
from there, the absolute values are there because the equations can't be less than 0?
merry depot
**Zybikron**
So this ^^^^ shows that |f(x) - L| < epsilon
and in the process, you chose your delta value
balmy schooner
**Zybikron**
are 2 delta < epsilon and 2 delta = epsilon the same thing?
🥹
come baack
lmfaoo
also wondering how we know that the |x-4| < DELTA rather than epsilon (or even that we knew we needed to have the comparison in the first place
am i allowed to ping 🥴
peak bough
Maybe it is useful to examplify the enemy-scenario. Let us say that a function f is defined with the rule f(x) = 2x + 3 and I say that f(x) --> 7 for x --> 2. First we can think of what this actually means. We are saying that, as x goes closer and closer to 2 our function f(x) goes closer and closer to 7.
An enemy approaches and says he doesn't believe us! He says that if we are so certain in our claim, then we should be able to give him a distance from x to what x approaches (x --> 2, so the distance |x - 2|) such that we can always guarantee that f(x) and 7 are a distance of at most 5 (so ε = 5 in this scenario) from eachother. We take the challenge and start our way around it. We know that we want
|f(x) - 7| < 5 <==> |2x + 3 - 7| = |2x - 4| < 5
We can do a bit of workaround here, to show that
|2x - 4| = |2(x - 2)| = |2||x - 2| < 5 <==> |x - 2| < 5/2
Looking at this, we can see that if we chose δ (our weapon against our enemies challenge) as δ = 5/2 we can by working backwards in the above, actually show that |x - 2| < δ = 5/2 will always lead to |f(x) - 7| < 5 like our enemy wanted.
Now say that our enemy is very stubborn. He thinks he made the challenge too easy, and now wants us to instead guarantee us, that if he choose ε = 1 instead of ε = 5 (so he makes the distance smaller, that is, f(x) closer to 7). We can walk through the same argument as above (try it yourself!), to conclude that if we choose δ = 1/2 we can again combat the enemy's challenge.
This game can get a bit tiresome. The enemy chooses a number, we combat this number, they choose a new one and the cycle repeats. What we actually want, is to always, no matter the choise of ε always be able to combat this with a δ. So instead of the enemy saying we should be able to find a value that guarantees f(x) being a distance of 5, 1 of maybe even 0.00001 close to 7, he now says for any value of ε, no matter how small, we should be able to find a corresponding δ, if our claim is correct.
Long text 
balmy schooner
@merry depot i'm not sure if i'm allowed to ping you or if you're just busy atm but i need help 😅 no rush if you're busy i totally get it
peak bough
Maybe it is useful to examplify the enemy-scenario. Let us say that a function f...
To finish this (I ran out of space), we can follow the same line of thought as before. We have that
|f(x) - 7| < ε <==> |2x + 3 - 7| = |2x - 4| < ε
Doing the same as before, we get that
|2x - 4| = |2(x - 2)| = |2||x - 2| < ε <==> |x - 2| < ε/2
So if we choose δ = ε/2, we can always combat the enemy instantly with a distance (δ) that x needs to be close to 2, no matter how close a distance (ε) he wants us to show that f(x) is to 7
lone heartBOT
@balmy schooner Has your question been resolved?
frigid hatch
Hello, I'm doing some elastic collision problems basically all of them have systems of equations like this one:
How can I solve for v_1f and v_2f?
frigid hatch
$\m_1=1055kg\ m_2=715kg\ v_{1i}=0\ v_{2i}=2.55\frac{m}{s}\\ \left{\begin{array}{@{}l@{}} m_2\cdot v_{2i} = m_1\cdot v_{1f} + m_2\cdot v_{2f}\ m_2\cdot (v_{2i})^2 = m_1\cdot (v_{1f})^2 + m_2\cdot (v_{2f})^2 \end{array}\right.$
ocean sealBOT
Thunder7
frigid hatch
What I'd do is isolate either v_1f or v_2f from the first and substitute in the second, solving the quadratic equation in v_1f or v_2f and substituting back to find the other one
but this process takes a lot of time, I was trying to algebraically manipulate the system in such a way that it was simpler but wasn't able to
lone heartBOT
@frigid hatch Has your question been resolved?
raw cairn
here you’ll have ready formulae: http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html
raw cairn
but this process takes a lot of time, I was trying to algebraically manipulate t...
if you show your attempt, I may help you with deriving the formulae
but the main hint is to square the first eq
frigid hatch
if you show your attempt, I may help you with deriving the formulae
I would if the masses didn't square too
raw cairn
but that’s not a problem, you just multiply eq 2 by m2
thin light
Can somebody help me understand how to answer this?
lone heartBOT
Can somebody help me understand how to answer this?
thin light
i have an example from a classmate about how my prof likes to solve this and i really want to know how 
lone heartBOT
@thin light Has your question been resolved?
lone heartBOT
@thin light Has your question been resolved?
lone heartBOT
@thin light Has your question been resolved?
rigid pollen
<@&286206848099549185> In this context what does that equation where summation and intergal does mean? Why those are added? Why does it imply f(u) is distribution of U?
midnight parcel
I’m not sure how to start these problems at all
lone heartBOT
I’m not sure how to start these problems at all
midnight parcel
I think is how you start it off but i’m not sure
solemn juniper
you're right, now differentiate both sides
midnight parcel
solemn juniper
,rccw
ocean sealBOT
solemn juniper
now it's just a matter of isolating y' and doing some algebra to clean up the final answer
midnight parcel
ok, bc i know that i’d multiply by y to just get dy/dx n that i can sub in y for the original equation right? it’s just the cleaning up at this point.
solemn juniper
yep
median python
Hey could anyone help me, with a problem
I need to calculate molarity of vinegar, 10 grams of vinegar in 10 grams of water, molar mass of acetic acid is 60.052g/mol, I somehow got 17M but I don’t think that correct
torpid shell
can someone explain to me how to do part c like the min/max type problems step by step?
median oar
not you
median python
Hey could anyone help me, with a problem
I need to calculate molarity of vineg...
<@&286206848099549185> mind helping me with this
alpine sable
Hey could anyone help me, with a problem
I need to calculate molarity of vineg...
Ive not much experience with these kinds of calculations, but I think by “liters of solution” you would include the volume that is vinegar
median python
Ive not much experience with these kinds of calculations, but I think by “liters...
Oh I accidentally miss typed the wrong unit
It’s 10 ml of vinegar
But idk how to get grams from it, so I just assumed it’s 10 grams 🤷♀️
alpine sable
Do you know how to use density?
median python
No
alpine sable
vinegar bas a density of 0.96g/mL
Which means for every millimeter of vinegar we have 0.96 grams
to find the total mass we would do 0.96g/mL * 10mL
median python
Which is 9.6 grams?
alpine sable
yes
median python
That still gives 15M?
median python
9.6 mols / 60.052 molar mass = mols
Mols / 0.01 L of water
alpine sable
As I said earlier, I think you include the volume in the vinegar in the solutions total volume
median python
Adding the 10 ml would drop it to abt 7M which is still way to high for molarity
median python
Idk, it’s supposed to be a titration lab and im stuck on the first problem of finding the molarity of the vinegar that we used
alpine sable
As I said, im not very knowledgable about chemistry, not much I can do to help beyond this
signal lichen
what happens to the (-1)^k here? Is it because of the absolute value?
hard vortex
oak pelican
I dont get the last two steps... shouldnt it be y = sqrt(rest of equation)
lone heartBOT
I dont get the last two steps... shouldnt it be y = sqrt(rest of equation)
@oak pelican Has your question been resolved?
lone heartBOT
@oak pelican Has your question been resolved?
late agate
ewie, cosine rule
burnt olive
I know how to do the question, but I have no clue express it in exact value
late agate
do you know your cosine rule?
burnt olive
yeah
late agate
I don't really see the problem then
burnt olive
ok, how would you convert -0.6186404848
burnt olive
oh ok
well you arcsin the part a fraction
due to the angle is an obtuse,
apply 180 and use subtraction
employ its cos value
burnt olive
ok, how would you convert -0.6186404848
and thus
burnt olive
do you have an alternative?
late agate
yea why don't you just make a triangle out of the 5sqrt2 over 9
hold on let me show you
burnt olive
oh i see what you mean
late agate
find cosP by phythag theorem and you should get an exact value
or maybe this doesn't work at all
burnt olive
i agree with this
but
check the mark scheme
i wonder where the negative value derives from
late agate
do you know what PQ is for the original triangle?
late agate
yes
burnt olive
as in the length?
late agate
yes
late agate
because how I'd do it is find PQ and put it in the cosine formula
burnt olive
right
late agate
but I have an inkling of a doubt that wouldn't work
burnt olive
lol
late agate
no it wouldn't work because theres no sqrt
burnt olive
i think our first approach was good
late agate
hold on
burnt olive
to be honest the mark scheme is a bit sus sometimes
i've been doing exercises since the start of easter and i've encountered 3 mark scheme errors
but i'm more inclined to think that we missed something in the question
late agate
well good thing is I found the sqrt31
high rapids
sin p = cos (pi/2 - p)
late agate
dang raidants
no my method 100% works
ok
see this
find the missing side
phythag theorem
keep it in exact value
no decimals
high rapids
Also negative because obtuse angle > 90 degrees and cos is neg in 2nd quadrant
late agate
you don't need theta
burnt olive
guess that's in the second quadrant
burnt olive
no as in to determine the positive/negative nature of the trig expression
late agate
oh I see
late agate
oh you just confused why it was negative
burnt olive
im back
this one is quite funny
its not difficult at all, but
i wonder if my second solution works
basically
convert tan2x=sin2x/cos2x
rearrange and you get 1=4cos2x
hence 1-4cos2x=0
therefore whatever value it times gives you 0, including sin2x
alpine sable
Simply plug the values into the function
Let me give you an example:
For x=0, you have: f(0)=2*0-1=-1
Notice I simply replaced x with 0
I'll let you do the rest
@alpine sable Sorry, I didn't see your messages
So the notation they're using is gonna need some explaining
You see how, like, a function can be graphed?
Like, for x, at x=0, the curve would pass through y=-1
Hence (0,-1) is a point of f
Similarly, you can interpret (1,7) as meaning f(1)=7
Or, in other words, that at x=1, y is 7 for function f
The domain is all the inputs you can put into the function
Notice how (0,...) is not included
That means f(0) is undefined
Hence, 0 is not part of the domain
The range is every possible output of the function
Notice how (...,6) is not included
That means the function can't output 6
Hence, 6 is not part of the range
@alpine sable Everything clear?
@alpine sable So, any questions?
Sure
So let's look at the functions points
More specifically their x-values, a.k.a. inputs, a.k.a. the first number
We see 1, 2, 3, and 4
Hence the domain (or list of valid inputs to the function) is {1,2,3,4}
For the range (or list of possible outputs of the function), you do the same but with the y-values, a.k.a. outputs
Pretty much
The domain being the list of possible inputs and the range being the list of possible outputs, that is indeed what you do
Well at least for finitely many points
Yep, with the curly brackets around it
The range being a set, duplicates and order doesn't matter
So while that is technically correct we would rather expect {7}
I just said order doesn't matter
Preferably ascending order just because it looks nice
lone heartBOT
alpine sable
Preferably ascending order just because it looks nice
I just realised this sounds contradictory
Like any order is technically correct and it's no big deal but ascending order is best
lone heartBOT
Available help channel!
modest osprey
lone heartBOT
modest osprey
cannot get the domain in the form of n+1/n
i tried putting the function inside the log>0 but that just made it more complex
someone please help me
<@&286206848099549185>
<@&286206848099549185>
please help
<@&286206848099549185>
anyone?
<@&286206848099549185>
<@&286206848099549185>
burnt olive
try sector 2
there are a lot of people active there
see if they can come around to assist
lone heartBOT
@modest osprey Has your question been resolved?
modest osprey
try sector 2
what is sector 2?
burnt olive
i meant help-2
modest osprey
they said go back to your channel and see what you get
modest osprey
so what now?
burnt olive
sorry but i can't provide help to your question
thick quartz
help
zenith lava
Hey there, can someone help me with the difference between a prediction and confidence interval
zenith lava
I've been told the difference is that CI find the average response whereas prediction intervals will find a particular response, but surely as you are using the predictor x0(what we are predicting for) to calc the CI the CI is also for that particular response?
lone heartBOT
@zenith lava Has your question been resolved?
amber plover
new to this topic so not sure what to answer
lone heartBOT
new to this topic so not sure what to answer
amber plover
what i've got so far is the first one the 5th one and the 6th answer
only one im kinda sure is not true is the third one
other ones i have no idea
echo socket
In 2nd it's given that f(x) -> inf, so what would 1/f(x) approach?
amber plover
In 2nd it's given that f(x) -> inf, so what would 1/f(x) approach?
uhh negative inf?
echo socket
No
amber plover
not very sure since i only got a very brief introduction to this topic
echo socket
The bigger the denominator gets, the smaller will be the fraction
So 1/f(x) would approach 0
amber plover
yeah i figured that
echo socket
4th is false, because it says that (0, f(0)) lies on both y = f(x) and y = g(x)
Meaning g(0) = f(0)
Or 1/f(0) = f(0)
But the solutions to that are f(0) = 1 and f(0) = -1
amber plover
alright nice thanks
echo socket
And the last one is also false because it implies that f(0) < 0, meaning g(0) < 0
I.e., y-intercept of y = g(x) would be below the x-axis as well
amber plover
And the last one is also false because it implies that f(0) < 0, meaning g(0) < ...
thanks a lot it was correct
have a good rest of your day
.close
alpine sable
I'm confused on the bottom part
alpine sable
.close
copper sequoia
Can someone help me with part (b)? I cant figure out how to start.
lone heartBOT
Can someone help me with part (b)? I cant figure out how to start.
@copper sequoia Has your question been resolved?
lone heartBOT
@copper sequoia Has your question been resolved?
true pumice
What did I do wrong here?
lone heartBOT
What did I do wrong here?
plain flame
$\sqrt{x^2} = |x|$
ocean sealBOT
heavy
true pumice
.close
winter ore
ouch
wary stream
what the problem?
Please read #❓how-to-get-help
You need to open your own channel as batata open this themself
shadow scarab
oh
lone heartBOT
@winter ore Has your question been resolved?
lone heartBOT
lone heartBOT
Available help channel!
patent wedge
Number 15
lethal belfry
Number 15
this is physics but I think I can give a few hinits
first, recall $F=\frac{dp}{dt}$
ocean sealBOT
physicsrocks
lethal belfry
use $v^2-u^2=2ad$
ocean sealBOT
physicsrocks
patent wedge
Yes
lethal belfry
v isn't 9.8 as that's accleration
not velcoty
I mean that teh accleration is 9.8
patent wedge
V = V node + at
lethal belfry
or s=ut+1/2at^2
patent wedge
If I use v2=v²node + 2as I get 2.8
lethal belfry
V = V node + at
use v^2-u^2=2ad
lethal belfry
If I use v2=v²node + 2as I get 2.8
vnode?
patent wedge
V node is like the staring of v
lethal belfry
you don't know t though
impact time isn't travel time
irrespetive vfinal =2.8
so F=2.8/(1/7)
ok, gtg
patent wedge
Ovhy
lethal belfry
bye
patent wedge
You got it,right?
I'm gonna try to solve for force multiple t
Wait nvm that is impulse
lethal belfry
I'm gonna try to solve for force multiple t
Yes , use delta p/delta T
Mass is given, isn’t it?
lone heartBOT
@patent wedge Has your question been resolved?
winter elbow
lone heartBOT
winter elbow
How would I go about actually checking to see this solution isn’t + and -
Plugging in x = -5/2 + root 13 seems like such a convoluted headache
Oh wait a minute
No it shouldn’t actually be that bad right? Because it’s all squared? Unless someone wants to just offer me a correction or clarification
But the root 13 isn’t squared by the 5(x)
last ether
What's the original equation
winter elbow
Root x + 1 minus 3/root x + 1 = 1
last ether
Oh it's because -5/2 - root(13)/2 < 0
Which is out of the domain of root(x)
Although -5/2 + sqrt(13)/2 is also less than 0...
last ether
Root x + 1 minus 3/root x + 1 = 1
You mean $\sqrt{x+1} - \frac 3{\sqrt{x+1}} = 1$?
Or is the 1 inside of the root
winter elbow
Yeah it’s inside the root
last ether
If that's the case you gotta use paranthesis to make that clearer
winter elbow
My bad, next time I’ll grab the original
ocean sealBOT
Umbraleviathan
last ether
So this
winter elbow
Yeah
last ether
Yeah what's the domain of sqrt(x+1)
winter elbow
The input?
last ether
No
The domain of sqrt(x+1) is x >= -1
If you input any value that is less than -1, you'd get a complex number
last ether
Let's analyze the solutions we got from the steps
winter elbow
It just can’t equal zero
last ether
Well typically they want real number solutions
I'm assuming that's why they omitted the extraneous solution
winter elbow
Yeah that’s kind of what’s messing me up
Half the other answers are complex
So I’m confused as to why that wouldn’t count
last ether
If we look at -5/2 - sqrt(13)/2 + 1, that's -3/2 - sqrt(13)/2 which is obviously less than -1 so it won't work
last ether
Half the other answers are complex
Like they contain iota and all that?
last ether
Was that for a quadratic or a square root
winter elbow
Quadratic
winter elbow
Even though it becomes a quadratic?
last ether
If you plug in your solutions here, the final answer will be a complex number, not 1
last ether
Even though it becomes a quadratic?
Always test it in the original question
When we turned it into a quadratic we made extraneous solutions
winter elbow
Yeah that’s what’s giving me a hard time actually
last ether
You always have to analyze the answers
last ether
If we look at -5/2 - sqrt(13)/2 + 1, that's -3/2 - sqrt(13)/2 which is obviously...
Hence the analysis here
winter elbow
Plugging in those huge fractions with root 13 is bugging me out
last ether
Oh no no don't plug in directly
Just use numerical analysis
Here let's go back to the extraneous solution, -5/2 - sqrt(13)/2
If we plug it into sqrt(x+1), we know that in order for the root to kick out a real number, the argument must be greater than 0
$-\frac52 - \frac{\sqrt{13}}{2} + 1 = -\frac32 - \frac{\sqrt{13}}{2} < 0$
ocean sealBOT
Umbraleviathan
winter elbow
OHHH and because this whole thing is only going to return a negative it literally cant work
winter elbow
Yeah it’s positive
I gotta say the way you break things down and explain this is awesome, seriously
I’ve had some people on here that dance around teaching me where the problem areas are and you’re great
That makes total sense and opened me up to seeing it a completely different way, I cannot thank you enough and I seriously hope I can get you again for any future issues
alpine sable
#1089953713449349131 i need help with this
lone heartBOT
<#1089953713449349131> i need help with this
@alpine sable Has your question been resolved?
Available help channel!
warped sentinel
How do I explain this the best I can? This is for an assignment
warped sentinel
sorry AJ
pallid scarab
So you're solving for x. Easiest way is to get rid of the fractions, so multiply by the LCD (lowest common denominator)
quasi vector
you want to eliminate the fractions
warped sentinel
So you're solving for x. Easiest way is to get rid of the fractions, so multiply...
I think I already solved for x, I need to explain it though.
warped sentinel
you want to eliminate the fractions
yes
pallid scarab
Well, proceed by equivalences
"Let x be a real number.
(x+1)/2 + (x+2)/3 = x/4 <=>..."
warped sentinel
ok
pallid scarab
You will end up with " <=> x = -2"
warped sentinel
ok
pallid scarab
So by your reasoning, -2 is the only solution to the original equation.
gray blade
hello im new and i need help on something cause im having trouble on understanding on how to like solve it
warped sentinel
So by your reasoning, -2 is the only solution to the original equation.
for my assignment, im going to have to break it down
warped sentinel
hello im new and i need help on something cause im having trouble on understandi...
uh-
pallid scarab
hello im new and i need help on something cause im having trouble on understandi...
#❓how-to-get-help, this channel is taken
lethal belfry
hello im new and i need help on something cause im having trouble on understandi...
I suggest you open this in #help-12 , or an open chat
or that's taken
any open char
gray blade
oh ok sorry
pallid scarab
for my assignment, im going to have to break it down
Meaning...? You want to justify at each step?
warped sentinel
Meaning...? You want to justify at each step?
yes
its like a equation breakdown
like how you solve a question using the quadratic formula
pallid scarab
"Multiplication by 12. Adding ... on both sides. Dividing by ..."
warped sentinel
hmmm
pallid scarab
That's easy isn't it?
warped sentinel
not for me
pallid scarab
Show me what you did for your calculations
warped sentinel
I just jump straight into conclusions and never thought about its steps and how it is solved
pallid scarab
"Let x be a real number.
(x+1)/2 + (x+2)/3 = x/4 <=>..."
Well, reason by equivalences. I'll do the whole thing for you just so you can see that not much is required
warped sentinel
I go with x equaling a negative number since x/4 is smaller than x/3 + x/2, plus if x was positive that would mean the number would be bigger than x/2 + x/3 alone.
warped sentinel
Well, reason by equivalences. I'll do the whole thing for you just so you can se...
ok
since +1 and +2 is being added to the equation
pallid scarab
"Let x be a real number.
(x+1)/2 + (x+2)/3 = x/4
<=> 6(x+1) + 4(x+2) = 3x
<=> 6x + 6 + 4x + 8 = 3x
<=> 7x = -14
<=> x = -2"
pallid scarab
Np
warped sentinel
now allI need to do is make a script for this and record it
quasi vector
sounds like a weird assignment
warped sentinel
its a Teaching Video assignment
you are supposed to record yourself solving a question of your choosing (well from the list I got)
time to close the channel
.close
floral snow
Let f : [a, b] → R be a function.
• Construct an example of a function f : [0, 1] → R such that f is continuous at every
point of x ∈ [0, 1] with the possible exception of x = 1/2 and such that f is not bounded.
pallid scarab
whit this task
Hello, this help channel is taken, I suggest you write your request on an empty channel such as #help-20
austere fulcrum
okay thanks
pallid scarab
Let f : [a, b] → R be a function.
• Construct an example of a function f : [0, 1...
Well, you can guess that f will not be bounded at 1/2, what functions can you think of that go to infinity in a single point?
pallid scarab
Im thinking 1/x, but that will be continuous at 1/2
Well 1/x explodes at 0, so in order for it to explode at 1/2, what should we do?
lone heartBOT
@floral snow Has your question been resolved?
ocean portal
can someone explain how this part is obtained? https://math.stackexchange.com/questions/3750607/prove-that-if-the-graph-on-n-vertices-has-at-least-n-22-1-edges-then
worn fox
ocean portal
was trying amgm inequality or cauchy schwartz
pallid scarab
was trying amgm inequality or cauchy schwartz
Continuing from the last line, keep it written as $\frac{k² + (n-k)²}{4} + \frac{k(n-k)}{2}$
ocean sealBOT
rafilou2003
pallid scarab
I would even suggest keeping everything under the same denominator
pallid scarab
**rafilou2003**
What would be really cool is if we could write this as (something) + k(n-k)
Because we're comparing this quantity to k(n-k)
Try to do it
lone heartBOT
@ocean portal Has your question been resolved?
ocean portal
What would be really cool is if we could write this as (something) + k(n-k)
ohh i think this is it
basically i start by doing -2k(n - k)/4 + 2k(n - k)/4
tall path
lone heartBOT
tall path
So do they want me to find the limit as x approaches 0?
alpine sable
open perch
i need help
lone heartBOT
i need help
open perch
.close
lone heartBOT
jagged cobalt
last ether
LOL
open perch
Amazing
I got help from my friend sorry for the waste of time
wont speak here again
.close
.close
vapid shuttle
the channel is closed don't worry, it will shut itself soon
lone heartBOT
Available help channel!
alpine sable
hey !
high rapids
Ask
alpine sable
um so i need help learning basic quantum mechanics
tacit arch
um so i need help learning basic quantum mechanics
ask a math question
or go to #old-network for physics server
alpine sable
kk
lone heartBOT
@sacred hare Has your question been resolved?
stone garden
Hi my name is Nami and i need help in math i am 18 years old and i go to college. I don't understand integral calculation. I'm going to show a picture in a couple of minuts of my classtest that i didn't understand but i tried so hard to learn for it and got a 5+.
naive valley
5+ out of what? is that good or bad?
small panther
I don't think she'd be here if it was good
pseudo ice
Hi my name is Nami and i need help in math i am 18 years old and i go to college...
Long time no see 
high rapids
Long time no see <:catlove:1073406411570163753>
Do you know them?
pseudo ice
Do you know them?
Helped them out some time ago before the new year!
stone garden
Wow i completly forgot you remember me lol
The translation: Task 1:
(40 points)
After you finish school, you take a job in a wood workshop in Gelsenkirchen Hassel.
One day you receive an order from your boss to use a new saw to cut wood into a certain shape. Since they are not yet familiar with the device, they first have the device carry out a few test cuts on discarded sheet iron and
the test log can be output. During the entire process, the feed rate fin is logged in minutes as a function of time x in minutes. The test protocol for the test cut is shown on the right. The feed rate can be described mathematically by the following functional equation:
f(x)=0.01 x 0.3x² + 2.25 x
a) Determine the feed rate at time x = 7 min.
(3 points)
b) Calculate a domain of definition for f(x) that makes sense in the context.
(5 points)
(7 points)
c) Determine the maximum feed rate.
d) Calculate the time when the feed rate is strongest
(5 points)
decreased. e) Draw the function graph of f(x) in a suitable coordinate system.
(10 points)
f) Determine a possible antiderivative for f(x) and explain why there are infinitely many more correct antiderivatives of f(x). (5 points)
g) Determine the distance traveled by the saw in the first 10 minutes and find the average feed rate over this period:
(5 points)
Hint: You can use without proof that the mean of the function values of a continuous function f in the interval [a, b] is given by:
771= a f(x)dx
22/SOP
5+
stone garden
5+ out of what? is that good or bad?
Very bad 1 = really good 2 = good 3 = satisfying 4 = enough 5 = defective 6 = is not enough
pseudo ice
Mean formula seems to have come out wrong but there’s a picture of it at least 
tacit arch
1 is better than 6? what a strange grading scale
stone garden
I understand the first task a) but then by b) it starts i understand literaly nothing 🥲
stone garden
1 is better than 6? what a strange grading scale
It's in Germany the scale🥲
1 is the best note and 6 is the worse note
5 is defective a bad grade*
tacit arch
ah that explains it. sorry for interrupting
stone garden
ah that explains it. sorry for interrupting
Np
pseudo ice
Oh yea also how do you feel with differentiation at the moment? Are you at least happy(/happier) with it?
stone garden
I'm disapointed about myself but i don't give up no matter how bad i am until i have a good grade and understand math and what the math teacher says
pseudo ice
That’s the spirit, you just sometimes have to work hard, might be pain but you’ll get there! 
What was your answer for part a?
stone garden
I answered in part a)
f(x) = 0,01×^3 - 0,3x^2 + 2,25x
f(x) = 0,01 × 7^3 - 0,3 × 7^2 + 2,25 × 7 = 4,48 mm/min
And i answered in german that it is 4,48 in and anwsered sentence
the feed rate is 4.48 mm/min
That's the only that the teacher said it was right lmao the rest was false
stone garden
Thanks
in part b) i understand nothing
I don't know where to start and how and what to learn and to understand it is really complicated
pseudo ice
For part b, you want to consider the real world basically
You don’t want the model to “talk about the past”
As in x is the the time in minutes since you started the process, you agree?
stone garden
Yes
pseudo ice
Cool, so if x was, say, -5, your model would be “describing the feed rate 5 minutes before the process started”, do you agree with that?
stone garden
Yes i read it few more times to understand more
I have a little question when it is 10- is it the same then? The model describing the feed rate 10 minuts before the process started?
And when it is 5 without the minus is it then describing zje feed rate 5 minuts after the process started?
Oh i understand it
pseudo ice
Cool 
stone garden
pseudo ice
But of course if you haven’t started the process, you can’t really talk too much about the feed rate with that model, right?
stone garden
Yes
I'm really sorry rn i am back in 20 minuts because i need to pray the nightprayer in islam (named ischaa) before it end i am in like 20 oder 15 mimuts back you can still write
pseudo ice
Okay cool, see you soon!
stone garden
Oki
pseudo ice
But of course if you haven’t started the process, you can’t really talk too much...
We said here that our model shouldn’t (really) talk about things before the process started, and that x represents the time in minutes from when the process started
So with all that said, can you think of a restriction that we need to place on x?
stone garden
How do we place a restriction on x?
pseudo ice
As in, do you want the value of x to be greater than something? Less than something?
pseudo ice
I have a little question when it is 10- is it the same then? The model describin...
Think about what we said here
stone garden
As in, do you want the value of x to be greater than something? Less than someth...
Hmmm.... i don't know
stone garden
Think about what we said here
Hmmm i think we need to place x here after the 7 minuts
Do you mean that when not i'm lost fr
pseudo ice
Cool, so if x was, say, -5, your model would be “describing the feed rate 5 minu...
Alright, do you agree that you don’t want x to be -5, for example?
pseudo ice
Hmmm i think we need to place x here after the 7 minuts
No no, not that
stone garden
No it's okay that -5 is the example for it
So it is -5 and we need i think to place x before the process starts right??
pseudo ice
Hmmmm
Alright; remember that when x=0, the process hasn’t started but it’s about to
Like just almost about to
Buuuttttt
pseudo ice
No it's okay that -5 is the example for it
x=-5 talks about before the process started
Why do you think x being -5 is alright?
pseudo ice
But of course if you haven’t started the process, you can’t really talk too much...
Remember that we made this statement here
stone garden
So i think we need first to start and then to calculate it
I think maybe we need to calculate the zeros?
pseudo ice
That is something we will need to do for sure, so let’s come back to what we’re thinking here later
Are you able to find the zeroes for me?
stone garden
I will try 😌
pseudo ice
(Self reference, we have $f(x) = 0.01 x^{3} - 0.3 x^{2} + 2.25x$)
ocean sealBOT
@pseudo ice
stone garden
Hmm i think we need to calculate first 0,01x^3 and the rest and it is then
f(x) = 0,01x^3 - 0,3x^2 + 2,25x
f(x)' = 0,03x^2 - 0,60x^1 + 2,25
f(x)'' = 0,06x^1 - 0,06
f(x)''' = 0,06
Or i think
0 = 0,01x^3 - 0,3x^2 + 2,25x
0 = 0,01x^3 x (0,60x^1 + 2,25x)
0 = 0,03x^2 - 0,60x + 2,25 | : 0,03
0 = 74,40
pseudo ice
Hmm i think we need to calculate first 0,01x^3 and the rest and it is then
f(x)...
I am happy you know how to differentiate from that, and we will need (some of) those later
pseudo ice
Or i think
0 = 0,01x^3 - 0,3x^2 + 2,25x
0 = 0,01x^3 x (0,60x^1 + 2,25x)
0 = 0,0...
The very first one is what we need for now
stone garden
Oki
pseudo ice
Find where 0.01x^3 - 0.3x^2 + 2.25x = 0