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# A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that BAL = ACB.Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
A triangle ABC is right angled at A. L is a point on BC such that AL BC. Prove that BAL = ACB.
Solution:
In ΔCAB,
∠B + ∠C + ∠CAB = 180°.............(i)
In ΔBLA,
∠B + ∠BLA + ∠BAL = 180°.............(ii)
now, from (i) and (ii) , we get,
∠B+∠C+ ∠CAB = ∠B + ∠BLA + ∠BAL
∠C +∠CAB = ∠BLA + ∠BAL
∠ACB = ∠BAL [∠BLA =
∠BAC = 90°]
Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
Solution:
Given: Two lines are respectively perpendicular to two parallel lines.
now, let the slope of parallel line be k.
as it is given that, two lines are perpendicular to two parallel lines
therefore,
slope of first line which is perpendicular to the parallel line = -1/k....(i)
and
slope of second line which is perpendicular to the parallel line = -1/k....(ii)
from (i) and (ii) ,
its clear that , slope of the perpendicular lines is equal.
hence, they are parallel to each other.
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You are on page 1of 34
Introduction to
Cryptographic Currencies
Claudio Orlandi
cs.au.dk/~orlandi
Thanks to: Jon K. Srensen and Peter S. Nordholt
Leave while you can!
Politics
Economics
Coming up next:
Algorithms
Cryptography
Outline
Part 0: a little history
Part 1: TheoryCoin
How to create coins
How to transfer coins
How to store coins
Part 2: diff( , )
Part 3: Problems and issues
crypto currency
The 1990s
David Chaum and anonymous ecash
The difference between
and well-developed digital cash
will determine whether
we will have a dictatorship
or a real democracy
(attributed to Chaum)
Anonymous payments
withdraw
withdraw
M or L?
Chaums anonymous e-cash
anonymous
secure (no double-spending)
only transfer (no creation/storage)
and bankrupted in 1999
2009: Bitcoin announced by Satoshi Nakamoto
Pseudonym for person or group of person
2009-2011: slow start
End 2013: Bitcoin price skyrockets
and the world notices!
Outline
Part 0: a little history
Part 1: TheoryCoin
How to create coins
How to transfer coins
How to store coins
Part 2: diff( , )
Part 3: Problems and issues
TheoryCoin:
How to create money
1. Everyone
tries to solve a puzzle
2. The first one to solve
the puzzle gets 1 TC
3. The solution of puzzle i
defines puzzle i+1
TheoryCoin:
How to create money
H
L {0,1}* R {0,1}*
T {0,1}
d
SolvePuzzle(L){
repeat{
R = my_name || i++
T = H(L,R)
}while(T 0
d
)
return R
}
The puzzle:
given L, find R
such that T=0
d
(a random function)
* aka Proof-of-Work
TheoryCoin: (coins to ppl)
How to create money
H
x
0
= Start!
x
1
=(P
1
, i
1
)
000000
x
2
=(P
2
, i
1
)
H
000000
x
3
=(P
3
, i
3
)
H
000000
P
3
P
1
P
2
x
1
x
1
x
2
x
2
x
3
x3
* aka the blockchain
x
7
=(P
3
, i
7
) x
6
=(P
3
, i
6
)
x
5
=(P
5
, i
5
)
x
0
=Start! x
1
=(P
1
, i
1
) x
2
=(P
2
, i
2
)
x
3
=(P
3
, i
3
)
x
4
=(P
4
, i
4
)
TheoryCoin:
How to create money
* aka the 51% attack
TheoryCoin:
How to create money
Recap:
Solve the next puzzle get a coin
To solve puzzle i find x
i
s.t H(x
i-1
,x
i
)=0
d
The longest chain defines next puzzle
The name in block x
i
gets coin i.
Outline
Part 0: a little history
Part 1: TheoryCoin
How to create coins
How to transfer coins
How to store coins
Part 2: diff( , )
Part 3: Problems and issues
TheoryCoin:
How to transfer money
(Digital) Signatures
Only you can sign
Everyone can verify
You cannot deny
Give coin 3 to Jesper
Claudio
TheoryCoin:
How to transfer money
Gen
Sign Verify
message
message, signature accept/reject
secret key public key
P
3
P
1
m=P3 gives coin 3 to P1
s=Sig(sk3,m)
If
Ver(pk3,m,s) = accept
and
P3 owns coin 3
then
return accept
TheoryCoin:
How to transfer money
P
3
P
1
P
2
accept
accept
TheoryCoin:
How to transfer money
m1=P3 gives coin 3 to P1
s1=Sig(sk3,m1)
m2=P3 gives coin 3 to P2
s2=Sig(sk3,m2)
* aka double spending
P
3
P
1
TheoryCoin:
How to transfer money
...
(m1,s1)
...
(m2,s2)
...
(m4,s4)
m1 = P3 gives coin 3 to P1
s1 = Sig(sk3,m1)
m2 = P3 gives coin 3 to P2
s2 = Sig(sk3,m2)
write
(m1,s1)
write
(m2,s2)
(m1,s1)
P
2
(m2,s2)
accept
reject
P
4
m4 = P1 gives coin 3 to P4
s4 = Sig(sk1,m4)
write
(m4,s4)
(m4,s4)
Outline
Part 0: a little history
Part 1: TheoryCoin
How to create coins
How to transfer coins
How to store coins
Part 2: diff( , )
Part 3: Problems and issues
TheoryCoin:
How to store money
Main Idea:
Record transfers in the blockchain
x
4
=(P4, (m,s), i
4
)
P
1
TheoryCoin:
How to store money
P
3
P
2
P
4
(m,s)
(m,s)
(m,s)
SolvePuzzle(L,...){
repeat{
R = my_name||(m,s)|| i++
T = H(L,R)
}while(T 0
d
)
return R
}
Outline
Part 0: a little history
Part 1: TheoryCoin
How to create coins
How to transfer coins
How to store coins
Part 2: diff( , )
Part 3: Problems and issues
diff( , )
How is money created in Bitcoin?
New block every ~10 mins
H = 2-SHA2
Initial reward: 50 BTC
Halved every ~4 years (now 25 BTC)
diff( , )
How is money transferred in Bitcoin?
P1 gives 14 to P1
Transaction fee 1
Example: P1 wants to give 60 to P2
... gives 50 to P1
gives 25 to P1
P1 gives 60 to P2
P2 gives 42 to P3
P2 gives 17 to P2
Transaction fee 1
diff( , )
How is money stored in Bitcoin?
Transaction in orphaned blocks are invalid
Wait 6 blocks (~1 hour) before accepting transaction.
Checkpoints to prevent complete history rollback.
All transaction are stored in the blockchain
(Currently ~14 GB)
Outline
Part 0: a little history
Part 1: TheoryCoin
How to create coins
How to transfer coins
How to store coins
Part 2: diff( , )
Part 3: Problems and issues
Anonymity?
Problem:
Every transaction ever made is recorded forever
Solution?
Use new identity for each transaction
But:
Heuristics allow to cluster identities
Anonymous alternatives:
Zerocoin, Zerocash
Users?
(and their devices)
Unfortunate property of DSA
1HKywxiL4JziqXrzLKhmB6a74ma6kxbSDj
probably stole ~250000kr this way
(due to bug in Android Java based random generator)
Extractor
Sig(sk,m1,r)
Sig(sk,m2,r)
sk
Programmable money?
Bitcoin uses a scripting system for transactions. Forth-like,
Script is simple, stack-based, and processed from left to
right. It is purposefully not Turing-complete, with no loops.
E.g., P1 gives 1 BTC to P2 if at least
2 out of (P1,P2,P3) sign this transaction
Functionality: more than money?
Security: malware payments?
Mining pools
Solving puzzles (mining) is hard!
Miners join pools and share work/reward
How to optimally split work?
Mechanism design?
rational miner?
how to allocate reward?
A final word
Distributed currencies:
for the good guys or the bad guys?
But sometimes governments are bad too!
Thanks! Questions?
Sources:
Story of Chaum and DigiCash (to be taken with a grain of salt)
http://cryptome.org/jya/digicrash.htm
Bitcoin paper and announcement
http://article.gmane.org/gmane.comp.encryption.general/12588/
http://www.mail-archive.com/[email protected]/msg10142.html
This pizza cost 750,000 usd
http://motherboard.vice.com/blog/this-pizza-is-worth-750000
Lily Allen turns down btcs
Signature attack
http://eprint.iacr.org/2013/734
Deanonymizing
http://cseweb.ucsd.edu/~smeiklejohn/files/imc13.pdf
http://eprint.iacr.org/2012/584
Zerocoin/Zerocash
http://zerocoin.org/
Graphs, stats etc
www.blockchain.info
Comparison with Altcoins
http://www.coinwarz.com/cryptocurrency
Bitcoin stolen from TV
http://nymag.com/daily/intelligencer/2013/12/bloomberg-anchors-christmas-bitcoin-gets-stolen.html
Visa/Mastercard vs Wikileaks
http://www.forbes.com/sites/andygreenberg/2010/12/07/visa-mastercard-move-to-choke-wikileaks/
Not in the talk, but very interesting:
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0:58 AM
Introductory Lecture on Statics: the Catenary and the Arch
# 1. Introductory Lecture on Statics: the Catenary and the Arch
### What? Statics?
Why begin a course on classical dynamics with a statics example? The reason is that the most important underlying principle of classical dynamics, and much of the rest of physics, is Hamilton’s Principle of Least Action: a system’s time development can always be describes as a path through some multidimensional parameter space (positions, velocities of all its component parts) and the path it actually follows is the one that (on varying the path) minimizes the integral of a certain function, called the action, along that path. A generalization of this principle by Dirac and Feynman (the "sum over paths") led to a fruitful reformulation of quantum mechanics and quantum field theory, now widely used.
The mathematical machinery necessary to minimize an integral along a path by varying the path is called the calculus of variations, so that’s what we have to master. To see how it works, we’ll first apply it to a simple example: finding the curve describing a uniform rope hanging between two points under gravity. The “action” for this statics problem is of course just the potential energy. And, to be sure we’ve found the right answer, we’ll first solve the problem by more traditional statics techniques. So this is where we begin.
### The Catenary
What is the shape of a chain of small links hanging under gravity from two fixed points (one not directly below the other)? The word catenary (Latin for chain) was coined as a description for this curve by none other than Thomas Jefferson! Despite the image the word brings to mind of a chain of links, the word catenary is actually defined as the curve the chain approaches in the limit of taking smaller and smaller links, keeping the length of the chain constant. In other words, it describes a hanging rope. A real chain of identical rigid links is then a sort of discretization of the catenary.
We're going to analyze this problem as an introduction to the calculus of variations. First, we're going to solve it by a method you already know and love -- just (mentally) adding up the forces on one segment of the rope. The tension is pulling at both ends, the segment's weight acts downwards. Since it's at rest, these three forces must add to zero. We'll show that writing down the balance of forces equation gives sufficient information to find the curve of the chain, meaning heightabove ground as a function of horizontal position. So we understand the mechanics of the problem.
A Cambridge mathematician, Thomas Whewell, famously stated in unconscious rhyme,
And so no force, however great,
can pull a string, however fine,
into a horizontal line,
that shall be absolutely straight.
(Trivial fact: Whewell had a way with words -- he even invented some of the words you use every day, for example the words scientist, physicist, ion, cathode, anode, dielectric, and lots more.)
### A Tight String
Let's take the hint and look first at a piece of uniform string at rest under high tension between two points at the same height, so that it's almost horizontal.
Each little bit of the string is in static equilibrium, so the forces on it balance. First, its weight acting downwards is , being the uniform mass per unit length. Second, the tension forces at the two ends don't quite balance because of the small change in slope.
Representing the string configuration as a curve, the balance of forces gives
so, and , taking the lowest point of the string as the origin.
So the curve is a parabola.
### Not So Tight
However, there are hidden approximations in the above analysis -- for one thing, we've assumed that the length of string between and is but it's really where the distance parameteris measured along the string. Second, we took the tension to be nonvarying. That's a pretty good approximation for a string that's almost horizontal, but think about a string a meter long hanging between two points 5 cm apart, and it becomes obvious that both these approximations are only good for a near-horizontal string.
Obviously, with the string nearly vertical, the tension is balancing the weight of string below it, and must be close to zero at the bottom, increasing approximately linearly with height. Not to mention, it's clear that this is no parabola, the two sides are very close to parallel near the top. The constant approximation is evidently no good -- but Whewell solved this problem exactly, back in the 1830's.
What he did was to work with the static equilibrium equation for a finite length of string, one end at the bottom.
If the tension at the bottom is , and at a distance s away, measured along the string, the tension is , and the string's angle to the horizontal there is (see diagram), then the equilibrium balance of force components is
,
from which the string slope
where we have introduced the constant which sets the length scale of the problem.
So we now have an equation for the catenary, in terms of distance along the string. What we want, though is an equation for vertical positionin terms of horizontal position the function for the chain.
Now we've shown the slope is
and the infinitesimals are related by so putting these equations together
that is,
Taking the square root and rearranging
which can be integrated immediately with the substitution
,
to give just
,
which integrates trivially to , with a constant of integration, or
,
choosing the origin at , which makes .
But of course what we want is the curve shape , not . We need to eliminate in favor of That is, we need to writeas a function of , then substitute .
Recall one of our first equations was for the slope and putting that together with gives
,
integrating to
.
This is the desired equation for the catenary curve
We've dropped the possible constant of integration, which is just the vertical positioning of the origin.
Question: is this the same as the curve of the chain in a suspension bridge?
(Notice the vertical ropes are uniformly spaced horizontally.)
### Ideal Arches
Now let's consider some upside-down curves, arches. Start with a Roman arch, an upside-down U.
Typically, Roman arches are found in sets like this, but let's consider one free standing arch. Assume it's made of blocks having the same cross-section throughout. What is the force between neighboring blocks? We'll do an upside-down version of the chain force analysis we just did.
Equating the pressure forces on the arch segment colored dark in the figure, we see pressure on the lowest block in the segment must have a horizontal componentto balance the forced at the top point, so the cement between blocks is under shear stress, or, for no cement, there's a strong sideways static frictional force. (So a series of arches, as shown in the photograph above, support each other with horizontal pressure.)
A single Roman arch like this is therefore not an ideal design -- it could fall apart sideways.
Let's define an ideal arch as one that doesn't have a tendency to fall apart sideways, outward or inward. This means no shear (sideways) stress between blocks, and that means the pressure force between blocks in contact is a normal force -- it acts along the line of the arch. That should sound familiar! For a hanging string, obviously the tension acts along the line of the string.
Adding to our ideal arch definition that the blocks have the same mass per unit length along the entire arch, you can perhaps see that the static force balance for the arch is identical to that for the uniform hanging string, except that everything's reversed -- the tension is now pressure, the whole thing is upside down.
Nevertheless, apart from the signs, the equations are mathematically identical, and the ideal arch shape is a catenary. Of course, some actual constructed arches, like the famous one in St. Louis, do not have uniform mass per unit length (It's thicker at the bottom) so the curve deviates somewhat from the ideal arch catenary.
Here's a picture of catenary arches -- these arches are in Barcelona, in a house designed by the architect Gaudi.
In fact, Gaudi designed a church in Barcelona using a web of strings and weights to find correct shapes for the arches -- of course, the actual building would have a shape like the reflection in a horizontal mirror above the strings -- here's how he did it:
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Newton's Laws of Motion
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by
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on 17 November 2013
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Transcript of Newton's Laws of Motion
Newton's Laws of Motion
Betsy Olson
Period 2
Physics
11/16/2013
Newton's First Law
When something is moving, it will keep moving unless something stops it. When something is not moving, it will not move until something from outside it pushes it.
In this video, the dominoes are not moving until the finger pushes one domino. The rest of the dominoes still aren't moving until the next domino touches each one. Finally, the ground stops each domino from moving after it falls.
Newton's Second Law
In order for something to accelerate, a force has to push or pull on it. The greater the mass of the thing being pushed, the greater the force needed.
In this picture, the snowball accelerated because the woman's hand pushed on it. It wasn't a very large snowball, so it didn't require much force to accelerate.
Photo by Erika Sitton
Newton's Third Law
All "actions" have "reactions" that are equal and opposite.
The hand pulling on the rope has an equal and opposite force pulling back on it.
Photo from Olson family
Lab Idea
Let's test Newton's Second Law: the greater the mass, the greater the force needed to accelerate it
Get an air-filled balloon, a bouncy ball, and a basketball. Hold each type of ball in one palm, and lightly push on it with a finger from your other hand. Which ball accelerates the easiest?
We'd expect the balloon to accelerate most and the basketball to accelerate least.
Full transcript
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1. Inverse Laplace
$
\frac {2x^{2s-1}}{s^3(2s-1)}
$
Well i know what the inverse of
$
\frac {2}{s^3(2s-1)}
$
is but i Dont know what to do with the $x^{2s-1}$
Any guidance would be very appreciated
Thank you
2. Originally Posted by gconfused
$
\frac {2x^{2s-1}}{s^3(2s-1)}
$
Well i know what the inverse of
$
\frac {2}{s^3(2s-1)}
$
is but i Dont know what to do with the $x^{2s-1}$
Any guidance would be very appreciated
Thank you
I will take a shot.
Assume that $x \ne 0$ and x is NOT a function of s then
$
\frac{2}{x} \cdot \frac {x^{2s}}{s^3(2s-1))}=\frac{1}{x} \cdot \frac{e^{s(2\ln(x))}}{s^3(2s-1)}
$
Now this is in the form of a Heaviside function
I hope this helps.
$\frac{1}{x}\mathcal{U}(t+2\ln(x) \mathcal{L}^{-1}\left( \frac{1}{s^3(2s-1)}\right) \bigg|_{t \to t+2\ln(x)}$
Where
$\mathcal{U}(t-a)=\begin{cases} 0, t < a \\ 1,t>a \end{cases}$
3. Oh wow thanks, i get it now, you helped me heaps
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# Question 07316
Sep 13, 2015
$p H = 0.35$
#### Explanation:
First we find how many moles of $H C l$ we have initially.
$c = \frac{n}{v}$
So: $n = c \times v$
So initial moles $H C l = 25.0 \times 0.723 = 18.075 \text{mmol}$
(its mmol since there are 1000ml in 1L)
The alkali neutralises the acid:
$H C {l}_{\left(a q\right)} + K O {H}_{\left(a q\right)} \rightarrow K C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$
This tells use that for every mole of $K O H$ added, 1 mole of $H C l$ is neutralised.
The no. moles $K O H$ added $= 10 \times 0.273 = 2.73 \text{mmol}$
So the no. moles $H C l$ remaining must be:
$18.075 - 2.73 = 15.34 \text{mmol"=15.34xx10^(-3)"mol}$
The new volume $= 25.0 + 10.0 = 35.0 \text{ml"=35xx10^(-3)"L}$
$c = \frac{n}{v}$
So the concentration of $H C l$ is now:
$\frac{15.34 \times \cancel{{10}^{- 3}}}{35 \times \cancel{{10}^{- 3}}}$
$= 0.44 \text{mol/L}$
$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right]$
So:
$p H = - \log \left[0.44\right]$
$p H = 0.35$
Sep 13, 2015
The pH will be $0.358$.
#### Explanation:
You're titrating hydrochloric acid, $\text{HCl}$, a strong acid, with potassium hydroxide, $\text{KOH}$, a strong base, which means that the pH of the resulting solution will be equal to $7$ at the equivalence point.
So right from the start, you know that if the neutralization is complete, the pH of the solution will be equal to $7$.
But is the equivalence point reached or not?
The balanced chemical equation for this neutralization reaction is
${\text{HCl"_text((aq]) + "KOH"_text((aq]) -> "KCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$
The $1 : 1$ mole ratio that exists between hydrochloric acid and potassium chloride tells you that the equivalence point is reached when the number of moles of the acid is equal to the number of moles of the base.
If you have fewer moles of base, the solution will remain acidic, which means that you can expect to have $\text{pH < 7}$.
If you have more moles of base, the solution will become basic, which corresponds to $\text{pH > 7}$.
Use the molarities and volumes of the two solution to find the number of moles of each
$n = \frac{C}{V} \implies n = C \cdot V$
${n}_{H C l} = \text{0.723 M" * 25.0 * 10^(-3)"L" = "0.01808 moles HCl}$
and
${n}_{K O H} = \text{0.273 M" * 10.0 * 10^(-3)"L" = "0.002730 moles KOH}$
This means that the reaction will completely consume the added potassium hydroxide, leaving you with
${n}_{H C l} = 0.01808 - 0.002730 = \text{0.01535 moles HCl}$
Since hydrochloric acid is a strong acid, it dissociates completely in aqueous solution to produce hydrogen ions, ${\text{H}}^{+}$, and chloride ions.
This means that the number of moles of hydrogen ions you have in solution will be equal to the number of moles of hydrochloric acid ($1 : 1$ mole ratio again).
The total volume of the solution will be
${V}_{\text{total}} = {v}_{H C l} + {V}_{K O H}$
${V}_{\text{total" = 25.0 + 10.0 = "35.0 mL}}$
The molarity of the hydrogen ions will be
["H"^(+)] = "0.01535 moles"/(35.0 * 10^(-3)"L") = "0.4386 M"
The pH of the solution will thus be
pH_"sol" = -log(["H"^(+)]) = - log(0.4386) = color(green)(0.358)#
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53499 in Hex
Welcome to 53499 in hex, our article explaining the 53499 decimal to hex conversion; hex is short for hexadecimal, and for decimal we sometimes use the abbreviation dec. 53499 decimal is usually denoted as 5349910, and the result in hexadecimal notation is commonly denoted in subscript 16.
Both, the denary (decimal) as well as the hex numeral for 53499 are place-value notations, aka positional notation numerations. Read on to find all about 53499 in hexadecimal.
53499 to Hex
53499 to hex is a base 10 to base 16 conversion which produces the following result:
5349910 = D0FB16
53499 in hex = D0FB
53499 decimal to hex = D0FB
Proof: 13×16^3 + 0x16^2 + 15×16^1 + 11×16^0 = 53499.
Note that D0FB16 means the same as 0xD0FB, the former notation is more common in math, whereas the later with the prefix 0x can frequently be seen in programming.
Conversions similar to 53499 base 10 to base 16, include, for example:
In the next part of this post we show you how to obtain 53499 in hex.
How to convert 53499 Decimal to Hexadecimal?
For the 53499 to hex conversion we employ the remainder method explained on our home page:
1. Divide 53499 by 16, record the integer part of the result as new quotient
2. Write down the remainder of 53499 over 16 in hexadecimal notation
3. Proceed the two steps above with the quotient until the result is 0
4. The result of 53499 to hex is the RTL sequence of the remainders: D0FB
If you like to convert a base 10 number different from fifty-three thousand, four hundred and ninety-nine to hexadecimal, then use our converter below. Simply insert your number, the result is calculated automatically.
Change Dec to Hex
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You have reached the final part of fifty-three thousand, four hundred and ninety-nine decimal in hex. In this article we have answered the following questions:
• How to convert 53499 to hex?
• What is 53499 in hexadecimal?
• How to convert 53499 base 10 to hexadecimal?
If you have a question about 53499 dec hex, or if you like to give us a feedback, then don’t hesitate filling in the comment form at the bottom, or getting in touch by email.
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https://roboticsknowledgebase.com/wiki/planning/frenet-frame-planning/
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# Trajectory Planning in the Frenet Space
The Frenet frame (also called the moving trihedron or Frenet trihedron) along a curve is a moving (right-handed) coordinate system determined by the tangent line and curvature. The frame, which locally describes one point on a curve, changes orientation along the length of the curve.
There are many ways to plan a trajectory for a robot. A trajectory can be seen as a set of time ordered state vectors x. The Frenet frame algorithm introduces a way to plan trajectories to maneuver a mobile robot in a 2D plane. It is specifically useful for structured environments, like highways, where a rough path, referred to as reference, is available a priori.
## Trajectory Planning in the Frenet Space
There are many ways to plan a trajectory for a robot. A trajectory can be seen as a set of time ordered state vectors x. The following algorithm introduces a way to plan trajectories to maneuver a mobile robot in a 2D plane. It is specifically useful for structured environments, like highways, where a rough path, referred to as reference, is available a priori.
The Frenet frame (also called the moving trihedron or Frenet trihedron) along a curve is a moving (right-handed) coordinate system determined by the tangent line and curvature. The frame, which locally describes one point on a curve, changes orientation along the length of the curve.
More formally, the Frenet frame of a curve at a point is a triplet of three mutually orthogonal unit vectors {T, N, B}. In three-dimensions, the Frenet frame consists of [1]: The unit tangent vector T, which is the unit vector in the direction of what is being modeled (like velocity), The [unit normal] (https://www.statisticshowto.com/unit-normal-vector/) N: the direction where the curve is turning. We can get the normal by taking the derivative of the tangent then dividing by its length. You can think of the normal as being the place the curve sits in [2]. The unit binormal B = T x N, which is the cross product of the unit tangent and unit normal.
The tangent and normal unit vectors span a plane called the osculating plane at F(s). In four-dimensions, the Frenet frame contains an additional vector, the trinormal unit vector [3]. While vectors have no origin in space, it’s traditional with Frenet frames to think of the vectors as radiating from the point of interest.
More details:here
## Algorithm
1. Determine the trajectory start state [x1,x2,θ,κ,v,a] The trajectory start state is obtained by evaluating the previously calculated trajectory at the prospective start state (low-level-stabilization). At system initialization and after reinitialization, the current vehicle position is used instead (high-level-stabilization).
2. Selection of the lateral mode Depending on the velocity v the time based (d(t)) or running length / arc length based (d(s)) lateral planning mode is activated. By projecting the start state onto the reference curve the the longitudinal start position s(0) is determined. The frenet state vector s,s˙,s¨,d,d′,d′′ can be determined using the frenet transformation. For the time based lateral planning mode, d˙,d¨ need to be calculated.
3. Generating the lateral and longitudinal trajectories Trajectories including their costs are generated for the lateral (mode dependent) as well as the longitudinal motion (velocity keeping, vehicle following / distance keeping) in the frenet space. In this stage, trajectories with high lateral accelerations with respect to the reference path can be neglected to improve the computational performance.
4. Combining lateral and longitudinal trajectories Summing the partial costs of lateral and longitudinal costs using J(d(t),s(t))=Jd(d(t))+ks⋅Js(s(t)) , for all active longitudinal mode every longitudinal trajectory is combined with every lateral trajectory and transformed back to world coordinates using the reference path. The trajectories are verified if they obey physical driving limits by subsequent point wise evaluation of curvature and acceleration. This leads to a set of potentially drivable maneuvers of a specific mode in world coordinates.
5. Static and dynamic collision check Every trajectory set is evaluated with increasing total costs if static and dynamic collisions are avoided. The trajectory with the lowest cost is then selected.
6. Longitudinal mode alternation Using the sign based (in the beginning) jerk a(0), the trajectory with the strongest decceleration or the trajectory which accelerates the least respectivel
“Frenet Coordinates”, are a way of representing position on a road in a more intuitive way than traditional (x,y) Cartesian Coordinates. With Frenet coordinates, we use the variables s and d to describe a vehicle’s position on the road or a reference path. The s coordinate represents distance along the road (also known as longitudinal displacement) and the d coordinate represents side-to-side position on the road (relative to the reference path), and is also known as lateral displacement. In the following sections the advantages and disadvantages of Frenet coordinates are compared to the Cartesian coordinates.
##Frenet Features
The image below[frenet path] depicts a curvy road with a Cartesian coordinate system laid on top of it, as well as a curved (continuously curved) reference path (for example the middle of the road).
The next image shows the same reference path together with its Frenet coordinates.
The s coordinate represents the run length and starts with s = 0 at the beginning of the reference path. Lateral positions relative to the reference path are are represented with the d coordinate. Positions on the reference path are represented with d = 0. d is positive to the left of the reference path and negative on the right of it, although this depends on the convention used for the local reference frame. The image above[frenet path] shows that curved reference paths (such as curvy roads) are represented as straight lines on the s axis in Frenet coordinates. However, motions that do not follow the reference path exactly result in non straight motions in Frenet coordinates. Instead such motions result in an offset from the reference path and therefore the s axis, which is described with the d coordinate. The following image shows the two different representations (Cartesian vs Frenet) To use Frenet coordinates it is required to have a continouosly smooth reference path. The s coordinate represents the run length and starts with s = 0 at the beginning of the reference path. Lateral positions relative to the reference path are are represented with the d coordinate. Positions on the reference path are represented with d = 0. d is positive to the left of the reference path and negative on the right of it, although this depends on the convention used for the local reference frame. The image above shows that curved reference paths (such as curvy roads) are represented as straight lines on the s axis in Frenet coordinates. However, motions that do not follow the reference path exactly result in non straight motions in Frenet coordinates. Instead such motions result in an offset from the reference path and therefore the s axis, which is described with the d coordinate. The following image shows the two different representations (Cartesian vs Frenet)
## Reference Path
Frenet coordinates provide a mathematically simpler representation of a reference path, because its run length is described with the s axis. This reference path provides a rough reference to follow an arbitrary but curvature continuous course of the road. To avoid collisions, the planner must take care of other objects in the environment, either static or dynamic. Such objects are usually not avoided by the reference path.
A reference path can be represented in two different forms although for all representations a run length information, which represents the s axis, is required for the transformation.
1. Polynome
2. Spline (multiple polynomes)
3. Clothoid (special polynome)
4. Polyline (single points with run length information)
Clothoid x(l)=c0+c1∗l
Polyline
## Transformation
The transformation from local vehicle coordinates to Frenet coordinates is based on the relations. Given a point PC in the vehicle frame search for the closest point RC on the reference path. The run length of RC, which is known from the reference path points, determins the s coordinate of the transformed point PF. If the reference path is sufficiently smooth (continuously differentiable) then the vector PR→ is orthogonal to the reference path at the point RC. The signed length of PR→ determines the d coordinate of PF. The sign is positive, if PC
lies on the left along the run lenght of the reference path.
The procedure to transform a point PF from Frenet coordinates to the local vehicle frame in Cartesian coordinates is analogous. First, the point RC, which lies on the reference path at run length s. Next, a normal unit vector d⃗ is determined, which, in this point, is orthogonal to the reference path. The direction of this vector points towards positive d values and therefore points to the left with increasing run length s. Therefore, the vector d⃗ depends on the run length, which leads to: PC(s,d)=RC(s)+d⋅d⃗ (s)(2)
#### Images and Video
Images and embedded video are supported.
## Summary
The given article describes in detail what is Frenet Frame and how robot motion planning is done. The importance and relevance of frenet frame with path planning in systems engineering is also highlighted in the given article.
## References
https://fjp.at/posts/optimal-frenet/#frenet-coordinates
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https://www.homeworkmarket.com/content/1the-math-grades-final-exam-varied-greatly-using-scores-below-how-many-scores-were-within-on
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# 1.The math grades on the final exam varied greatly. Using the scores below, how many scores were within one standard deviation of the mean?
1.The math grades on the final exam varied greatly. Using the scores below, how many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean?
99 34 86 57 73 85 91 93 46 96 88 79 68 85 89
2.The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%?
3.If you know the standard deviation, how do you find the variance?
4.To get the best deal on a stereo system, Louis called eight appliance stores and asked for the cost of a specific model. The prices he was quoted are listed below:
\$216 \$135 \$281 \$189 \$218 \$193 \$299 \$235
Find the standard deviation.
5.A company has 70 employees whose salaries are summarized in the frequency distribution below.
Salary Number of Employees
5,001–10,000 8
10,001–15,000 12
15,001–20,000 20
20,001–25,000 17
25,001–30,000 13
a.Find the standard deviation.
b.Find the variance.
6. Calculate the mean and variance of the data. Show and explain your steps. Round to the nearest tenth.
14, 16, 7, 9, 11, 13, 8, 10 7
.Create a frequency distribution table for the number of times a number was rolled on a die. (It may be helpful to print or write out all of the numbers so none are excluded.)
3, 5, 1, 6, 1, 2, 2, 6, 3, 4, 5, 1, 1, 3, 4, 2, 1, 6, 5, 3, 4, 2, 1, 3, 2, 4, 6, 5, 3, 1
8.Answer the following questions using the frequency distribution table you created in No. 7.
a.Which number(s) had the highest frequency?
b.How many times did a number of 4 or greater get thrown?
c.How many times was an odd number thrown?
d.How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown?
9.The wait times (in seconds) for fast food service at two burger companies were recorded for quality assurance. Using the data below, find the following for each sample.
a.Range
b.Standard deviation
c.Variance
Lastly, compare the two sets of results.
Company Wait times in seconds
Big Burger Company 105 67 78 120 175 115 120 59
The Cheesy Burger 133 124 200 79 101 147 118 125
10.What does it mean if a graph is normally distributed? What percent of values fall within 1, 2, and 3, standard deviations from the mean?
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# Turán's theorem
In graph theory, Turán's theorem bounds the number of edges that can be included in an undirected graph that does not have a complete subgraph of a given size. It is one of the central results of extremal graph theory, an area studying the largest or smallest graphs with given properties, and is a special case of the forbidden subgraph problem on the maximum number of edges in a graph that does not have a given subgraph.
An example of an ${\displaystyle n}$-vertex graph that does not contain any ${\displaystyle (r+1)}$-vertex clique ${\displaystyle K_{r+1}}$ may be formed by partitioning the set of ${\displaystyle n}$ vertices into ${\displaystyle r}$ parts of equal or nearly equal size, and connecting two vertices by an edge whenever they belong to two different parts. The resulting graph is the Turán graph ${\displaystyle T(n,r)}$. Turán's theorem states that the Turán graph has the largest number of edges among all Kr+1-free n-vertex graphs.
Turán's theorem, and the Turán graphs giving its extreme case, were first described and studied by Hungarian mathematician Pál Turán in 1941.[1] The special case of the theorem for triangle-free graphs is known as Mantel's theorem; it was stated in 1907 by Willem Mantel, a Dutch mathematician.[2]
## Statement
Turán's theorem states that every graph ${\displaystyle G}$ with ${\displaystyle n}$ vertices that does not contain ${\displaystyle K_{r+1}}$ as a subgraph has a number of edges that is at most
${\displaystyle {\frac {r-1}{r}}\cdot {\frac {n^{2}}{2}}=\left(1-{\frac {1}{r}}\right)\cdot {\frac {n^{2}}{2}}.}$
The same formula gives the number of edges in the Turán graph ${\displaystyle T(n,r)}$ , so it is equivalent to state Turán's theorem in the form that, among the ${\displaystyle n}$ -vertex simple graphs with no ${\displaystyle (r+1)}$ -cliques, ${\displaystyle T(n,r)}$ has the maximum number of edges.[3]
## Proofs
Aigner & Ziegler (2018) list five different proofs of Turán's theorem.[3]
Turán's original proof uses induction on the number of vertices. Given an ${\displaystyle n}$ -vertex ${\displaystyle K_{r+1}}$ -free graph with more than ${\displaystyle r}$ vertices and a maximal number of edges, the proof finds a clique ${\displaystyle K_{r}}$ (which must exist by maximality), removes it, and applies the induction to the remaining ${\displaystyle (n-r)}$ -vertex subgraph. Each vertex of the remaining subgraph can be adjacent to at most ${\displaystyle r-1}$ clique vertices, and summing the number ${\displaystyle (n-r)(r-1)}$ of edges obtained in this way with the inductive number of edges in the ${\displaystyle (n-r)}$ -vertex subgraph gives the result.[1][3]
A different proof by Paul Erdős finds the maximum-degree vertex ${\displaystyle v}$ from a ${\displaystyle K_{r+1}}$ -free graph and uses it to construct a new graph on the same vertex set by removing edges between any pair of non-neighbors of ${\displaystyle v}$ and adding edges between all pairs of a neighbor and a non-neighbor. The new graph remains ${\displaystyle K_{r+1}}$ -free and has at least as many edges. Repeating the same construction recursively on the subgraph of neighbors of ${\displaystyle v}$ eventually produces a graph in the same form as a Turán graph (a collection of ${\displaystyle p}$ independent sets, with edges between each two vertices from different independent sets) and a simple calculation shows that the number of edges of this graph is maximized when all independent set sizes are as close to equal as possible.[3][4]
Motzkin & Straus (1965) prove Turán's theorem using the probabilistic method, by seeking a discrete probability distribution on the vertices of a given ${\displaystyle K_{r+1}}$ -free graph that maximizes the expected number of edges in a randomly chosen induced subgraph, with each vertex included in the subgraph independently with the given probability. For a distribution with probability ${\displaystyle p_{i}}$ for vertex ${\displaystyle v_{i}}$ , this expected number is ${\displaystyle \textstyle \sum _{(v_{i},v_{j})\in E(G)}p_{i}p_{j}}$ . Any such distribution can be modified, by repeatedly shifting probability between pairs of non-adjacent vertices, so that the expected value does not decrease and the only vertices with nonzero probability belong to a clique, from which it follows that the maximum expected value is at most ${\displaystyle (r-1)/2r}$ . Therefore, the expected value for the uniform distribution, which is exactly the number of edges divided by ${\displaystyle 1/n^{2}}$ , is also at most ${\displaystyle (r-1)/2r}$ , and the number of edges itself is at most ${\displaystyle n^{2}(r-1)/2r}$ .[3][5]
A proof by Noga Alon and Joel Spencer, from their book The Probabilistic Method, considers a random permutation of the vertices of a ${\displaystyle K_{r+1}}$ -free graph, and the largest clique formed by a prefix of this permutation. By calculating the probability that any given vertex will be included, as a function of its degree, the expected size of this clique can be shown to be ${\displaystyle \textstyle \sum 1/(n-d_{i})}$ , where ${\displaystyle d_{i}}$ is the degree of vertex ${\displaystyle v_{i}}$ . There must exist a clique of at least this size, so ${\displaystyle \textstyle r\geq \sum 1/(n-d_{i})}$ . Some algebraic manipulation of this inequality using the Cauchy–Schwarz inequality and the handshaking lemma proves the result.[3] See Method of conditional probabilities § Turán's theorem for more.
Aigner and Ziegler call the final one of their five proofs "the most beautiful of them all"; its origins are unclear. It is based on a lemma that, for a maximal ${\displaystyle K_{r+1}}$ -free graph, non-adjacency is a transitive relation, for if to the contrary ${\displaystyle (u,v)}$ and ${\displaystyle (v,w)}$ were non-adjacent and ${\displaystyle (u,w)}$ were adjacent one could construct a ${\displaystyle K_{r+1}}$ -free graph with more edges by deleting one or two of these three vertices and replacing them by copies of one of the remaining vertices. Because non-adjacency is also symmetric and reflexive (no vertex is adjacent to itself), it forms an equivalence relation whose equivalence classes give any maximal graph the same form as a Turán graph. As in the second proof, a simple calculation shows that the number of edges is maximized when all independent set sizes are as close to equal as possible.[3]
## Mantel's theorem
The special case of Turán's theorem for ${\displaystyle r=2}$ is Mantel's theorem: The maximum number of edges in an ${\displaystyle n}$ -vertex triangle-free graph is ${\displaystyle \lfloor n^{2}/4\rfloor .}$ [2] In other words, one must delete nearly half of the edges in ${\displaystyle K_{n}}$ to obtain a triangle-free graph.
A strengthened form of Mantel's theorem states that any Hamiltonian graph with at least ${\displaystyle n^{2}/4}$ edges must either be the complete bipartite graph ${\displaystyle K_{n/2,n/2}}$ or it must be pancyclic: not only does it contain a triangle, it must also contain cycles of all other possible lengths up to the number of vertices in the graph.[6]
Another strengthening of Mantel's theorem states that the edges of every ${\displaystyle n}$ -vertex graph may be covered by at most ${\displaystyle \lfloor n^{2}/4\rfloor }$ cliques which are either edges or triangles. As a corollary, the graph's intersection number (the minimum number of cliques needed to cover all its edges) is at most ${\displaystyle \lfloor n^{2}/4\rfloor }$ .[7]
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# sudoku /
Filename Size Date modified Message
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# sudoku!
A history of free-time sudoku development. Created to squash the urge to ever do
a sudoku by hand. Sudoku's rules are simple.
See Peter Norvig's Solving Every Sudoku Puzzle
for a general idea of how to solve them programmatically.
## Limitations
• If a puzzle is unsolvable, there's no feedback as to why.
• If there is more than one solution, the solver stops after finding the first.
## Evolution
### sudoku-bigmodel-cloned.html
My first attempt included a literal translation of the possible sudoku outcomes -
81 cells with the nine possible values marked true (still possible) or false
(eliminated).
To mark a value x, every other value in its cell is eliminated. x is
then eliminated in every other cell in the row, column, and section/block/region.
```function markImpl(row,col,val,m){
var v = (1 * val) - 1;
for(var i=0;i<9;i++){
//everything else in the cell
if(v != i){
m[row][col][i] = false;
}
//everything in the row
if(col != i){
markOne(row,i,v,m);
}
//everything in the column
if(row != i){
markOne(i,col,v,m);
}
}
m[row][col].marked = 1;
//everything in the 'section'
var srow = Math.floor(row / 3) * 3;
var scol = Math.floor(col / 3) * 3;
for(var i = 0; i < 3; i++){
for(var j = 0; j < 3; j++){
if(srow + i != row && scol + j != col){
markOne(srow + i,scol + j,v,m);
}
}
}
}
```
To solve the puzzle, create a model and eliminate the first non-eliminated
value in the first unsolved cell. The remaining cells are solved by recursively
cloning the model and eliminating the first non-eliminated value in the next
unsolved cell. If no valid elimination remains, the recursion unwinds and tries
the next non-eliminated value.
As you might imagine, this approach is slow. At its deepest, the model is cloned
80 times. That's 81 cells * 9 values * 81 models, or 59049 booleans to solve a
simple puzzle. In addition, marking a single value involves three loops and five
comparisons. I added small optimizations to avoid locking IE:
• track the possible values remaining (marked) to avoid scanning the
array on each operation.
• when eliminating values, do a secondary elimination when only one value remains
in a value array (versus waiting for the recursive step)
Regardless, this approach stinks.
### sudoku-smallmodel-cloned.html
My first improvement came from an article on the
eight queens puzzle.
The author (I can't recall the source) noticed that students usually modeled the
problem with a literal and naive two dimensional array where each cell represents
a square on the chess board. Since each column and row can contain only one queen,
you can shrink the model by using a column array and row array where each cell
indicates the queen's position in that column or row. (that's 8 + 8 data items versus
8 * 8 in the two-dimensional array). Dijkstra went one better and
showed you can represent positions with a single
array
(that's 8 items) where x[i] is the column in which the queen at row i resides.
Since I was using a literal and naive model, I figured I could shrink it. The
easiest reduction was to use three arrays, one per possible values in a row,
column, and section/block/region. To avoid scanning arrays, I also added a two
dimensional array to represent a cell's current value. All together, I could
represent the current state with 324 data items versus 512. Not great, but better.
Marking a value x doesn't require iteration. Simply flip the boolean per row,
column, and section. Then set the solved cell to x.
```model.solved[row][col] = x;
model.rows[row][x] = false;
model.cols[col][x] = false;
model.sect[sect][x] = false;
```
Solving works as before: recursively eliminate possibilties in cloned
representations until you find a solution.
```function brute(index, m){
if(index == 81){
return m;
}
var i = Math.floor(index / 9);
var j = index % 9;
if(m.solved[i][j] > -1){
return brute(index + 1,m);
}
var sect = section(i,j);
for(var k=0;k<9;k++){
if(m.rows[i][k] && m.cols[j][k] && m.sect[sect][k]){
var clone = \$.extend(true, {}, m);
mark(i,j,k+1,clone);
var result = brute(index + 1,clone);
if(result != null){
return result;
}
}
}
return null;
}
```
This approach is better - marking values is much faster and the model is smaller
- but it still leaves me cold.
### sudoku-smallmodel.html
The next improvement was indirectly inspired by Donald Knuth's
algorithm. It turns out there's a cottage industry dedicated to solving sudoku
explanation
is excellent.
While studying Dancing Links and Exact Cover
problems, I realized a key property of Dancing Links is easy reversibility.
This allows solution searching and backtracking on a single model. An operation
on my first model is not easily reversible because you have to scan rows, columns,
and sections to guarantee a value is not eliminated by more than one constraint.
This isn't a problem on my latest model. As long as you're careful not to mark
a value that's already marked, when x is marked it can be easily reversed:
```//rollback changes
model.solved[row][col] = -1;
model.rows[row][x] = true;
model.cols[col][x] = true;
model.sect[sect][x] = true;
```
With reversibility, the recursive search can be re-written without cloning the
model. This has several positive side-effects:
• Code is simplified
• The space required to compute a solution drops from a worst case 59049 data
items to 324
• It's fast
Dancing Links boasts impressive performance because of its unique representation
of options and constraints. Still, in the typical "hard" sudoku puzzle with
26 values pre-determined, my simple model and recursive search performs well
enough. It has the added benefit of being extremely easy to follow and use.
### sudoku.js, sudokuUI.js
My final product (sudoku isn't actually that interesting, so I don't imagine
continued development) fixes a few glaring problems in my prototypes:
• Eliminates the one big ugly HTML file with embedded JavaScript
• Separates the solver from the UI
• Reduces JavaScript global namespace pollution
• Fixes a bug that allowed an invalid state to be loaded (the same value in a
row, column, or section). This caused the solver to hang as it futility searched
for a solution with the remaining options.
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# Acceleration In One Dimension
1. Sep 25, 2004
### Faiza
****acceleration In One Dimension***
Hi everyone, I need some major help right now, I dun understand some concepts, im a first time physics student. Can an object have a negative acceleration and be speeding up?
2. Sep 26, 2004
### Jovaro
Yeah, if it is going in the negative direction. E.g backwards
3. Sep 26, 2004
### modmans2ndcoming
it is all based on your frame of reference. that means that if you call the direction that the object is accelerating in the negative direction, then it will have a negative acceleration and be increasing its speed.
frames of reference are important parts of setting up Physics problems. it is best to think about the problem before hand and set it up so that its final destination or the direction is is mostly moving in (the ground, point b, etc) is in a positive direction. it makes the problems much easier to do.
4. Sep 26, 2004
### recon
I don't mean to hijack the thread, but I also have a similar question to ask: What is the difference between deceleration and negative acceleration? Are they the same thing?
5. Sep 26, 2004
### Faiza
Deceleration means slowing down
Acceleration means speeding up
I think lol im a first time physics student i dun know :(
6. Sep 26, 2004
### Jovaro
I have never seen deceleration in my physics book. Always acceleration whether it is positive or negative.
7. Sep 26, 2004
### arildno
The pair acceleration/deceleration are "layman definitions" in that the pair is not used as a technical distinction, but as a fancy way of saying speeding up/slowing down
(a pair I prefer to ac/dec).
8. Sep 26, 2004
### Tom McCurdy
deceleration means you are in fact slowing down it is more of a broad term
negitive acceleration just means you are speeding up in the direction opposite to the one defined as positive. But you could I guess say that you were decelerating from -25 m/s^2 to -20 m/s^2
deceleration really can mean different things based upon how you use them. I generally think of it as something that means your speed is slowing down
aka closer to 0 m/s
9. Sep 26, 2004
### Faiza
Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone
straight down with a speed of 12.0 m/s. He throws another pebble straight
upward with the same speed so that it misses the edge of the bridge on the way
back down and falls into the river. For each stone find (a) the velocity as it
reaches the water and (b) the average velocity while it is in flight.
Note: Ignore the affects of air resistance.
(a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ?????
Vf2 = Vi2 + 2a (ΔX)
Vf2 = (12.0 m/s)2 + 2(9.80 m/s2) (20 m)
Vf2 = (144 m2/s2) + (19.6 m/s2) (20 m)
Vf2 = (144 m2/s2) + (392 m2/s2)
Vf2 = (536 m2/s2)
Vf = 23.15167381 m/s
Vf = 23 m/s
Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height.
(b) *AVERAGE VELOCITY*= ΔX/Δt
STONE
Xf = 20 m Vxf = 23 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????
Vxf = Vxi + ax t Vavg= ΔX/Δt
23 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20m/1.12244898 s
23 m/s – 12.o m/s = (9.80 m/s2) t Vavg= 17.81818181 m/s
11 m/s = (9.80 m/s2) t Vavg= 17.81 m/s
t = 11 m/s .
9.80 m/s2
t = 1.12244898 s
t = 1.12 s
(b) *AVERAGE VELOCITY*= ΔX/Δt
PEBBLE
Xf = 20 m Vxf = 12.0 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????
Vxf = Vxi + ax t
12.0 m/s = 12.0 m/s + (9.80 m/s2) t
12.0 m/s – 12.0 m/s= (9.80 m/s2) t
0 m/s = (9.80 m/s2) t
0 m/s = t
9.80 m/s2
t = 0 m/s
CHECK MY ASNWER???
10. Sep 26, 2004
### Faiza
Acceleration vs. Deceleration
How can you tell if the object is speeding up (acceleration) or slowing down (deceleration)? Speeding up means that the magnitude (the value) of the velocity is increasing. For instance, an object with a velocity changing from +3 m/s to + 9 m/s is speeding up. Similarly, an object with a velocity changing from -3 m/s to -9 m/s is also speeding up. In each case, the magnitude of the velocity (the number itself, not the sign or direction) is increasing; the speed is getting larger.
Given this fact, an object is speeding up if the line on a velocity-time graph is changing from a location near the 0-velocity point to a location further away from the 0-velocity point. That is, if the line is moving away from the x-axis (the 0-velocity point), then the object is speeding up. Conversely, if the line is moving towards the x-axis, the object is slowing down.
11. Sep 26, 2004
### arildno
If
$$\vec{v}\cdot\vec{a}\geq{0}$$
then the object "accelerates", otherwise it "decelerates".
($$\vec{v}$$ velocity, $$\vec{a}$$ rate of change of velocity (the proper meaning of acceleration)
12. Sep 26, 2004
### robphy
Some synonyms:
deceleration, "slows you down" , "reduces your speed"
symbolically (and without reference to coordinates): $$\vec a\cdot \vec v < 0$$.
Proof:
$$0>\vec a\cdot \vec v =\frac{d \vec v}{dt}\cdot \vec v=\frac{1}{2} \frac{d (\vec v \cdot \vec v)}{dt}=\frac{1}{2}\frac{d (v^2)}{dt}= \frac{1}{2}\frac{d}{dt}((\text{speed})^2)$$
So,
$$0>\frac{d}{dt}(\text{speed})$$
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## Reduction Formulae
We can derive iterative formulae for certain integrals to express am integration in terms of a simpler integration. By doing this repeatedly an integral can often be reduced to a very simple integral.
Example:
$I= \int^{\infty}_0 x^ne^{-x} dx$
.
Let
$I_n= \int^{\infty}_0 x^ne^{-x} dx$
. Integration by parts gives
\begin{aligned} I_n &= \int^{\infty}_0 x^ne^{-x} dx \\ &= [-nx^{n-1} e^{-x} ]^{\infty}_0 -(-n \int^{\infty}_0 x^{n-1}e^{-x} dx \\ &= n \int^{\infty}_0 x^{n-1}e^{-x} dx \\ &=nI_{n-1} \end{aligned}
.
Repeatedly application of integration by parts gives
$I_n= \int^{\infty}_0 x^ne^{-x} dx=n! \int^{\infty}_0 e^{-x}= n! [-e^{-x} ]^{\infty}_0 = n!$
.
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## How Mortgage Loans Work
Excluding property taxes and insurance, a traditional fixed-rate mortgage payment consist of two parts: (1) interest on the loan and (2) payment towards the principal, or unpaid balance of the loan.
Many people are surprised to learn, however, that the amount you pay towards interest and principal varies dramatically over time. This is because mortgage loans work in such a way that the early payments are primarily in interest, and the later payments are primarily towards the principal.
In the beginning... you pay interest
To help calculate monthly payments for loans based on different interest rates, lenders long ago developed what are known as "amortization tables." These tables also make it fairly easy to calculate how much money of each payment is interest, and how much goes towards the principal balance.
For example, let`s calculate the principle and interest for the very first monthly payment of a 30-year, \$100,000 mortgage loan at 7.5 percent interest. According to the amortization tables, the monthly payment on this loan is fixed at \$699.21.
The first step is to calculate the annual interest by multiplying \$100,000 x .075 (7.5 %). This equals \$7,500, which we then divide by 12 (for the number of months in a year), which equals \$625.
If you subtract \$625 from the monthly payment of \$699.21, we see that:
• \$625 of the first payment is interest
• \$74.21 of the first payment goes towards the principal
Next, if we subtract \$74.21 (the first principal payment) from the \$100,000 of the loan, we come up with a new unpaid principal balance of \$99,925.79. To determine the next month`s principal and interest payments, we just repeat the steps already described.
Thus, we now multiply the new principal balance (99,925.79) times the interest rate (7.5%) to get an annual interest payment of \$7,494.43. Divided by 12, this equals \$624.54. So during the second month`s payment:
• \$624.54 is interest
• \$74.67 goes towards the principal.
Equity
As you can see from the above example, even though you pay a lot of interest up front, you`re also slowly paying down the overall debt. This is known as building equity. Thus, even if you sell a house before the loan is paid in full, you only have to pay off the unpaid principal balance--the difference between the sales price and the unpaid principle is your equity.
In order to build equity faster--as well as save money on interest payments--some homeowners choose loans with faster repayment schedules (such as a 15-year loan).
Time versus savings
To help illustrate how this works, consider our previous example of a \$100,000 loan at 7.5 percent interest. The monthly payment is around \$700, which over 30 years adds up to \$252,000. In other words, over the life of the loan you would pay \$152,000 just in interest.
With the aggressive repayment schedule of a 15-year loan, however, the monthly payment jumps to \$927-for a total of \$166,860 over the life of the loan. Obviously, the monthly payments are more than they would be for a 30-year mortgage, but over the life of the loan you would save more than \$85,000 in interest.
Bear in mind that shorter term loans are not the right answer for everyone, so make sure to ask your lender or real estate agent about what loan makes the best sense for your individual situation.
Call Denise Christinzio for more help and understanding more about mortgages and get pre-qualfied for a mortgage at 856-227-1122 or 856-589-4848 x2005..cell 609-206-6186 or email me at: [email protected]
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0
# What are the common multiples of 16 and 48?
Updated: 4/28/2022
Wiki User
8y ago
The common multiples of 16 and 48 are any multiple of 48. Thus, the common multiples are 48, 96, 144, 192, 240, etc.
Wiki User
8y ago
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Related questions
### What is the least common multiples 12 and 16?
Least common multiples 12 and 16 is 48.
### What are the common multiples of 3 and 16?
The common multiples of 3 and 16 are all multiples of their LCM, which is 48. Thus, their common multiples are 48, 96, 144, 192, 240, 288, etc.
### What are some common multiples of 4 and 16?
All multiples of 16 are common. 16, 32, 48 etc
### What is the least common multiples for 6 and 16?
48 8*6 = 48 3*16 = 48
### What are the multiples of 16 and 24?
The multiples of 16 are 16, 32, 48, 64, 80, 96, 112, 128, 144, . . . The multiples of 24, 48, 72, 96, 120, 144, 168, 192, 216, . . . The common multiples are any multiples of 48.
### Common multiple of 12 and 16?
48 is the lowest common multiple. Other common multiples will be the multiples of 48: 2 x 48 = 96, 3 x 48 = 144, etc.48 and every multiple of 48.
### What are the three least common multiples of 12 and 16?
The LCM is 48. Other common multiples include 96 and 144.
### What are the multiples of 12 and 16?
The multiples of 12 are 12, 24, 36, 48 and keep on adding 12.The multiples of 16 are 16, 32, 48, 64 and keep on adding 16.If you look at the two of these, you can see there is a least common multiple. A least common multiple is the smallest multiple that are shared by two (or more) numbers. If you list the multiples down, you can see 48 are in both, so 48 is the least common multiple for 12 and 16.
48 and 96
### What are common multiples of 12 and 16?
Any multiple of 48.
### What are the common multiples of 12 and 16?
48 is a common multiple48, 96, 144 and so on.
### What are common multiples of 2 and 16?
16, 32, 48, 64 and so on.
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15,991,686 members
Articles / General Programming / Algorithms
Tip/Trick
# Levenshtein Algorithm in Visual Basic
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3 Jan 2015CPOL1 min read 28.1K 6 6
How to calculate the distance between two strings according to Levenshtein algorithm
## Introduction
Levenshtein's distance measures the minimum number of character edits required to change a word into another. In this tip, we'll see a simple implementation of the Levenshtein algorithm in Visual Basic. It will be useful in several situations, when managing - for example - large amount of text, and we are in need of fast and massive modifications in our data, to spot and correct typos, or to match strings in terms of similarity, and not simply in terms of equal/not equal. Levenshtein's algorithm allows to compute a score regarding string similarity. We'll see how in a moment.
## Standard Levenshtein Algorithm
Here follows the standard Levenshtein implementation in VB.NET, according to the algorithm as shown at Wikipedia.
VB.NET
Public Function Levenshtein(ByVal s As String, ByVal t As String) As Integer
Dim n As Integer = s.Length
Dim m As Integer = t.Length
Dim d(n + 1, m + 1) As Integer
If n = 0 Then
Return m
End If
If m = 0 Then
Return n
End If
Dim i As Integer
Dim j As Integer
For i = 0 To n
d(i, 0) = i
Next
For j = 0 To m
d(0, j) = j
Next
For i = 1 To n
For j = 1 To m
Dim cost As Integer
If t(j - 1) = s(i - 1) Then
cost = 0
Else
cost = 1
End If
d(i, j) = Math.Min(Math.Min(d(i - 1, j) + 1, d(i, j - 1) + 1), d(i - 1, j - 1) + cost)
Next
Next
Return d(n, m)
End Function
## Examples
Using the above function(s) is trivial: it's sufficient to call it by passing two strings, plus the optional boolean flag for case-sensitivity check. Some examples could be:
VB.NET
MsgBox(Levenshtein("Inwards", "inwards").ToString) ' Returns 1
MsgBox(Levenshtein("Inwards", "inwards").ToString) ' Returns 1
MsgBox(Levenshtein("towards", "towards").ToString) ' Returns 0
MsgBox(Levenshtein("dinner", "breakfast").ToString) ' Returns 9
MsgBox(Levenshtein("breakfast", "braekfast").ToString) ' Returns 2
MsgBox(Levenshtein("efficient", "sufficient").ToString) ' Returns 2
MsgBox(Levenshtein("grandma", "anathema").ToString) ' Returns 5
MsgBox(Levenshtein("aunt", "ant").ToString) ' Returns 1
The results (1, 0, 0, 9, 2, 2, 5, 1) are the Levenshtein's distances between given strings, i.e., a score regarding strings similarity. The lower the score, the nearer are the two entities. A value of zero means, obviously, a total convergence of the two. We could use a function like this (with predetermined conditions to be satisfied, like "the score must not exceed 3", for example) to correct typos (as in the "breakfast - braekfast" example), or to search for differences in hypothetical data (like "new York - New York"), and so on.
## History
• 2015-01-06 Added standard algorithm, revised text, revised code
• 2015-01-03 First release for CodeProject
Written By
Software Developer
Italy
Working in IT since 2003 as Software Developer for Essetre Srl, a company in Northern Italy.
I was awarded in 2014, 2015 and 2016 with Microsoft MVP, for Visual Studio and Development Technologies expertise. My technology interests and main skills are in .NET Framework, Visual Basic, Visual C# and SQL Server, but i'm proficient in PHP and MySQL also.
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My vote of 1 Etter Frédéric5-Jan-15 21:43 Etter Frédéric 5-Jan-15 21:43
Wrong code - Sorry Etter Frédéric5-Jan-15 21:42 Etter Frédéric 5-Jan-15 21:42
Re: Wrong code - Sorry Emiliano Musso5-Jan-15 22:21 Emiliano Musso 5-Jan-15 22:21
Re: Wrong code - Sorry Etter Frédéric6-Jan-15 2:46 Etter Frédéric 6-Jan-15 2:46
Re: Wrong code - Sorry Emiliano Musso6-Jan-15 6:02 Emiliano Musso 6-Jan-15 6:02
My vote of 1 Etter Frédéric5-Jan-15 21:36 Etter Frédéric 5-Jan-15 21:36
Wrong Algorithm Code contains error
Last Visit: 31-Dec-99 18:00 Last Update: 9-Sep-24 1:36 Refresh 1
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https://www.bytelearn.com/math-algebra-1/worksheet/find-slope-from-two-points
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# Find Slope From Two Points Worksheet
## 6 problems
Finding slope from two points is a fundamental concept in algebra and geometry. It involves calculating the slope of a line given two points on the line. The slope is defined as the ratio of the rise (change in y-coordinates) to the run (change in x-coordinates). This skill is essential for solving problems involving linear equations and graphing lines.
Algebra 1
Linear Relationship
## How Will This Worksheet on "Find Slope from Two Points" Benefit Your Student's Learning?
• Reinforces understanding of slope through regular practice.
• Develops critical skills in identifying coordinates and performing arithmetic operations.
• Enhances problem-solving and analytical abilities.
• Improves geometric interpretation of slope and line direction.
• Prepares students for advanced mathematical concepts
## How to Find Slope from a Graph?
• Choose two points on the line whose coordinates are
Show all
## Solved Example
Q. Find the slope of the line that passes through $(1, 5)$ and $(3, 6)$.$\newline$Write your answer in its simplest form.
Solution:
1. Identify Coordinates: Identify the coordinates of the two points.$\newline$Point $1$: $(x_1, y_1) = (1, 5)$$\newline$Point $2$: $(x_2, y_2) = (3, 6)$$\newline$We will use these coordinates to calculate the slope of the line.
2. Recall Slope Formula: Recall the formula for the slope $m$ of a line given two points: $m = \frac{y_2 - y_1}{x_2 - x_1}$. We will plug the coordinates of the two points into this formula to find the slope.
3. Substitute Coordinates: Substitute the coordinates into the slope formula. $m = \frac{6 - 5}{3 - 1}$
4. Perform Subtraction: Perform the subtraction in the numerator and the denominator.$\newline$$m = \frac{1}{2}$
5. Simplify Fraction: Simplify the fraction if necessary.$\newline$In this case, the fraction $\frac{1}{2}$ is already in its simplest form.
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# SAT Math : Perpendicular Lines
## Example Questions
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
The equation of line p is y= 1/4x +6. If line k contains the point (3,5) and is perpendicular to line p, find the equation of line k.
y = 1/4x + 17
y = -4x + 17
y = 4x - 17
y = 3x + 5
y = -4x + 17
Explanation:
Using the slope intercept formula, we can see the slope of line p is ¼. Since line k is perpendicular to line p it must have a slope that is the negative reciprocal. (-4/1) If we set up the formula y=mx+b, using the given point and a slope of (-4), we can solve for our b or y-intercept. In this case it would be 17.
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
In the xy-coordinate plane, a line A contains the points (0,0) and (3,1). If the line B is perpendicular to A at (3,1), what is the equation of the line?
y = -3x + 1
y = 3x + 1
y = -3x + 10
y = -1/3x + 10
y = 1/3x + 1
y = -3x + 10
Explanation:
First, you need to obtain the equation of the first line, A. Its slope is given by:
(y2 - y1) / (x2 - x1) = (1 - 0) / (3 - 0) = 1/3 = slope of A.
Remember that the slope of a perpendicular line to a given line is -1 times the inverse of its slope. Thus the slope of B:
(-1) x 1 / (1/3) = -3
Thus with y = mx + b, m = -3. Now the line must include (3,1). Thus:
with y = -3x + b:
1 = -3(3) + b;
1 = -9 + b; add 9 to both sides:
10 = b
### Example Question #31 : Lines
What line is perpendicular to the line 2x + 3y = 6 through (4, 1)?
2x + 3y = –5
–3x + 2y = –10
–2x + 3y = 5
–x + 4y = 8
3x + 2y = 10
–3x + 2y = –10
Explanation:
The given equation is in standard form, so it must be converted to slope-intercept form: y = mx + b to discover the slope is –2/3. To be perpendicular the new slope must be 3/2 (opposite reciprocal of the old slope). Using the new slope and the given point we can substitute these values back into the slope-intercept form to find the new intercept, –5. In slope-intercept form the new equation is y = 3/2x – 5. The correct answer is this equation converted to standard form.
### Example Question #41 : Coordinate Geometry
The endpoints of line segment AB are located at (5, –2) and (–3, 10). What is the equation of the line that is the perpendicular bisector of AB?
2x – 3y = –10
2x + 3y = 14
2x – 3y = –20
3x – 4y = –4
3x – 4y = –13
2x – 3y = –10
Explanation:
We are asked to find the equation of the line that is the perpendicular bisector of AB. If we find a point that the line passes through as well as its slope, we can determine its equation. In order for the line to bisect AB, it must pass through the midpoint of AB. Thus, one point on the line is the midpoint of the AB. We can use the midpoint formula to determine the midpoint of AB with endpoints (5, –2) and (–3, 10).
The x-coordinate of the midpoint is located at (5 + –3)/2 = 1.
The y-coordinate of the midpoint is located at (–2 + 10)/2 = 4.
Thus, the midpoint of AB is (1, 4).
So, we know that the line passes through (1,4). Now, we can use the fact that the line is perpendicular to AB to find its slope. The product of the slopes of two line segments that are perpendicular is equal to –1. In other words, if we multiply the slope of the line by the slope of AB, we will get –1.
We can use the slope formula to find the slope of AB.
slope of AB = (10 – (–2))/(–3 – 5) = 12/–8 = –3/2.
Since the slope of the line multiplied by –3/2 must equal –1, we can write the following:
(slope of the line)(–3/2) = –1
If we multiply both sides by –2/3, we will find the slope of the line.
The slope of the line = (–1)(–2/3) = 2/3.
Thus, the line passes through the ponit (1, 4) and has a slope of 2/3.
We will now use point-slope form to determine the line's equation. Let's let m represent the slope and (x1, y1) represent a ponit on the line.
y – y = m(x – x1)
y – 4 = (2/3)(x – 1)
Multiply both sides by 3 to get rid of the fraction.
3(y – 4) = 2(x – 1)
Distribute both sides.
3y – 12 = 2x – 2
Subtract 3y from both sides.
–12= 2x – 3y – 2
–10 = 2x – 3y.
The equation of the line is 2x – 3y = –10.
The answer is 2x – 3y = –10.
### Example Question #41 : Coordinate Geometry
A line passes through (2, 8) and (4, 15). What is a possible equation for a line perpendicular to this one?
y = (–2/7)x + 4
y = (2/7)x + 4
y = (–7/2)x + 4
y = (8/3)x + 4
y = (–2/7)x + 4
Explanation:
Remember, perpendicular lines have opposite-reciprocal slopes; therefore, let's first find the slope of our line. That is found by the equation: rise/run or y2 – y1/x2 – x1
Substituting in our values: (15 – 8)/(4 – 2) = 7/2
The perpendicular slope is therefore –2/7.
Since ANY perpendicular line will intersect with this line at some point. We merely need to choose the answer that has a line with slope –2/7. Following the slope intercept form (y = mx + b), we know that the coefficient of x will give us this; therefore our answer is: y = (–2/7)x + 4
### Example Question #43 : Coordinate Geometry
Line p is given by the equation y = –x + 4. Which of the following equations describes a line that is perpendicular to p?
x + y = –4
y = –x – 4
y = –4
y = x + 4
x + y = 4
y = x + 4
Explanation:
The equation of line p is given in the form y = mx + b, where m is the slope and b is the y-intercept. Because the equation is y = –x + 4, the slope is m = –1.
If two lines perpendicular, then the product of their slopes is equal to –1. Thus, if we call n the slope of a line perpendicular to line p, then the following equation is true:
m(n) = –1
Because the slope of line p is –1, we can write (–1)n = –1. If we divide both sides by –1, then n = 1. In short, the slope of a line perpendicular to line p must equal 1. We are looking for the equation of a line whose slope equals 1.
Let's examine the answer choices. The equation y = –x – 4 is in the form y = mx + b (which is called point-slope form), so its slope is –1, not 1. Thus, we can eliminate this choice.
Next, let's look at the line x + y = 4. This line is in the form Ax + By = C, where A, B, and C are constants. When a line is in this form, its slope is equal to –A/B. Therefore, the slope of this line is equal to –1/1 = –1, which isn't 1. So we can eliminate x + y = 4. Simiarly, we can eliminate the line x + y = –4.
The line y = –4 is a horizontal line, so its slope is 0, which isn't 1.
The answer is the line y = x + 4, because it is the only line with a slope of 1.
The answer is y = x + 4.
### Example Question #44 : Coordinate Geometry
Give the equation of a line perpendicular to .
Explanation:
The slope of a perpendicular line is the opposite reciprocal, therefore we are looking for a line with a slope of .
is the only answer choice that satisfies this criteria.
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
Find the equation of the line that is perpendicular to and passes through (5, 6).
Explanation:
We know that the slope of the original line is
Thus the slope of the perpendicular line is the negative reciprocal of , or –2.
Then we plug the slope and point (5, 6) into the form , which yields
When we simplify this, we arrive at
### Example Question #1 : How To Find The Equation Of A Perpendicular Line
What line is perpendicular to through ?
Explanation:
We need to find the slope of the given equation by converting it to the slope intercept form:
The slope is and the perpendicular slope would be the opposite reciprocal, or
The new equation is of the form and we can use the point to calculate . The next step is to convert into the standard form of .
### Example Question #47 : Coordinate Geometry
What line is perpendicular to and passes through ?
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https://homeoflearning.in/what-is-zero-correlation/
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# What is Zero correlation?
The degree or the extent of correlation between two variables is described by the value of the correlation coefficients. The degrees of correlation are: Perfect Correlation, Zero Correlation, Limited Degree of Correlation
## Zero correlation
If there is no relation between two variables, i.e. change in one variable has no effect on the change in the other, then the variable lacks correlation. The value of the correlation coefficient for zero correlation equals zero.
Note:Â A zero correlation between any two variables does not mean that there is no relationship at all between them. In fact, it should be interpreted that the 2 variables are not linearly related. However, it may be possible that the two variables may be nonlinearly related to each other.
#### Positive and Negative Correlation
A positive correlation between two variables exists when both of them move in the same direction And the negative correlation between two variables
#### What is Correlation?
Correlation is a statistical tool that measures the quantitative relationship between different variables. It studies the degree and intensity of the connection between the two variables. The relationship between two variables is studied with the help of a statistical tool i.e. ‘Correlation’.
#### Measures of Dispersion in Statistics
Dispersion measures the extent to which the different items tend to disperse away from the central tendency. In other words, while central tendency indicates a representative value, the measures of dispersion indicate the divergence of the values of different items from the central value.
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https://gowanusballroom.com/which-step-is-the-rate-determining-step-in-the-iodine-clock-reaction/
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## Which step is the rate determining step in the iodine clock reaction?
Thus the first reaction is the rate determining step. It is the slowest of the reactions in the overall reaction mechanism and ultimately determining the rate of the reaction. Thus, increasing the concentration of iodide, hydrogen peroxide, or acid (it neutralizes the hydroxide ion) will accelerate the reaction.
## What is the order of reaction for iodine clock?
The rate of reaction is first-order in potassium iodine. For the qualitative option, the details of the mechanism are not revealed to the students in order to have the students focus on the kinetics concepts of changing the concentration of one reactant versus time or reaction.
Why is starch used in iodine clock reaction?
The starch solution serves as an indicator of the end of the reaction by forming a deep-blue colored starch–iodine complex. The reaction time can thus be measured by not- ing the time until the appearance of the blue color for each trial.
### How do I find a first-order?
To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction.
### How do you know if a reaction is first order?
Which of the following decides order of reaction?
The correct answer is (D)Mechanism Of Reaction As Well As Relative Concentration Of Reactants.
## How do you calculate the rate of a reaction?
Reaction rate is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period Δt. The rate of reaction can be observed by watching the disappearance of a reactant or the appearance of a product over time.
## How do you calculate reaction rate with iodine concentration?
2 O 8 -) decreasesand the concentration of the products 2(I 2 and SO 4 -) increasesaccordingly. In this experiment the Reaction Rate will be calculated by dividing the experimentally determined increase in concentration of one of the products (elemental iodine, I 2 ), by the corresponding time interval: Δ I 2
What is the reaction order?
mis the Reaction Order with respect to Reactant A, and n: is the Reaction Order with respect to Reactant B. The values of the reaction orders (“m” and “n”) determine the dependence of the reaction rate on concentration of the respective reactants. Reaction orders commonly have one of the following values: 0, 1, -1, or 0.5.
### How can the presence of iodine in starch be monitored?
), during a corresponding time interval, can be easily monitored, since the presence of even small amounts of iodine can be detected by virtue of the intensely blue colored complex formed between iodine and starch. 2 I-(aq) + S 2 O 8 2(aq) I 2 (aq) + 2 SO
### What is rate constant and reaction order?
where: k: is a proportionality constant called the Rate Constant, mis the Reaction Order with respect to Reactant A, and n: is the Reaction Order with respect to Reactant B. The values of the reaction orders (“m” and “n”) determine the dependence of the reaction rate on concentration of the respective reactants.
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# How do you solve sin(x+60°) + cos(x+30°) = 1/2 in the range 0° < x < 360°?
May 2, 2015
$x = {73.22}^{\circ}$ and $x = 360 - 73.22 = {286.78}^{\circ}$
#### Explanation:
Use the trig identities:
$\sin \left(a + b\right) = \sin a . \cos b + \sin b . \cos a$
$\cos \left(a + b\right) = \cos a . \cos b - \sin a . \sin b$
$\sin \left(x + 60\right) = \left(\frac{1}{2}\right) . \sin x + \left(\frac{\sqrt{3}}{2}\right) . \cos x$
$\cos \left(x + 30\right) = \cos x . \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right) . \sin x$
Adding the 2 expressions, we get:
$\left(\sqrt{3}\right) . \cos x = \frac{1}{2} \to \cos x = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6}$
->$x = {73.22}^{\circ}$ and $x = 360 - 73.22 = {286.78}^{\circ}$
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# Cal
posted by on .
A plane flying horizontally at an altitude of 3 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 6 mi away from the station. (Round to the nearest whole number.)
• Cal - ,
Use Pythagoras theorem.
x=horizontal distance from station
L(x)=oblique distance from station
L(x)=√(480²+x²)
dx/dt = horizontal speed = 480 mph
Find dL(x)/dx by differentiation.
Then
dL(x)/dt
= dL(x)/dx * dx/dt
• Cal - ,
that makes no sense
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# Day 21: Super Bear
Geometry wrapped up Sam’s lesson on Conditional Statements and went through a gallery walk.
After we talked about what puzzled them and questions they had. One area students struggled was seeing relationships between the posters, they were hesitant to draw any conclusions which I see as a good thing. We debriefed with an example then went through some practice statements.
In Algebra students worked through Super Bear which is another of my favorites.
Here is a quick run down of the lesson:
I had students fold a scratch sheet of paper into 6 regions which will later be used for the following:
I then played the clip and asked students to write down the first question that came to their mind. They shared that with a neighbor then we put them up on the board. After each question I asked for a show of hands for anyone who also thought the question was interesting and put + that number at the end. (This helps me focus on the big idea – brain storming – some questions are not very relevant and that is OK.)
Typo… oh well.
Then students estimated how many mini and regular bears it would take and we threw those up on the board too with names attached.
After this students brainstormed what info they needed to answer the question.
Then I gave the info I had to ’em and set them loose to solve the problem.
We put our calculated answers on the board too, not so sure how productive this was. Students really wanted to just get at the answer and it felt like I was making way too many lists. I also went back to the estimates and we looked at who was the closest.
After I talked a little about rates and how it tied into this problem then students named the activity.
Once we wrapped this up I went back to the list of questions and addressed each of them, which brought some great closure to the period.
# Day 20: If… Then…
I spent a lot of time today talking about college and telling stories about my crazy roommates. It is important for students to see that I am not just a mean-lean math machine. Plus it was college day so every teacher was sharing out.
Algebra spent the rest of the time practicing absolute value and multi-step equations.
In Geometry we picked back up on Sam’s lesson. Students created big and little posters, we will gallery walk tomorrow.
A few take-aways:
• I did not emphasize that the black statement card is true; therefore their first statement needs to be an If Then version of it.
• Students started off slow but quickly picked up on what was happening.
• I am really excited for the gallery walk and giving students time tomorrow to work through and discuss which statements are logically equivalent.
• Some students mixed up the order of the statements, this will also be interesting.
# Day 19: Age Estimation
Introducing absolute value is one of my favorite lessons of the year.
Crazy.
Once again credit for this lesson goes out to Dan.
I pretty much followed his structure, just a few things are different which I will mention after we get through the good stuff:
Copy this down in your notes, you will have 21 rows.
For the first entry; students write down the name and their guess. I provide no structure or hints, we have practiced estimating a ton, let things happen.
Team Edward; how old do you think this guy is?
We get through all 21 slides then go through another 21 with their ages
After ask students to total how much they were off for each celebrity then add all those together; the person who was off by the least gets a homework pass. Let them run into problems, don’t take that away from them.
After a while students realize that there is one serious issue; under estimating an age. Say one student guessed Willie was 42 and another 44. Both off by 2, how should we deal with this?
No big deal; drop the negative.
We find the totals again; I made it a competition between my four algebra classes. The lowest 5 went up on the wall and will stay there for the rest of the year.
Finally, we moved into absolute value, smooth transition for students.
In Geometry students started Sam’s lesson on conditional statements. So much good happened here, we only made it through the sketching and first discussion. I couldn’t help but laugh at some of the statements in the slide show. More on this to come tomorrow; here are a few finished products in the mean time.
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A328975 Numbers whose trajectory under repeated application of the map in A053392 increases without limit. 4
1496, 1497, 1498, 1499, 1587, 1588, 1589, 1691, 1692, 1693, 1694, 1695, 1696, 1697, 1698, 1699, 1719, 1728, 1729, 1783, 1784, 1785, 1786, 1787, 1788, 1789, 1791, 1792, 1793, 1794, 1795, 1796, 1797, 1798, 1799, 1819, 1867, 1868, 1869, 1874, 1875, 1876, 1877, 1878, 1879, 1880, 1881, 1882, 1883, 1884, 1885 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Computed by Hans Havermann, Nov 01 2019. There is a simple technique, due to Hans Havermann, that proves that many terms that in this sequence blow up: find a term in the trajectory that has an internal substring of three 9's. See A328974 for the case 1496. The same reasoning shows that almost all numbers belong to the sequence. From Scott R. Shannon, Nov 23 2019: (Start) Although many of the numbers in this sequence eventually reach a term that contains three 9's, some do not. The first such example is 4949 which leads to 44444 after two steps, and starts a sequence of values that leads back to a much larger all-4's number every nine steps. No 9's appear in any of the intermediate values. Similar series found which lead to limitless loops are values with all 4's with a final 5, or all 3's with a final digit of 1 to 6. Of the first 33909 staring values that increase without limit 209 lead into one of these non-9 loops. (End) From Scott R. Shannon, Nov 24 2019: (Start) One can show that strings of repeated digits longer than a certain length will increase without limit by calculating the number of digits when a new value containing only the same digit appears again in the iterative sequence. Below shows the details of this for digits 1 to 9. The number of cycles is the number of iterations of A053392 required before a new value containing only the starting digit is seen. The minimum starting value is the smallest same-digit number such that subsequent iterations will produce another value with only the same digit of equal or longer length. The number of digits in the subsequence reoccurrence shows the number of digits in this value given the starting value has n digits. All sufficiently long single digit numbers, with digits 1 to 8, increase 8-fold in length, minus a constant, after 9 iterations. . digit | # of cycles | min start value | # digits in reoccurrence 1 | 9 | 1111111 | 8*n - 45 2 | 9 | 222222 | 8*n - 38 3 | 3 | 33333 | 2*n - 5 4 | 9 | 44444 | 8*n - 31 5 | 9 | 55555 | 8*n - 33 6 | 3 | 6666 | 2*n - 4 7 | 9 | 77777 | 8*n - 27 8 | 9 | 8888 | 8*n - 28 9 | 2 | 999 | 2*n - 3 (End) LINKS Scott R. Shannon, Table of n, a(n) for n = 1..33909 (terms 1..109 from Hans Havermann). Hans Havermann, Proof of correctness of first 109 terms CROSSREFS Cf. A053392, A328974. Sequence in context: A067841 A237968 A062912 * A328974 A236732 A184078 Adjacent sequences: A328972 A328973 A328974 * A328976 A328977 A328978 KEYWORD nonn,base AUTHOR N. J. A. Sloane, Nov 02 2019 STATUS approved
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# The Time Constant of an RC Circuit
1 Objectives
1. To determine the time constant of an RC Circuit, and
## 2. To determine the capacitance of an unknown capacitor.
2 Introduction
What the heck is a capacitor? Its one of the three passive circuit elements: the resistor
(which weve already met), the capacitor (which stores energy in electric fields), and the
inductor (which stores energy in magnetic fields, and is the main subject a few weeks from
now). Also known as condensers, capacitors store energy an electric field by separating
positive and negative charges on opposing terminals known as plates - which may or may
not actually be plate-like. Capacitors trace their lineage to the middle of the Eighteenth
century and the invention of the Leyden jar. For forty years, the biggest names in natural
philosophy spent time in the study of charge storage and improving capacitors, including
Volta (for whom we name the volt) and Franklin (of American Founding Father fame).
In this lab, we will study the behavior of capacitors while we add or remove charge from
the plates. Although we wont touch on it in this lab, RC circuits form the foundation of the
modern electronics that underlie a whole host of technologies, including radio, computers,
environmental and medical sensors, and - not least of all - the touch screens in cell phones
and the touch pads of laptops.
3 Theory
We normally write Ohms Law for the resistor as V (t) = I(t)R. But what is that current
I(t)? In the circuits here (see Figure 1), its just the rate at which charge passes through
the resistor. Since all the charge that runs through the resistor ends up going from or to the
capacitor, q(t), it must also be the case that
dq(t)
I(t) = ,
dt
where q(t) is the time dependent charge on the capacitor, with the sign depending on whether
the charge on the capacitor is increasing or decreasing. For this circuit, we could instead
1
Figure 1: The switching circuit used to discuss charging and discharging a capacitor.
## write Ohms Law in the form
dq(t) V (t)
= .
dt R
In words, a resistor is a passive device where the applied voltage causes charge to flow
through the device, while a capacitor is a passive device where the applied voltage causes
charge storage in the device; in equation form
q(t) = CV (t) .
The constant C is called the capacitance, and is measured in Farads (named for Michael
Faraday, whose research output we will learn about later this semester) where F = C/V.
Since capacitors are used to store charge, we must find a way to change the charge state
on the capacitor. Lets discuss the circuit in Figure 1. This circuit has three states: 1. The
steady state, where the switch is open, no current flows in the resistor, and the charge state
on the capacitor is constant, 2. the charging state, where the battery or power supply is
connected to the capacitor and adds charge to the capacitor, and 3. the discharging state,
where the battery is disconnected, the two plates of the capacitor are connected to each
other through the resistor, which removes charge from the capacitor.
We can analyze the dynamic states of the circuit using Kirchoffs Rules1 :
1. The Loop rule (energy conservation) requires the sum of all voltages drops around a
closed loop to vanish, and
2. The Junction rule (charge conservation) requires the sum of all currents into a junction
to vanish.
Lets apply these rules, first to the discharging and then to the charging state:
1. Discharging: In the discharging case, the current will flow off the capacitor in a counter
clockwise direction (why?). When we close that switch, Kirchoffs rules become
Vc VR = 0
dq(t) q(t)
=0.
dt RC
1
As long as the frequencies arent so high that we cant use the lumped element approximation; talk to
your instructor if you want to know more.
2
Again, the rate is negative, because the capacitor is discharging, not charging. The
method of solution for this equation is given in Appendix A:
## Vc (t) = VR (t) = Vs et/RC ,
assuming that the initial voltage across the capacitor is Vs . This discharge curve is
plotted in Figure 2.
2. Charging: In the charging case, the current flows clockwise from the battery (at voltage
Vs ), through the resistor (at VR ), and across the capacitor Vc ; we want to solve for Vc
as a function of time. Kirchoffs Rules tell us
Vs VR Vc = 0
q
Vs IR
C
But here, the current through the resistor depends on the rate at which charge is chang-
ing on the capacitor; they have the same sign here since the charge on the capacitor is
increasing
dq(t) q(t)
Vs R =0
dt C
dq(t) q(t) Vs
+ = .
dt RC R
We can solve this differential equation too for the time dependent voltage profile across
the capacitor; see Appendix A. If we start with a completely discharged capacitor, the
voltages across the resistors and capacitors vary as
Vc (t) = Vs 1 et/RC
VR (t) = Vs et/RC .
As the capacitor charges, the voltage across - and hence the charge on - the capacitor
rises exponentially, while the voltage across - and hence the current through - the
resistor will fall exponentially; see Figure 2.
The quantity RC - which appears in the argument of the exponential - is known as the
time constant of the system; it has units of time (hence the name), and determines the time
interval over which voltages, charges, and currents change in the circuit. The time constant
can be tuned by modifying either R or C. In practice, more resistor than capacitor values
are commercially available, so its usually easier to tune the resistor values.
Given a known resistance we can measure the time constant of the RC circuit, and
algebraically determine the capacitance C. When discharging the capacitor,
Vc (t) = Vs et/RC .
3
Capacitor charging and discharging curves
1
Discharging
Charging
0.8
0.6
Vc(t)/Vs
0.4
0.2
0
0 1 2 3 4 5
t/RC
Figure 2: The capacitor charging and discharging curves. The vertical blue line is the half
life point of the charging and discharging timeline.
We can easily measure and use the half-life T1/2 of the discharge: T1/2 is the time it takes for
the voltage to fall by half. Substituting into the above equations and solving for T1/2 :
Vs
= Vs eT1/2 /RC
2
1 T1/2
ln =
2 RC
T1/2 = (C ln 2) R .
If we allow R to vary, then this equation defines a line with abscissa R, ordinate T1/2 , and
slope C ln 2. If we measure multiple (R, T1/2 ) pairs, we can plot those data points, fit them
to a line, and extract the slope - and hence the capacitance.
4 Procedures
You should receive an oscilloscope, a function generator, a multimeter, one unknown capac-
## 4.1 Time Constant of an RC Circuit
In this part of the experiment, instead of a DC voltage and a mechanical switch, we apply
a square wave signal to the capacitor as shown in Figure 3. An ideal square wave has two
values: high and low (here Vs and 0), and it switches between them instantaneously. The
capacitor will charge when the voltage of the square wave is Vs ; the capacitor will discharge
when the voltage of the square wave is zero. The oscilloscope traces of the charging and
discharging of the capacitor are also shown in Figure 3.
4
Prototypical Oscilloscope Trace
Trace A
Trace B
1
0.8
0.6
Vc(t)/Vs
0.4
0.2
0 1 2 3 4 5
t/RC
## (a) Measurement Circuit (b) Oscilloscope Display
Figure 3: The left-hand figure is the circuit used to measure the time constant of an RC
circuit, while the right-hand figure shows the Oscilloscope traces.
If the period of the square wave Ts is much less than the time constant = RC (Ts ),
then the capacitor will start discharging before it has sufficient time to acquire the maximum
charge q0 , making measurement of the time constant difficult. What will this look like on
the oscilloscope? If Ts is much more than (Ts ), you will not be able to fit both the
charging and discharging portions of the capacitor voltage on the oscilloscope trace at the
same time. Why?
1. Construct the circuit shown in Figure 3a.
2. Perform the initial turn-on procedure for the oscilloscope (from the Oscilloscope Lab).
To start, set both channels to 0.2 V/div. Make sure that the ground of both traces is
at the same position on the screen.
3. Start by setting the frequency of the square wave to about 100 Hz and the output
voltage to 0.5 V.
4. Now we have to set R and C. Adjust R to about 8 k, and C to about 0.1 F. What
is the time constant of this combination?
5. Adjust the frequency of the square wave, and the time and voltage settings of the
oscilloscope until you obtain traces similar to Figure 3b. Measure and record R, C, f ,
V , T1/2 , and the oscilloscope settings.
6. Measure the voltage of the decay curve above ground at a number of points using the
oscilloscope cursors; similarly, measure the height of the charging curve at a number
of points. You will use these to map out the charging/discharging curves, and fit for
the time constant.
7. Repeat for a second set of R and C.
## 4.2 Determine the Capacitance
Here, well use the same techniques we just did to determine the value of an unknown
capacitor.
5
1. Replace the known capacitor with the unknown capacitor in the circuit.
2. Set the resistance to about 4 k, and make the necessary adjustments to the oscillo-
scope settings to again obtain the appropriate display on the oscilloscope.
## 3. Measure and record R, f , V , the oscilloscope settings, and T1/2 .
4. Repeat for four or five values of R; you need enough points to fit a line through your
(R, T1/2 ) pairs.
5. Use your multimeter to directly measure the capacitance, for comparison to your ex-
tracted value.2
A Derivation of Solutions
The differential equations in Section 3 are the simplest of all differential equations to solve:
first order, linear equations with constant coefficients. Here, well solve a slightly more
complicated equation (a first order linear equation with non-constant coefficients) in complete
generality, and use this solution to find the solutions were interested in. Consider the
following differential equation:
dy(x)
+ P (x)y(x) = Q(x) .
dx
This equation looks very much like the equations we found in Section 3 using Kirchoffs Rules.
We can solve this in the following way. First, well multiply both sides of this equation by a
new function, M (x), and well first solve for M (x).
dy(x)
M (x) + M (x)P (x)y(x) = M (x)Q(x) .
dx
Lets now assume that the left hand side is the expansion of the product rule of M (x)y(x):
d
(M (x)y(x)) = M (x)Q(x) .
dx
What must be true for this to hold? Well,
## d dy(x) dM (x) dy(x)
(M (x)y(x)) = M (x) + y(x) = M (x) + M (x)P (x)y(x) .
dx dx dx dx
This is true if and only if
dM (x)
= M (x)P (x) .
dx
2
The multimeter uses an automated version of exactly this procedure to calculate C, based on a time
constant measurement across a precision internal resistor.
6
This is a trivial problem to solve: just separate and integrate:
R R R
M (x) = e P (x)dx+c
= ec e P (x)dx
= c0 e P (x)dx
,
where c is the constant of integration for this indefinite integral. Now, go a few lines back
up the page:
d
(M (x)y(x)) = M (x)Q(x) .
dx
We can solve this one trivially, too, just by integrating; lets make this a definite integral
between x0 and x. We get the left hand side from the Fundamental Theorem of Calculus
Z x
M (x)y(x) M (x0 )y(x0 ) = M (x)Q(x)dx ,
x0
where, since this is an indefinite integral, requires an integration constant c. Finally, then,
we can solve the general problem:
Z x
1
y(x) = M (x) M (x0 )y(x0 ) + M (x)Q(x)dx .
x0
Note that the integration constant in the definition of M (x) cancels here, so it doesnt really
matter what it was; lets just choose c0 = 1.
In the case of the charging RC circuit with zero initial charge, x = t, y(x) = q(t), x0 = 0,
y(x0 ) = 0, P (t) = 1/RC, and Q(t) = Vs /R. Integrating gives us M (t) = et/RC , giving us the
solution:
Z t
t/RC Vs
q(t) = e et/RC dt
0 R
Vs
= RCet/RC et/RC 1
R
= Vs C 1 et/RC ,
where weve used the integration constant to get the t = 0 value right (here, no charge on
the capacitor). Dividing both sides by C gives us the voltage across the capacitor at time t
Vc (t) = Vs 1 et/RC .
## In the case of the discharging circuit, we have
q(t) = Vs Cet/RC ,
where we assume the voltage across the capacitor at t = 0 is Vs . Again, divide by C to get
the voltage profile
Vc (t) = Vs et/RC .
7
Pre-Lab Exercises
Answer these questions as instructed on Blackboard; make sure to submit them before your
lab session!
## 2. If it takes 5 s for a capacitor to charge to half the battery voltage, through a 10 k
resistor, what is the capacitance C?
## 3. A given RC circuit charges through a particular resistor R, and capacitor, C. If I
double the capacitance, what happens to the time constant? What if I double the
resistance?
8
Post-Lab Exercises
1. Determine the time constant of the RC circuit with known capacitance. Do this by
plotting your data of time versus the amplitude of the signal. You should fit this data
to an exponential, and extract . There are a number of ways to do this: (a) use a
mathematics package capable of sophisticated statistical analysis, such as R, Octave,
Matlab, Mathematica, Root, etc., or (b) using a spreadsheet, fit the data using an
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# How can I Intercept object with a circular motion
I am creating a 2d space game and need to make the spaceship intercept a planet. I have working code for straight line intercepts but cannot figure out how to calculate the planets location in a circular orbit.
The game is not scientifically accurate so I am not worried about inertia, gravity, elliptical orbits, etc.
I know the spaceships location and speed and also the planets orbit (Radius) and speed
-
No I am trying to calculate the angle the ship needs to move in order to intercept the planet. – Ausa May 14 '14 at 12:01
This would probably work better in math.stackexchange.com.. – Jari Komppa May 14 '14 at 12:32
Is your ship able to change speed and direction, or are those constant? Also, this question about avoiding having missiles circle a target might be helpful. – thegrinner May 14 '14 at 14:53
To clarify, is the situation? given for the planet: orbit centre, orbit radius, angular speed, current location; for the ship: current location, current speed; determine direction of motion for ship in order to intercept planet – AakashM May 14 '14 at 15:45
As an interesting historical note: planets usually rotate in the same direction as their orbit, which is therefore also anticlockwise as seen from above the northern hemisphere. From this fact we can deduce that sundials were invented in the northern hemisphere. Had sundials been invented in the southern hemisphere then clockwise would be the other way. – Eric Lippert May 14 '14 at 22:41
An analytic solution to this is difficult, but we can use binary search to find a solution to within the required accuracy.
The ship can reach the closest point on the orbit in time t_min:
``````shipOrbitRadius = (ship.position - planet.orbitCenter).length;
t_min = shortestDistance/ship.maxSpeed;
``````
The ship can reach ANY point on the orbit in time less than or equal to t_max:
(Here, for simplicity, I assume the ship can drive through the sun. If you want to avoid this then you will need to switch to non-straight-line paths for at least some cases. "Kissing circles" may look nice and orbital mechanics-y, without changing the algorithm by more than a constant factor)
``````if(shipOrbitRadius > planet.orbitRadius)
{
t_max = planet.orbitRadius * 2/ship.maxSpeed + t_min;
}
else
{
t_max = planet.orbitRadius * 2/ship.maxSpeed - t_min;
}
``````
Now we can use binary search between these extremes, t_min and t_max. We'll search for a t-value that gets the error close to zero:
``````error = (planet.positionAtTime(t) - ship.position).squareMagnitude/(ship.maxSpeed*ship.maxSpeed) - t*t;
``````
(Using this construction, error @ t_min >= 0 and error @ t_max <= 0, so there must be at least one intercept with error = 0 for a t-value in-between)
where, for completeness, the position function is something like...
``````Vector2 Planet.positionAtTime(float t)
{
angle = atan2(startPosition - orbitCenter) + t * orbitalSpeedInRadians;
return new Vector2(cos(angle), sin(angle)) * orbitRadius + orbitCenter;
}
``````
Note that if the planet's orbital period is very short compared to the ship's speed, this error function may change signs several times over the span from t_min to t_max. Just keep track of the earliest +ve & -ve pair you encounter, and continue searching between them until the error is close enough to zero ("close enough" being sensitive to your units and gameplay context. The square of half the frame duration may work well - that ensures the interception is accurate to within a frame)
Once you have a nice error-minimizing t, you can just point the ship at planet.positionAtTime(t) and go full throttle, confident that the planet will reach that point at the same time you do.
You can always find a solution within Log_2((2 * orbitRadius/ship.maxSpeed)/errorThreshold) iterations. So for example, if my ship can traverse the orbit in 60 frames, and I want an intercept accurate to within one frame, I'll need about 6 iterations.
-
Lots of good answers here, also some interesting alternative options but from what I already had this solutions looks the best for my instance. I have created a little JavaScript demo of my results. Demo – Ausa May 15 '14 at 12:10
Let's not over-complicate this. This is not a "perfect" solution but should work for most games and any imperfection should be invisible to the player.
``````if(!OldTargetPoint)
TargetPoint = PlanetPosition;
else
TargetPoint = OldTargetPoint;
Distance = CurPosition - TargetPoint;
TimeNeeded = Distance / Speed;
TargetPoint = PlanetPositionInFuture(TimeNeeded);
SteerTowards(TargetPoint);
[...repeat this every AI update, for example every second...]
``````
1. Calculate the time needed to reach the target point.
2. Calculate at what position the planet will be at the calculated time.
3. Move towards the calculated point.
4. Repeat
This works because the nearer the spacecraft gets towards the lower the error becomes. So the calculation becomes more stable over time.
The error is the difference between the calculated needed time to reach the planet (TimeNeeded) and the actual time needed to reach the planet (after taking into account the new TargetPoint).
-
You may want to run 2 iterations of this when starting an intercept course, otherwise you may see the ship flicker between two directions momentarily (the second guess can be much better than the first, and result in a very different heading - particularly if the ship is close to or within the planet's orbit) – DMGregory May 14 '14 at 19:36
@DMGregory Oh! We could just take the current Position of the Planet instead of the Orbit Center as the starting point. When we are close that is much closer, if we are far away it doesn't matter. – API-Beast May 14 '14 at 21:28
It's also worth noting that this works best when the planet moves slowly compared to the ship. If the planet's speed is comparable to or greater than the ship's, you may see oscillations in the ship's path. At pathological speed ratios, the ship can chase the planet forever on a concentric orbit. If your planets are fast and you notice this happening, you may want to plan your entire intercept course up-front rather than iterating mid-flight. – DMGregory May 14 '14 at 22:16
Let's start off by taking a look at the math behind the problem.
# Step 1:
Finding the intersection between a line and a shape is just a matter of inserting the equation of the line in the equation of the shape, which is a circle in this case.
Take a circle with center c and radius r. A point p is on the circle if
|p - c|^2 = r^2
With a line expressed as p = p0 + μv (where v is a vector, http://en.wikipedia.org/wiki/Euclidean_vector), you insert the line into the circle formula and get
|p0 + μv - c|^2 = r^2
The squared distance can be rewritten as a dot product (http://en.wikipedia.org/wiki/Dot_product).
(p0 + μv - c)•(p0 + μv - c) = r^2
Define a = c - p0 and rewrite to (μv - a)•(μv - a) = r^2
Perform the dot product and we get μ^2(v•v) - 2μ(a•v) + a•a = r^2
Assume that |v| = 1 and we have
μ^2 - 2μ(a•v) + |a|2 - r^2 = 0
which is a simple quadratic equation, and we arrive at the solution
μ = a • v +- sqrt((a • v)^2 *a^2 – r^2)
If μ < 0, the line of the ship in your case does not intersect with the planets orbit.
If μ = 0, the line of the ship will simply touch the circle in one point.
Otherwise, this gives us two μ-values that corresponds to two points on the orbit!
# Step 2:
So we can define a line for the ship, and out of that we get either 0, 1 or 2 μ-values. If we get 1 value, use that one. If we get 2, simply choose one of them.
What can we do with this? Well, we now know the distance the ship has to travel and what point it will end up in!
p = p0 + μv gives us the coordinate, and the μv-component gives us how far is will have to travel. Simply divide this last component with the speed of your ship to get how much time it will take for it to get there!
Now, all that is left to do is to calculate where the planet should be when the ship begins coming towards it's orbit. This is easily calculated with so called Polar coodinates (http://mathworld.wolfram.com/PolarCoordinates.html)
x = c + r*cos(θ)
y = c + r*sin(θ)
And since you had the speed of you ship, and we have the time it will take for the ship to reach the orbit, and where it will collide, we simply move the planet back t*angularVelocity degrees in it's orbit, and we are done!
# Summary
Choose a line for your ship, and run the math to see if it collides with the planets orbit. If it does, calculate the time it will take to get to that point. Use this time to go back in orbit from this point with the planet to calculate where the planet should be when the ship starts moving.
-
Good analysis, but it doesn't appear to answer the question (clarified in a comment): "No I am trying to calculate the angle the ship needs to move in order to intercept the planet." You are taking the ship's angle as a given and calculating the position of the planet, instead of the other way around. – Chaosed0 May 14 '14 at 14:53
Not going to downvote this because it's useful analysis, but I agree with @Chaosed0 that it does not answer the question. In your summary you say "Choose a line for your ship..." but choosing that line is exactly the hard part. – Drake May 14 '14 at 15:40
Ah, shoot... I missed that. You are correct. Feel free to edit it in any way. I don't have the time right now. – Tholle May 14 '14 at 15:46
If you dont' want to use polar coordinates, consider that the all the possible positions of the ship form a cone in (x, y, t)-space. The equation for this is
t*v = sqrt(x^2 + y^2)
where v is the the ship velocity. It is assumed the ship starts at zero.
The position of the planet in space and time can be parametrized by e.g.
x = x0 + r*cos(w*u + a)
y = y0 + r*sin(w*u + a)
t = u
where u goes from 0 upwards. w is the angular speed and a is the starting angle of the planet at time zero. Then solve where the ship and planet could meet in time and space. You get an equation for u to solve:
u * v = sqrt( (x0 + r*cos(w*u + a))^2 + (y0 + r*sin(w*u + a))^2 ) =>
u^2 * v^2 = (x0 + r*cos(w*u + a))^2 + (y0 + r*sin(w*u + a))^2 =>
u^2 * v^2 = x0^2 + y0^2 + r^2 + 2*x0*r*cos(w*u + a) + 2*y0*r*sin(w*u + a)
This equation needs to be solved numerically. It may have many solutions. By eyeballing it, it seems it always has a solution
-
Here are two slightly "out of the box" solutions.
The question is: given that the ship moves in a straight line at a given velocity, and the planet moves in a circle of given radius at a given angular velocity, and the starting positions of the planet and ship, determine what direction vector the ship's straight line should be in to plot an intercept course.
Solution one: Deny the premise of the question. The quantity that is "slippable" in the question is the angle. Instead, fix that. Point the ship straight at the center of the orbit.
• Calculate the position at which the ship will encounter the planet; that's easy.
• Calculate the distance from the ship to the intercept position ; also easy.
• Calculate the time it will take until the planet next reaches the intercept position. Easy.
• Divide the distance from the ship to the intercept by the time until the planet gets to the intercept.
• If that is smaller than or equal to the maximum speed of the ship, you're done. Set the ship moving at that speed straight towards the sun.
• Otherwise, add the orbital period of the planet to the time and try again. Keep doing that until you get a speed that is within reason for the ship.
Solution two: Don't do it on autopilot at all. Make a mini-game where the player has to use thrusters to approach the planet, and if they hit it at too high a relative speed, they blow up, but they have limited fuel as well. Make the player learn how to solve the intercept problem!
-
Here's part of a solution. I didn't get to finish it in time. I'll try again later.
If I understand correctly, you have a planet's position & velocity, as well as a ship's position and speed. You want to get the ship's movement direction. I'm assuming the ship's and planet's speeds are constant. I also assume, without loss of generality, that the ship is at (0,0); to do this, subtract the ship's position from the planet's, and add the ship's position back onto the result of the operation described below.
Unfortunately, without latex, I can't format this answer very well, but we'll attempt to make do. Let:
• `s_s` = the ship's speed (s_s.x, s_s.y, likewise)
• `s_a` = the ship's bearing (angle of movement, what we want to calculate)
• `p_p` = the planet's initial position, global coords
• `p_r` = the planet's distance (radius) from the center of orbit, derivable from `p_p`
• `p_a` = the planet's initial angle in radians, relative to the center of orbit
• `p_s` = the planet's angular velocity (rad/sec)
• `t` = the time to collision (this turns out to be something we must calculate as well)
Here's the equations for the position of the two bodies, broken down into components:
``````ship.x = s_s.x * t * cos(s_a)
ship.y = s_s.y * t * sin(s_a)
planet.x = p_r * cos(p_a + p_s * t) + p_p.x
planet.y = p_r * sin(p_a + p_s * t) + p_p.y
``````
Since we want `ship.x = planet.x` and `ship.y = planet.y` at some instant `t`, we obtain this equation (the `y` case is nearly symmetrical):
`````` s_s.x * t * cos(s_a) = p_r * cos(p_a + p_s * t) + p_p.x
s_s.y * t * sin(s_a) = p_r * sin(p_a + p_s * t) + p_p.y
``````
Solving the top equation for s_a:
`````` s_s.x * t * cos(s_a) = p_r * cos(p_a + p_s * t) + p_p.x
=> s_a = arccos((p_r * cos(p_a + p_s * t) + p_p.x) / (s_s.x * t))
``````
Substituting this into the second equation results in a fairly terrifying equation that Wolfram alpha won't solve for me. There may be a better way to do this not involving polar coordinates. If anyone wants to give this method a shot, you're welcome to it; I've made this a wiki. Otherwise, you may want to take this to the Math StackExchange.
-
I would love to have TeX enabled for this site. It would make some graphics related stuff (e.g. vector, matrices, quaternions..) easier to represent. – mvw May 15 '14 at 11:53
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## 27171
27,171 (twenty-seven thousand one hundred seventy-one) is an odd five-digits composite number following 27170 and preceding 27172. In scientific notation, it is written as 2.7171 × 104. The sum of its digits is 18. It has a total of 3 prime factors and 6 positive divisors. There are 18,108 positive integers (up to 27171) that are relatively prime to 27171.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 27 thousand 171 twenty-seven thousand one hundred seventy-one
## Notation
Scientific notation 2.7171 × 104 27.171 × 103
## Prime Factorization of 27171
Prime Factorization 32 × 3019
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 9057 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 27,171 is 32 × 3019. Since it has a total of 3 prime factors, 27,171 is a composite number.
## Divisors of 27171
1, 3, 9, 3019, 9057, 27171
6 divisors
Even divisors 0 6 3 3
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 39260 Sum of all the positive divisors of n s(n) 12089 Sum of the proper positive divisors of n A(n) 6543.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 164.836 Returns the nth root of the product of n divisors H(n) 4.15247 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 27,171 can be divided by 6 positive divisors (out of which 0 are even, and 6 are odd). The sum of these divisors (counting 27,171) is 39,260, the average is 65,43.,333.
## Other Arithmetic Functions (n = 27171)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 18108 Total number of positive integers not greater than n that are coprime to n λ(n) 3018 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2975 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 18,108 positive integers (less than 27,171) that are coprime with 27,171. And there are approximately 2,975 prime numbers less than or equal to 27,171.
## Divisibility of 27171
m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 4 3 0
The number 27,171 is divisible by 3 and 9.
## Classification of 27171
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (27171)
Base System Value
2 Binary 110101000100011
3 Ternary 1101021100
4 Quaternary 12220203
5 Quinary 1332141
6 Senary 325443
8 Octal 65043
10 Decimal 27171
12 Duodecimal 13883
20 Vigesimal 37ib
36 Base36 kyr
## Basic calculations (n = 27171)
### Multiplication
n×i
n×2 54342 81513 108684 135855
### Division
ni
n⁄2 13585.5 9057 6792.75 5434.2
### Exponentiation
ni
n2 738263241 20059350521211 545032613011824081 14809081128144272104851
### Nth Root
i√n
2√n 164.836 30.0632 12.8389 7.70586
## 27171 as geometric shapes
### Circle
Diameter 54342 170720 2.31932e+09
### Sphere
Volume 8.40244e+13 9.27729e+09 170720
### Square
Length = n
Perimeter 108684 7.38263e+08 38425.6
### Cube
Length = n
Surface area 4.42958e+09 2.00594e+13 47061.6
### Equilateral Triangle
Length = n
Perimeter 81513 3.19677e+08 23530.8
### Triangular Pyramid
Length = n
Surface area 1.27871e+09 2.36402e+12 22185
## Cryptographic Hash Functions
md5 e3fe004c8a465494fcb14db3bb9f0ee1 48bbfeed1de065874a31e934000d62306b60f9b0 66f5acc7b6134602134f8534ecde80b1f325f0f4334522d07fccb351ac60aea4 b3dd51ee2f464cd48703069d1e32644ca362dd448625eee3f754b7d51df82ee7561fe88250744aea4a10f2f8132ec7ed755c464e40b7ff8a6477660c25a4f157 c0285123b3f86fbb8c27b55572d078f5a2b563a7
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Write your own article idiot!
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# PHYSICS: FORM THREE: Topic 7 - MEASUREMENT OF THERMAL ENERGY
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Schemes of Work 2024
TOPIC 7: MEASUREMENT OF THERMAL ENERGY
Heat capacity is the amount of heat required to raise the temperature of an object or substance by one degree. The temperature change is the difference between the final temperature ( Tf) and the initial temperature ( Ti).
The Factors which Determine Heat Quality of a Substance
Explain the factors which determine heat quality of a substance
Heat is a form of energy transferred between bodies due to difference in temperature between them. The energy possessed by the body due to its temperature is called the internal thermal energy. The heat content is due to the random motion of the particles that make up the body. The heat content of the substance depends on;
• mass of the substance
• temperature change
• specific heat capacity of the substance
The Heat Capacity
Determine the heat capacity
Heat capacity is the quantity of heat required to raise the temperature of an object by one degree Celsius.
Different objects have different heat capacities. That means, some can feel so hot than others for the same amount of heat energy supplied to them. For example, hot food on a metal plate feels very hot and difficult to touch than hot food on a plastic plate because a metal plate absorbs heat and raises its temperature faster than plastic plate. In this case we can say plastics have higher heat capacity than metals. In other words, it requires small amount of heat energy to raise temperature of metals than than in plastics.
If the object requires amount/quantity of heat energy Q to raise its temperature by one degree of centigrade, then the heat capacity C of the substance is obtained as;
Heat capacity can also be obtained mathematically by taking the product of mass of the substance and its specific heat capacity.
Example 1
Find the heat capacity of a lump of copper of mass 50kg. The specific heat capacity of copper is 420 J/ Kg ºc.
Data Given
Mass of copper, m= 50kg
The specific heat capacity of copper, c = 420J/KgºC
Required: To calculate heat capacity, C.
C = mc
= 50Kg x 420J/KgºC
= 21600J
=21KJ
Calculating a quantity of heat
• The quantity of heat required to change the temperature of a body with mass, mkg byθ degree Celsius is mcθ joules.
• In order to raise the temperature of a body, heat must be supplied to it.
• In order to lower its temperature, heat must be removed from it.
The Heat Equation is therefore written:
Heat Gained or Heat Lost = Mass X specific heat capacity X change in temperature
Change in temperature:
H =mcθ
Where
H=Heat gained / lost
m= Mass of the body
θ= change (Rise or fall) In Temperature of the body.
Example 2
Water of mass 3kg is heated from 26ºc to 96ºC. Find the amount of heat supplied to the water given that the specific heat capacity of water is 4.2 x 103 J / Kg ºc
Data Given
Mass of water, m = 3kg
Specific Heat capacity, c = 4.2 X 103 j / kgº C
Initial temperature, θi = 26 ºC
Final Temperature, θf = 96ºC
Required
The amount of heat, H
H= mcθ
c = H/mθ
H = mcθ
H = 3Kg x 4.2 x103 (96-26) ºC
The amount of heat energy is H= 882000J= 882KJ
The Specific Heat Capacity
Determine the specific heat capacity
Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.
The quantity of heat supplied to or taken away from a body depends on:
1. The mass of the body, M
2. The temperature different, ΔT
3. The thermal properties of the body.
Transfer of Heat
Heat energy tends to flow from High temperatures to Low temperatures
If you pick up a warm object, heat energy transfers from the object to your hands and your hands feel warm. If you pick up a cool object, heat energy transfers from hands to the object and your hands feel cold.
Determining specific Heat capacity
• Calorimeter – Is the special instrument or vessel used for measurement of Heat.
• Calorimeter is highly polished metal can usually made of copper or aluminium.
• It is flitted with an insulating cover in which there are two holes.
• Two holes allow a thermometer and a stirrer to be inserted.
• The stirrer is made of the same metal as that of the calorimeter.
Demonstration of the specific Heat capacity of a solid
Determining specific Heat capacity by Method of Calculation.
Heat lost by solid, Hs = Ms x Cs (Qs – Qf)
Heat Gained by Calorimeterand stirrer, Hc = Mc x Cc (Qf – Qi)
Heat Gained by Water, HW = Mw x Cw (Qf – Qi)
But the heat lost by the solid is equal to heat gained by the calorimeter and stirrer plus the heat gained by the water in the calorimeter.
Hs = Hc + Hw
But
Heat gained by a calorimeter and content equal to heat lost by the solid.
Thus Hc + Hl = Hs
Mc Cc (Qf – Qi) + Mi Ci ( Qf-Qi) = Ms Cs ( Qs-Qs)
Example 3
A piece of metal with a mass of 200g at a temperature of 100ºC is quickly transferred into 50g of water at 20ºC find the final temperature of the system ( specific Heat capacity of water Cw = 4200J/ Kg ºC specific Heat capacity of the metal Cm = 400J/KgºC.
Ms Cs (Qs-Qf) = Mc Cc (Qf-Q) Mm Cw (Qf-Ql)
Where:
Cs. Is the specific Heat capacity of the solids.
Determining the specific heat capacity of liquid, Cl
By calculation method;
Heat Gained by calorimeter and stirrer
Hc = Mc Cc (Qf – Qi)
Heat Gained by liquid
Hi = ML CL (Qf – Qi)
Heat lost by the solid
Hs = Ms Cs ( Qs-Qf)
Let
Q be the final Temperature of the system
If there are no heat Losses to the surroundings, then.
(Heat gained by water) = (Heat lost by metal)
210 (Q-20) = 80(100 – Q)
21(Q-20) = 8 (100-Q)
21Q – 420 = 800 – 8Q
21Q +8Q = (800+420)
29Q = 1220
Q = (1220/29)
Q = 42.1ºC
Change of state is the transformation of the condition of matter from one (state) to another caused by the change In temperature.
The Behaviour of Particles of Matter by Applying Kinetic Theory
Explain the behaviour of particles of matter by applying kinetic theory
The kinetic theory of matter (particle theory) says that all matter consists of many, very small particles which are constantly moving or in a continual state of motion. The degree to which the particles move is determined by the amount of energy they have and their relationship to other particles. The particles might be atoms, molecules or ions. Use of the general term 'particle' means the precise nature of the particles does not have to be specified.
Particle theory helps to explain properties and behaviour of materials by providing a model which enable us to visualise what is happening on a very small scale inside those materials. As a model, it is useful because it appears to explain many phenomena but as with all models it does have limitations.
Solids, liquids and gases
In solids the particles In liquids the particles In gases the particles are held tightly and packed fairly close together - they are strongly attracted to each other o are in fixed positions but they do vibrate are fairly close together with some attraction between them are able to move around in all directions but movement is limited by attractions between particles have little attraction between them are free to move in all directions and collide with each other and with the walls of a container and are widely spaced out
The model can be used to help explain:
1. the properties of matter
2. what happens during physical changes such as melting, boiling and evaporating
The properties of matter
SolidsLiquidsGases
• have a definite shape
• maintain that shape
• are difficult to compress as the particles are already packed closely together
• are often dense as there are many particles packed closely together
• do not have a definite shape
• flow and fill the bottom of a container. They maintain the same volume unless the temperature changes
• are difficult to compress because there are quite a lot of particles in a small volume
• are often dense because there are quite a lot of particles in a small volume
• do not have a definite shape
• expand to fill any container
• are easily compressed because there are only a few particles in a large volume
• are often low density as there are not many particles in a large space
The graph of temperature versus temperature for a Heated.
The Melting Point of a Substance from its Cooling Curve
Determine experimentally the melting point of a substance from its cooling curve
Melting is the process of change of the state of matter from solid into liquid e.g ice into water.
Melting point (M.P): It is the temperature at which solid substance tends to change into liquid.
Freezing: It is the process of change of the state of matter from liquid to solid e.g water into ice.
Freezing point: Is the temperature at which liquid change into solid. E.g water change into ice at OºC.
Evaporation:Is the process of change liquid substance into vapour (gas).
Sublimation: It is the change of state of matter from solid to gas and vice versa without passing through the liquid phase.e.g. Ammonium Chloride ( NH4CL) and Iodine tends to sublime.
Sublimation point is the temperature at which a solid tends to change into gas and vice versa without passing through liquid state.
Condensation:Is the change of state of gaseous state of matter into liquid state.e.g steam into water.
Deposition: Is the change of the state matter from gas into solid. e.g. Ammonium chloride vapour and Iodine vapour into solid (NH4CI) and (Iodine).
Demonstration of cooling and melting curves for (octadecanoic acid).
Melting point (m.p) table
Substance Melting point (ºC) Copper 1083 Glass 1000 – 1400 Iron 1450 Lead 327 Pitch 40 – 80 Mercury - 39 Platinum 1775 Tin 232 Tungsten 3377
The Effect of Impurities on the Freezing Point and the Boiling Point of a Substance
Demonstrate the effect of impurities on the freezing point and the boiling point of a substance
The effect of dissolved substances on the boiling point and melting point (M.P) means that the additional of impurities will result in increased (B.P) and (M.P).
Effect of impurities on Boiling Point
When an impurity is added to a substance its boiling point is elevated, i.e., its boiling point is increased.
The elevation in boiling point increases with increase in concentration of the solute because when adding the solute vapour pressure of the solution becomes lower than pure solvent. Thus the solution has to be heated more to make the vapour pressure equal to atmospheric pressure. Thus the boiling point gets elevated.
For example boiling point of water is 100oC under normal atmospheric pressure. If we add sugar or salt to this water its vapour pressure becomes lower and boiling point increases.
Generally, when 1 mole of any non electrolyte is dissolved in 1 litre of water the elevation of boiling point is 0.530.
Effect of impurities on freezing point
When an impurity is added its freezing point is lowered i.e. its freezing point decreases.
The depression in freezing point increases with the increase in concentration of the solute because on adding the solute the vapour pressure of solution becomes lower than that of pure solvent. Since freezing point is the temperature at which vapour pressure of liquid and solid phase are equal, therefore, for the solution, this will occur at a lower-temperature.
For example the freezing point of water is OoC under normal atmospheric pressure. If we add sugar or salt to this water its vapour pressure lowers and freezing point decreases.
Generally, when 1 mole of any non-electrolyte is dissolved in 1 litre of water the depression in freezing point of water is 1.860C.
Conclusion
1. The impurities present in a liquid pull its two fixed points away from each other i.e. the freezing point is lowered while the boiling point is raised.
2. The depression in freezing point and the elevation in boiling point increases with increase in the concentration of the solute or impurity i.e. these are the colligative properties that depends only on the no. of moles of the solute. They are independent of the nature of the solute.
The Effect of Pressure on the Boiling Point and Freezing Point of a Substance
Demonstrate the effect of pressure on the boiling point and freezing point of a substance
If a substance expands on solidifying, e.g., water, then the application of pressure lowers its melting point.
If a substance contracts on freezing, the pressure raises its melting point, e.g., paraffin wax.
The freezing point of water is lowered by 0 .007 ºC per atmosphere increase in pressure, whereas that of paraffin wax increases by 0.04 ºC per atmosphere increase in pressure.
When a is liquid heated, its temperature rises and eventually remains constant.
Boiling is the process of forming bubbles of vapour inside the body of a liquid. It rises to the surface of liquid. The process usually depends onexternal pressure above the liquid.
The Phenomenon of Regelation
Explain the phenomenon of regulation
Regelation is the Refreezing process which takes place when copper wire is passed through the Ice BLOCK
Regelation is the Refreezing process which takes place when the wire is observed to Cuts right through the ice block and falls on the floor.
The Concept of Boiling and Evaporation in Respect to the Kinetic Theory of Matter
Give the concept of boiling and evaporation in respect to the kinetic theory of matter
If a liquid is heated, the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. As the liquid gets warmer more particles have sufficient energy to escape from the liquid. Eventually, even particles in the middle of the liquid form bubbles of gas in the liquid. At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid except that they have more energy. At normal atmospheric pressure, all materials have a specific temperature at which boiling occurs. This is called the "boiling point" or boiling temperature. As with the melting point, the boiling point of materials vary widely, e.g., nitrogen -210°C, alcohol 78°C, and aluminium 459°C.
Any material with a boiling temperature below 20°C is likely to be a gas at room temperature. When liquids boil the particles must have sufficient energy to break away from the liquid and to diffuse through the surrounding air particles. As these particles cool down and lose energy they will condense and turn back to liquid. When steam is formed by water boiling at 100°C the particles quickly condense as the surrounding air temperature is likely to be much less that 100°C so the particles cool rapidly. In fact the "steam" coming out of a boiling kettle can only be seen because some of the gas particles have condensed to form small droplets of water.
Evaporating
Within a liquid some particles have more energy than others. These "more energetic particles" may have sufficient energy to escape from the surface of the liquid as gas or vapour. This process is called evaporation and the result of evaporation is commonly observed when puddles or clothes dry. Evaporation takes place at room temperature which is often well below the boiling point of the liquid. Evaporation happens from the surface of the liquid. As the temperature increases, the rate of evaporation increases. Evaporation is also assisted by windy conditions which help to remove the vapour particles from the liquid so that more escape.
Evaporation is a complex idea for children for a number of reasons. The process involves the apparent disappearance of a liquid which makes the process difficult for them to understand. It is not easy to see the water particles in the air. Also, evaporation occurs in a number of quite differing situations - such as from a puddle or bowl of water where the amount of liquid obviously changes, to situations where the liquid is less obvious - such as clothes drying or even those where there is no obvious liquid at all to start with - such as bread drying out. A further complication is that evaporation may be of a solvent from a solution e.g. water evaporating from salt water to leave salt. These situations are quite different yet all involve evaporation.
Evaporation may also involve liquids other than water e.g. perfume, petrol, air fresheners. The particle model can be used to explain how it is possible to detect smells some distance away from the source.
Latent Heat of Fusion and Vaporisation
Demonstrate latent heat of fusion and vaporisation
Latent Heat is the energy when is supplied in form of heat required to change the state of the Matter from one form into another.
Latent heat is not determined (detected) by using a thermometer. So latent heat is also called hidden heat.
Specific latent Heat is the energy supplied to a unit Mass and change Its state from one state of Matter to another state of matter.
Latent heat of Vaporization is the heat required to change a liquid into a gaseous state at constant temperature.
• Mass of Beaker = Mkg
• Mass of Beaker + Water = M2 kg
• Time taken to Boil =tMinutes
• Time taken to Boil away = tMinutes
• Specific latent heat of = L J / kg Vapor
• Heat gained by steam = (M2 – M1)L
Generally
Ttime taken to evaporate
T2 time taken to boil
In this experiment , the Heat gained by the Beaker may be Neglected.
Latent heat of fusion is the amount of heat required to change a substance from solid to liquid at constant temperature.
Example 4
Calculate the amount of Heat required to melts 800g of Ice at 0ºC The specific Latent of fusion of Ice 33400J/kg
Data given:
Mass of Ice , M = 800g (0.8kg)
Specific Heat of fusion, L = 33400 J/kg
Heat gained, H = ML
H= ( 0.8 x 33400J/ kg)
H = 267520J
Determination of the specific Latent Heat of fusion of Ice.
• Mass of Calorimeter + stirrer = M1
• Mass of calorimeter +Water =M2
• Mass of Calorimeter +Water = M3
• Initial Temperature of Water = Q1
• Final temperature of Water =Qf
• Mass of Water = ( M2 - M1 )
• Mass of Ice = ( M3 - M2 )
The Ice melts and forms Water at 0ºC .The Water formed warm up to Temperature Qf.Heat gained by ice during melting at 0ºC = (M3 - M2)L where L is the specific latent Heat of fussion.
Heat gained by the water formed = (M3 - M2) CW QF
Where
• CW is the specific heat capacity of water.
• Heat lost by the original water in the calorimeter = (M- M1) ( Q1 - QF ) Cw.
• heat lost by calorimeter and stirrer = M1 CC ( Qi - Qf ).
• Cc is the specific heat capacity of the material of the calorimeter.
Applying the heat equation:
(Heat gained by ice in Melting + Heat gained by the Water formed) =(Heat lost by calorimeter and stirrer + Heat lost by original Water)
( M3 - M2) L + (M3- M2) CW QF= M1 CC (Q1 - QF)+ (M2- M1 CW Q1 - QF)
Specific Latent heat of Vaporisation is the amount of heat required to change a unit Mass of liquid into gaseous state ( Vapour) at constant temperature.
Specific latent Heat of fusion is the amount of heat required to change a unit Mass of solid substance into liquid at constant temperature
SUBSTANCE SPECIFIC LATENT HEATOF FUSSION J/ kg Ice 334400 Naphthalene 146300 Lead 24662 Copper 179740 Aluminum 317680 Gold 66880
Example 5
0.6 kg of ice at - 10ºC is dropped into 2kg of Water 49ºC contained in a Copper calorimeter of mass 0. 15kg . If the final temperature of the Mixture is 20ºC fin d the specific latent Heat of fusion of ice.
Where
• Specific Heat capacity of ice = 2.1 x 103 J/ KgºC
• Specific Heat capacity of copper = 420 J/ Kg ºC
• Specific Heat Capacity of Water = 4200 j/ Kg º C
Solution
Heat gained by ice during warming up form - 10 ºC to 0ºC
= ( 0 . 6 ×2 . 1 X 103 ×10)
= 12600J
Heat gained when ice at 0ºC changes to water at 0ºC = 0.6L; where L is the latent heat of fusion of ice
Heat gained by cold Water in warming up from 0ºC to 20ºC
=( 0 . 6 ×4 . 2 × 103 × 20)
=50400 J
Heat lost by Water during cooling from 49ºC to 20ºC
= 0 . 15 ×420 ×29
= 1827 J
But
Total Heat gained = Total Heat lost
12600 + 0.6 L + 50400 = 243600 + 1827
L = 245427 - 6300\ O.6
L = 304045J/ Kg
The Mechanism of Refrigeration
Describe the mechanism of refrigeration
Refrigerator is a machine which can enable Heat to flow from a cold Region to a Hot region.The Basic principle used in Refrigeration is Cooling by absorption of latent Heat
How it Works
A volatile liquid such as freon, evaporates inside the copper coils surrounding the freezing cabinet or the refrigeration.
• The latent heat of Vaporization comes from the air surrounding the coil i.e. from the inside of the freezing g cabinet
• An eclectically driven pump P remove the vapor from A and force it into the heat exchanger C, which is made of copper coils.
• The coils of the heat exchanger are filled with cool fins F
• In the heat exchanger, vapor is compressed by the pump and condensed back to liquid.
• The conversion of vapour into liquid in (c) gives out the latent heat of vaporization, which is conducted away by the fins.
• The condensed liquid is then returned to the evaporator coil (A) through avalve (V) (in this way a continuous circulation of vapour and liquid is set up).
• The rate of evaporation and the degree of cooling is controlled by a thermostat, which switches the pumps motor on and off at intervals.
• The thermostat can be adjusted to give the desired low temperature inside the freezing cabinet where food is preserved.
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## virtus 3 years ago In triangle ABC, CD bisects the angle BCA and D lies on side AB. Angle c = 60 degrees, AC = 8cm and BC = 6cm
1. virtus
|dw:1342162593694:dw|
2. virtus
calculate the length of AB, giving your answer correct to one decimal place
3. shandelman
Are you allowed to use the law of cosines?
4. dpaInc
use the law of cosines.... \(\large (AB)^2=(AC)^2+(BC)^2-2(AC)(BC)cosC \)
5. shandelman
I'm not sure what DC has to do with anything...put there to confuse, I guess? I can't think of a better way to do solve it.
6. virtus
@dpaInc I tried to use cosine rule, but i don't think that is how you do it cause you get a whole no. as an answer and the question asks you to round off to 1 d.p Furthermore i checked the solution booklet and it says the answer is 7.2 cm
7. shandelman
...I didn't get a whole answer, but I also didn't get 7.2. Hmm...
8. shandelman
Oh, wait. Hold on.
9. shandelman
No, I did get 7.2, never mind. c^2 = 6^2 + 8^2 - 2(6)(8)cos(60)
10. virtus
oh i must have made a mistake then! THANKS
11. virtus
I get 13.8 for some reason
12. ganeshie8
put the calculator in degrees measure.. .
Find more explanations on OpenStudy
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Survey
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Document related concepts
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Transcript
```Probability and
Statistics in
Engineering
Philip Bedient, Ph.D.
Probability: Basic Ideas
Terminology:
Trial: each time you repeat an
experiment
Outcome: result of an experiment
Random experiment: one with random
outcomes (cannot be predicted exactly)
Relative frequency: how many times a
specific outcome occurs within the entire
experiment.
Statistics: Basic Ideas
Statistics is the area of science that deals with
collection, organization, analysis, and
interpretation of data.
It also deals with methods and techniques that
can be used to draw conclusions about the
characteristics of a large number of data points-commonly called a population- By using a smaller subset of the entire data.
For Example…
You work in a cell phone factory and are asked
to remove cell phones at random off of the
assembly line and turn it on and off.
Each time you remove a cell phone and turn it
on and off, you are conducting a random
experiment.
Each time you pick up a phone is a trial and the
result is called an outcome.
If you check 200 phones, and you find 5 bad
phones, then
relative frequency of failure = 5/200 = 0.025
Statistics in Engineering
Engineers apply physical
and chemical laws and
mathematics to design,
develop, test, and
supervise various
products and services.
Engineers perform tests
to learn how things
behave under stress, and
at what point they might
fail.
Statistics in Engineering
As
engineers perform experiments, they
collect data that can be used to explain
relationships better and to reveal
information about the quality of products
and services they provide.
Frequency Distribution:
Scores for an engineering class are as follows: 58, 95, 80,
75, 68, 97, 60, 85, 75, 88, 90, 78, 62, 83, 73, 70, 70, 85,
65, 75, 53, 62, 56, 72, 79
To better assess the success of the class, we make a
frequency chart:
Now the information can be better analyzed.
For example, 3 students did poorly, and 3 did
exceptionally well. We know that 9 students
were in the average range of 70-79. We can also
show this data in a freq. histogram (PDF).
Divide each no. by 26
Cumulative Frequency
The data can be further organized by calculating the
cumulative frequency (CDF).
The cumulative frequency shows the cumulative number
of students with scores up to and including those in the
given range. Usually we normalize the data - divide 26.
Measures of Central Tendency &
Variation
Systematic errors, also called fixed errors, are
errors associated with using an inaccurate
instrument.
These errors can be detected and avoided by properly
calibrating instruments
Random errors are generated by a number of
unpredictable variations in a given measurement
situation.
Mechanical vibrations of instruments or variations in
line voltage friction or humidity could lead to random
fluctuations in observations.
When analyzing data, the mean alone cannot signal
possible mistakes. There are a number of ways to define
the dispersion or spread of data.
You can compute how much each number deviates from
the mean, add up all the deviations, and then take their
average as shown in the table below.
As exemplified in Table 19.4, the sum of deviations
from the mean for any given sample is always zero.
This can be verified by considering the following:
n
1
x xi
n i1
di (xi x )
Where xi represents data points, x is the average, n
is the number of data points, and d, represents the
deviation from
the average.
n
n
n
n
d x x
d
i1
i1
i
i
i1
i1
i
nx nx 0
Therefore the average of the deviations from the
mean of the data set cannot be used to measure
the spread of agiven data set.
Instead we calculate the average of the absolute
values of deviations. (This is shown in the third
column of table 19.4 in your textbook)
For group A the mean deviation is 290, and Group
B is 820. We can conclude that Group B is more
scattered than A.
Variance
Another
way of measuring the data is by
calculating the variance.
Instead of taking the absolute values of
each deviation, you can just square the
deviation and find the means.
(n-1) makes estimate unbiased
n
v
i1
(x i x )
n 1
2
Taking
the square root of the variance
which results in the standard deviation.
n
s
The
i1
(x i x )
2
n 1
standard deviation can also provide
data set.
The mean for a grouped distribution is calculated
from:
(xf )
x
n
Where
x = midpoints of a given range
f =
frequency of occurrence of data in the range
n = f = total number of data points
The standard deviation for a grouped distribution is
calculated from:
2
(x x ) f
s
n 1
Normal Distribution
We could use the probability distribution from the figures
below to predict what might happen in the future. (i.e.
next year’s students’ performance)
Normal Distribution
Any probability distribution with a bell-shaped
curve is called a normal distribution.
The detailed shape of a normal distribution
curve is determined by its mean and standard
deviation values.
THE NORMAL CURVE
zi = (xi - x) / s
Using Table 19.11, approx. 68% of the data will
fall in the interval of -s to s, one std deviation
~ 95% of the data falls between -2s to 2s, and
approx all of the data points lie between -3s to 3s
For a standard normal distribution, 68% of the
data fall in the interval of z = -1 to z = 1.
AREAS UNDER THE NORMAL CURVE
z = -2 and z = 2 (two standard deviations below and
above the mean) each represent 0.4772 of the total area
under the curve.
99.7% or almost all of the data points lie between -3s
and 3s.
Analysis of Two Histograms
Graph A is class distribution of numbers 1-10
Graph B is class distribution of semester credits
Data for A = 5.64 +/- 2.6 (much greater spread than B)
Data for B = 15.7 +/- 1.96 (smaller spread)
Skew of A = -0.16 and Skew B = 0.146
CV of A = 0.461 and CV of B = 0.125 (CV = SD/Mean)
Frequency B
Frequency A
9
8
7
6
5
4
3
2
1
0
7
6
5
4
3
2
1
0
2
3
4
5
6
7
8
9
10
12
13
14
15
16
17
18
19
20
```
Related documents
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# 3090 in Words
3090 in words is three thousand and ninety. The number 3090 can be converted into English words using the place value system. Thus, we can say that the place value system plays a major role in writing the number names. In this article, let us discuss how to spell the number 3090 in words and its procedure in detail.
3090 in Words: Three Thousand and Ninety. Three Thousand and Ninety in Numerical Form: 3090.
## How to Write 3090 in Words?
Using the place values of 3090, we can easily convert the number 3090 into words.
Thousands Hundreds Tens Ones 3 0 9 0
The expanded form of 3090 is as follows:
= 3 × Thousand + 0 × Hundred + 9 × Ten + 0 × One
= 3 × 1000 + 0 × 100 + 9 × 10 + 0 × 1
= 3000 + 90
= 3090
= Three thousand and ninety
Hence, 3090 in words is three thousand and ninety.
## About the Number 3090
3090 in words – Three thousand and ninety
Is 3090 an odd number? – No
Is 3090 an even number? – Yes
Is 3090 a perfect square number? – No
Is 3090 a perfect cube number? – No
Is 3090 a prime number? – No
Is 3090 a composite number? – Yes
## Frequently Asked Questions on 3090 in Words
Q1
### Write 3090 in English words.
3090 in words is three thousand and ninety.
Q2
### Simplify 3000 + 90, and express it in words.
Simplifying 3000 + 90, we get 3090. Hence, 3090 in words is three thousand and ninety.
Q3
### Is 3090 a prime number?
No, 3090 is not a prime number.
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What is 20 percent of 189? - SupportMyMoto
# What is 20 percent of 189?
WRITTEN BY: supportmymoto.com STAFF
#### Answer for What’s 20 % of 189:
20 % *189 =
(20:100)*189 =
(20*189):100 =
3780:100 = 37.8
Now we have now: 20 % of 189 = 37.8
Query: What’s 20 % of 189?
Share resolution with steps:
Step 1: Our output worth is 189.
Step 2: We characterize the unknown worth with {x}.
Step 3: From step 1 above,{189}={100%}.
Step 4: Equally, {x}={20%}.
Step 5: This ends in a pair of easy equations:
{189}={100%}(1).
{x}={20%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand facet) of each
equations have the identical unit (%); we have now
frac{189}{x}=frac{100%}{20%}
Step 7: Once more, the reciprocal of either side offers
frac{x}{189}=frac{20}{100}
Rightarrow{x} = {37.8}
Due to this fact, {20%} of {189} is {37.8}
#### Answer for What’s 189 % of 20:
189 % *20 =
(189:100)*20 =
(189*20):100 =
3780:100 = 37.8
Now we have now: 189 % of 20 = 37.8
Query: What’s 189 % of 20?
Share resolution with steps:
Step 1: Our output worth is 20.
Step 2: We characterize the unknown worth with {x}.
Step 3: From step 1 above,{20}={100%}.
Step 4: Equally, {x}={189%}.
Step 5: This ends in a pair of easy equations:
{20}={100%}(1).
{x}={189%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand facet) of each
equations have the identical unit (%); we have now
frac{20}{x}=frac{100%}{189%}
Step 7: Once more, the reciprocal of either side offers
frac{x}{20}=frac{189}{100}
Rightarrow{x} = {37.8}
Due to this fact, {189%} of {20} is {37.8}
NOTE : Please do not copy - https://supportmymoto.com
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# Beauty of some numbers discovered by Ramanujan
I am a graduate PhD student and my topic is analytic number theory. I am also a mathematics teacher. I am planning to give a course to pupils in high school that motivates them to study arithmetic and see the beauty of numbers. For instance, I have enough informations about the golden ratio. I watched Professor Ken Ono on national geographic speaking about the beauty of Pi discovered by Ramanujan. I also heared about the story of the taxi cab number $1729.$ Could you help me by providing me with some references or informations about some discovered facts about some special numbers that Ramanujan did. I wanted my pupils to see the beauty and the genuis in Ramanujan's work.
• There are many beautiful formulae arising from the problems Ramanujan submitted to the Indian Mathematical Society: see math.uiuc.edu/~berndt/jims.ps. – Mark Wildon Jan 13 '17 at 12:15
Euler's formula $e^{\pi i}+1=0$ is everyone's favorite. In the same spirit, but to show the massive computational power of Ramanujan, here is special case from Entry 17, page 435, Part V, of the above-mentioned series. $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}k}{e^{k\pi\sqrt{3}}-(-1)^k}=\frac1{4\pi\sqrt{3}}-\frac1{24}.$$ Notice the interplay of the two famous constants $\pi$ and $e$.
In view of Robert Israel's reasonable comment, perhaps we could go for the modest expressions: $$\sqrt{2\left(1-\frac1{3^2}\right)\left(1-\frac1{7^2}\right)\left(1-\frac1{11^2}\right)\left(1-\frac1{19^2}\right)} =\left(1+\frac17\right)\left(1+\frac1{11}\right)\left(1-\frac1{19}\right)$$ found in S. Ramanujan, Notebooks of Srinivasa Ramanujan, Volume II, Tata Institute of Fundamental Research, Bombay, 1957. See pp. 309 and 363.
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# Thread: Induction question
1. ## Induction question
A brief induction question if no-one minds.
"Prove by induction, 7n - 1 is divisible by 6 for all n >= 1."
n = 1 -> 7 - 1 /6 = 1 so the first condition of induction is satisfied, but I honestly can't work out where to go next...
2. ## It is not true.
n=10
7*10-1=69!=k*6 when k is natural.
3. sorry - accidental double post.
4. Not true... hmm, I'll include what the book says.
"In the case n = 1, 7n - 1 is divisible by 6 so P(1) is true.
Assume now the 7k - 1 is divisible by 6 for some k >= 1.
Then,
7k + 1 = 7(7k) - 1
= 7(7k - 1) + 7 -1
= 7)7k - 1) + 6
Since 7k - 1 is divisible by 6 it follows that 7(7k - 1) + 6 is also divisible by 6."
5. ## but if...
you need to prove:
6|7^n - 1 for n>=1
so
we suppose that its true for some n and we prove for n+1.
7^(n+1) - 1 = 7^n*7 - 1 = 7^n(6+1) - 1 = 7^n*6 +7^n - 1
7^n-1 divides by 6 by the step of induction, 6*7^6 divides also by 6.
6. Originally Posted by Skeith
Not true... hmm, I'll include what the book says.
"In the case n = 1, 7n - 1 is divisible by 6 so P(1) is true.
Assume now the 7k - 1 is divisible by 6 for some k >= 1.
Then,
7k + 1 = 7(7k) - 1
= 7(7k - 1) + 7 -1
= 7)7k - 1) + 6
Since 7k - 1 is divisible by 6 it follows that 7(7k - 1) + 6 is also divisible by 6."
Make your post clearer next time and you will get a proper answer. As you can see, from your first post it looks like you are trying to prove that $\displaystyle 6 | (7\cdot n - 1)$ which is obviously wrong (take n=2, for example).
However I guess you are trying to prove that $\displaystyle 6|(7^n-1)$, which is correct. If this is the case, you have shown P(1) holds. Assume P(k), and now we need to prove that P(k+1) follows.
Note that:
$\displaystyle 7^{k+1}-1 = 7\cdot 7^k - 1 = (6+1)7^k - 1 = 6 \cdot 7^k + (7^k-1)$
Now, by the induction hypothesis (that is, $\displaystyle 6|7^k -1$), we get:
$\displaystyle 6 \cdot 7^k + (7^k-1) = 6\cdot 7^k + 6m$ for some $\displaystyle m \in \mathbb{N}$.
But $\displaystyle 6\cdot 7^k + 6m = 6(7^k + m)$ which is divisible by 6, therefore P(k+1) holds and we are done.
If you want to follow your book's way:
$\displaystyle 7^{k+1}-1 = 7\cdot 7^k-1$
But, $\displaystyle 7\cdot 7^k -1 = 7\cdot 7^k -7 +7 -1 =7 \cdot 7^k + 7\cdot (-1) + 7 -1= 7 \cdot (7^k-1) + 7 - 1 = 7 \cdot (7^k -1)+6$
Now, by the induction hypothesis, $\displaystyle 6|(7^k-1)$ and therefore $\displaystyle 7^k-1 = 6m$ for some $\displaystyle m \in \mathbb{N}$. Substitute and get:
$\displaystyle 7 \cdot (7^k-1)+6 = 7 \cdot 6m + 6 = 6(7m + 1)$ and obviously $\displaystyle 6 | 6(7m+1)$ therefore P(k+1) holds and we are done.
7. I'll be sure to do that in future Def. Thankyou all once again for your help, I understand the problem now.
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Stabilization of Curve Directions
Let me start from what I’d like to make:
I.e., we got two curves, between which I’d like to, for example, stretch Blend or do the previously described center curve procedure by spreading points in Ratio Mode, connect them with lines, and through center points of those lines lead Spline (our approximate center curve). Those two curves have opposite vectors (I’ll purposefully start from such example), meaning, when you indicate them using the tree, Blend won’t pass through without correcting vector of one of the curves… That will give us:
So the condition for the procedure is, to direct two curves so that they have the same direction. When marking out center curve, we’ll get false curve, because the straight lines will intersect each other. And something like that will always happen during any change in Parent orientation. How do we avoid that? We need to carry out a procedure, that will somehow connect vectors of those two curves, and keep them together, pointing in the same direction…
Here’s an example procedure:
– create a Spine curve between our two curves, it will define common direction for both of our curves
– next, create normal plane for the Spine:
Now let’s stabilize the first of our curves:
– create intersections between the normal plane for Spine and the one from our curves,
– using that intersection, create normal line for the plane, let’s say 20 mm long (that line will adopt the orientation of the plane),
– create a line outgoing from intersection and contacting our curve – keep the 20 mm length (the line will adopt the orientation of our curve),
– then, on both lines create a point in Ratio mode (on the ends) :
– then create inverse out of our curve, and parameterize it with the following formula:
the distance between two points > 20 mm
– do the same procedure with the other curve.
Now we got two, so to speak, “intelligent” inverses, which react to the change in Parent orientation, and are interconnected through the Spine curve. That will result in two curves with vectors pointing in the same direction… Now we can stretch Blend over, and it will never reverse.
Here is a clip showing that procedure:
In the end, they’ve developed a template for Blend – for a stable one. And what’s the advantage of that template?? You don’t have to mind the directions of inputs when inserting a template – all you need to do is blindly click them out, and the stabilization procedure will do the rest… A good early approach to templates, at the stage of their creation, i.e., by stabilizing all directions of input elements. An example of such template is already included in the entry about pipe axis – follow that procedure, because there’s a direction that’s been stabilized in a similar way… Soon, I’ll describe how to make a stable parallel curve, and finally, a spatial middle curve (having half of the job already done :-).
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## See explanation for result.
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##### Christina K.
Rutgers, The State University of New Jersey
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##### Jared E.
University of Winnipeg
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### Video Transcript
in these grass wonder to forces acting on the spacecraft. One force is exerted by the room on the spacecraft and another force is exerted by the earth on the spacecraft. The big question then is when does this happen? When is the force exerted by the year from the spacecraft is equals to the force exerted by the room on that spacecraft. So when those it happened to do that to some things equation, we have to remember how to tell plate the magnitude off the reputational force and remember that the equation is the following. The magnitude of the gravitational force is equals to the Newton. Constant times the mass off the massive body times the mass off the other body divided by the distance between these two bodies squared. So the force that is exerted by the earth on the spacecraft is given by the Newton constant times the mass off the earth times the mass off the spacecraft divided by the distance between the earth on the spacecraft squared. So are E s squared. On the other side of this equation, we have the gravitational force exerted by the moon on the spacecraft. In this case, we have the Newton constant times the mass off the moon times the mass off the spacecraft divided by the distance between Dimon. Unless spacecraft squared as the Newton Constant is the same, we have one simplification and at the same time the mass off the spacecraft is the same. So we have another simplification. Then we get the following relation. The mass off the earth divided by the distance between the earth on the spacecraft squared Is it close to the mass off the moon, divided by the distance between the room on the spacecraft squared. Then we can send the mass of the moon to the other side, dividing and send their distance between the earth and the spacecraft through the other side, multiplying together following the mass off the earth divided by the mass off the moon Is it cost too? The distance between the earth on the spacecraft squared divided by the distance between the moon on the spacecraft squared But note that the mass of the earth is equals to the mass off the moon Times 81.4 Therefore the mass off the earth divided by the mass off the moon. Is it close to a 21.4 true, 8 to 1.4 is a close to their the stents between the spacecraft and the earth squared, invited by the distance between the moon under spacecraft squared. Now to continue serving his equation. Let me organize the board now, not the following. We can write the distance between the spacecraft and the boom as follows. So this is the full distance between the earth from the moon, and this is the distance between a boom on the spacecraft. So if we pick up the full distance and subtract from the full distance, the distance between the spacecraft and hear what is left is the distance between this spacecraft and the moon. So we can write the distance between the spacecraft and the moon as the distance between the Earth and the moon. Minus that, these tests between this spacecraft on earth too 81.4 is equals to the distance between the spacecraft and the earth squared, divided by the four distance minus the distance between the spacecraft and your squared. Now we have to solve this equation for the distance between this spacecraft and the earth. How can we do that? We begin by standing this term to the other side, multiplying so 81.4 times Are you a M minus? R E s squared is equals two R s squared. Now we take the square it off both sides off this equation to get the following the square root off into 1.4 times are the M minus R E s is the close r e s the reform The square root off 8 to 1.4 times our a m minus x squared off 1.4 times are as easy close to R E s Now send this term to the other side together following square it off to 1.4 times r e m is equals to R E s times one plus square root off to 1.4. Now send this term to the other side Dividing to get spirit off 81.4 times R e m divided by one plus square root off 81.4 Easy of course to R. E s. This is exactly what we want to complete. Now let me organized import to finish the question. Finally, we substitute the value off R E M to get the following r E s is it goes to the square it off 81.4 times 3.85 times 10 38 divided by one plus the square root off 8 to 1.4 And this gives us our ES as approximately three 0.47 times 10 to the eight meters. So when the spaceship is that these these tests from the earth the magnitude of the gravitational force exerted on the spaceship by the moon is bigger than the magnitude off the reputational force exerted by the earth on the spaceship.
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##### Christina K.
Rutgers, The State University of New Jersey
LB
##### Jared E.
University of Winnipeg
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# Vector Subtraction is Explained In-detail
There are several parallels between vectors and scalars. Vector subtraction is no exception. Especially, vector subtraction is:
From the definition mentioned above, it is evident that vector subtraction suggests the addition of negative vectors. Therefore, it is essential to examine negative vectors before learning vector subtraction.
Moreover, a negative vector is obtained when multiplying an offered vector by -1. This reverses the direction of the vector.
Let’s say A is a vector aiming from entrusted to the right. Multiplying the vector A by -1 offers us -A, which is vector A’s negative. Although the size of the two vectors An and also– A will stay the same, the negative vector– A will direct from right to left.
In this subject, we will further talk about adhering to facets of vector subtraction:
### Exactly How to Subtract Vectors
Subtracting Vectors Graphically
How to Subtract Vectors
We know that two vectors, An and B, can be totalled using vector addition. Therefore, the resultant vector can be composed as R = A + B. At the same time, if we go for the subtraction of two vectors, An, as well as B, is expressed mathematically as:
R = A– B.
### Alternatively as.
R = A +(- B).
Therefore, subtracting both vectors coincides with adding vector An and vector B’s negative (i.e., B). The vectors B and– B will certainly have the same size. However, -B’s direction will undoubtedly be opposite to that of vector B.
Vector subtraction likewise functions when both vectors are given up element kind or as column vectors. If B = (bx1, by1) and A = (ax1, ay1), the disparity between the two is.
R = A– B.
You can share the horizontal and vertical parts of the resultant vector R.
Rx = ax1– bx1.
And.
Ry = ay1– by1.
Hence, you can compute the resultant vector by calculating the difference between the original vectors’ corresponding horizontal and vertical elements.
### Subtracting Vectors Graphically.
Graphically, the head-to-tail policy that uses vector addition can be adjusted for vector subtraction. For instance, look at the two vectors Q and P, as depicted in the diagram. Note that the vector– Q is acquired by reversing the direction of Q.
A vector can be specified as a quantity, measurement, or thing with both a magnitude and direction. It is one of the most essential and fundamental concepts in Physics that locates its applications in almost all the subject branches. A vector can be imagined geometrically as a directed line section. The length of the line segment represents the size of the vector, and the arrowhead represents the direction of the vector. The direction of any vector is defined from its tail to its head, where the arrow is.
Read Also: Area of a Polygon – Learn with Examples
Most typical examples of vector quantities include force, rate, velocity, displacement, energy, electrical area, and so on. You can put on fundamental algebraic operations vectors, yet they have their guidelines for these operations. We can not add or subtract vectors like we add or subtract numbers. Vectors also have directions, so this has to be considered.
### What is Vector Subtraction?
The vector subtraction of 2 vectors, an and b, is stood for by a – b and includes the vector b negative to the vector a., i.e., a – b = a + (- b). Hence, the subtraction of vectors involves the addition of vectors and also the negative of a vector. The outcome of vector subtraction is once again a vector. The following are the regulations for subtracting vectors:
You should execute it between two vectors (not between one vector and one scalar).
Both vectors in the subtraction ought to represent the same physical quantity.
Let us comprehend just how to subtract vectors graphically in the approaching sections.
### Vector Subtraction by Parallelogram Law
We mean that an and b are two vectors. How can we translate the subtraction of these vectors graphically? That is, what significance do we affix to a – b? To begin with, we consider that a – b will be a vector which, when contributed to b, must return a., i.e.,
( a – b) + b = a
However, how do we identify the vector a – b, given the vectors an and b? The following number reveals vectors an and b (we have attracted them to co-initial).
When employing the parallelogram law of vector addition, we can identify the vector as complies with. We analyze a – b as a + (- b), the vector sum of an and − b. Currently, we turn around vector b, and after that, including an as well as -b using the parallelogram law.
Graphically subtract the offered vectors An and B, shown in the image listed below, using the head-to-tail technique.
### Solution
We initially draw the vector B negative by reversing its direction, i.e., -B. Subsequently, we add the vectors An and– B using the head-to-tail method.
We top place the given vectors An and– B such that the tail of the vector -B links to the head of vector A as received in the image below. Next off, to locate their sum, we develop a resultant vector R. Hence, it links vector A tail to the head of a vector– B. Mathematically, the consequent can be shared as:
R = A + (- B).
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# Weight of a person on two planets with different masses
1. Jul 9, 2014
### Medgirl314
1. The problem statement, all variables and given/known data
A person standing on the surface of a planet has a weight of 250 N. Suppose he goes to another planet that is the same size(volume) but twice the mass(more dense). What would his weight be on the second planet?
2. Relevant equations
g=Gme/re2
3. The attempt at a solution
Hopefully, it is logical to assume that the planets are approximate spheres. In the problem, it is stated that the volume for the two problems is equivalent. Since the volume formula for a sphere is 4/3∏*3r, we can assume that the radii are equivalent. G is a constant. Logic would seem to dictate that the answer is twice the force on the first planet, so the answer would be 1300. However, the answer just doesn't feel right. I can't place it, but it seems like the answer is too simple. Could someone please confirm my answer, or if it's wrong, tell me the problem?
Thanks so much.
2. Jul 9, 2014
### ZetaOfThree
You're right. Good job.
3. Jul 9, 2014
Thank you!
4. Jul 9, 2014
### HallsofIvy
Staff Emeritus
Though the volume of a sphere is $\frac{4}{3}\pi r^3$ not $\frac{4}{3}\pi*3r$!
5. Jul 9, 2014
### Medgirl314
Thank you! I remember now, when I looked it up I must have misread it.
6. Jul 10, 2014
### Bandersnatch
Wait, how did you get 1300 from doubling 250 again?
7. Jul 10, 2014
### Medgirl314
I didn't. I somehow managed to type a 2 instead of a 6, even though the keys are nowhere near each other. Sorry for the confusion! Now it's a little concerning to me that others agreed when I typed the wrong number.
8. Jul 10, 2014
### Bandersnatch
No need. You've got it right.
9. Jul 10, 2014
### Medgirl314
Okay, thank you!
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Home | Math | Geometrical Representation of Complex Numbers
# Geometrical Representation of Complex Numbers
July 15, 2022
written by Azhar Ejaz
We have seen that there is a (1, -1) correspondence between the elements (ordered pairs) of the Cartesian plane R x R and the complex numbers.
Therefore, there is a (1, -1) correspondence between the points of the coordinate plane and the complex numbers. We can, therefore, represent complex numbers by points of the coordinate plane.
## How do we represent a complex number geometrically?
 In this representation, every complex number will be represented by one and only one complex number. The components of the complex number will be the coordinates of the points representing it.
In this representation, the x-axis is called the real axis and the y-axis is called the imaginary axis. The coordinate plane itself is called the complex plane or z-plane.
By way of illustration, several complex numbers have been shown below in fig.
The figure representing one or more complex numbers on the complex plane is called an argand diagram. Points on the x-axis represent real numbers whereas the points on the y-axis represent imaginary numbers.
x and y are the coordinates of a point.
It represents the complex number x + iy. The real number√(a2 + b2) is called the modulus of the complex number a + ib.
In the figure:
In the right-angled triangle OMA, we have, by Pythagoras theorem.
Thus |OA| bar represents the modulus of x + iy. In other words: the modulus of a complex number. The modulus of a complex number is generally denoted as |x + yi| or |(x,y)|. For convenience, a complex number is denoted by z.
If z = x + iy + (x,y), then
|z| = √x2 + y2
## The polar form of a Complex number
Consider adjoining representing the complex number z = x + i.
From the diagram, we see that x = rcosθ and y = rsinθ where r = |z| and θ is called arguments of z.
Hence
x + iy = rcosθ + irsinθ … (i)
where
r = √(x2 + y2) and θ = tan-1y/x
Equation (i) is called the polar form of the complex number z.
## Express the complex number 1 + i√3 in polar form.
Step-I:
put rcosθ = 1 and + rsinθ = √3
Step-II:
r2 = (1)2 + (√3)2
r2 = 1 + 3 = 4
r = 2
Step-III:
θ = tan-1√3/1
θ = tan-1√3 = 60o
thus
1 + i√3 = 2 cos 60o + i2sin60o
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## Yen's Proportion Problems
If 3 apples cost \$6, how much would 12 apples cost?
To make lemonade you need 6 lemons and 8 cups of water. How many lemons do you need it you have 24 cups?
Now it's your turn to solve!
Three posters cost \$9.60. At this rate, how many posters can you buy for \$48.
#### 1 comment:
1. Three posters cost \$9.60. At this rate, how many posters can you buy for \$48.
I did this question the answer is 15
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# Thread: Double summation - Binomial Theorem?
1. ## Double summation - Binomial Theorem?
Find the value of:
(i. nCj + j. nCi)
1≤ i < j ≤ n
NOTE: i. nCj is i multiplied by n choose j
2. Originally Posted by fardeen_gen
Find the value of:
(i. nCj + j. nCi)
1≤ i < j ≤ n
NOTE: i. nCj is i multiplied by n choose j
In that sum, each integer from 1 to n is multiplied by each binomial coefficient $\textstyle n\choose k$ (k going from 1 to n), except for the "diagonal" products $k{n\choose k}$. So the sum is equal to $\sum_{i=1}^ni\sum_{j=1}^n{n\choose j} - \sum_{k=1}^nk{n\choose k}$. But $\sum_{i=1}^ni = \tfrac12n(n+1)$, $\sum_{j=1}^n{n\choose j} = 2^n-1$ and $\sum_{k=1}^nk{n\choose k} = 2^{n-1}n$. Therefore $\mathop{\sum\sum}_{1\leqslant i.
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# the sum ∑∑(10Cj)(JCi) is equal to (0<=i<=j<=10)
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# Thread: Trig Help ( verify answers)
1. ## Trig Help ( verify answers)
Convert the angle given in degrees to radian measure in terms of .380°
i got 6.63 rad
to 4 sig.figs
1.018
Solve the equation for all nonnegative values of less than . Do by calculator, if needed, and give the answers to three significant digits in the order of increasing.
(not sure what to do)
A satellite is in a circular orbit 225 km above the equator of the earth. How many kilometres must it travel for its longitude to change by 86.3°?
Assume the radius of the earth equals 6400 kilometres.(round to the nearest whole number)
i got 9639.80 =9600
2. Originally Posted by rock candy
Convert the angle given in degrees to radian measure in terms of .380°
i got 6.63 rad
Unfortunately, I cannot see any of your attachments.
"6.63 rad" is the NUMERICAL value but the problem asked you give it in terms of $\displaystyle \pi$: Since there are 360 degrees or $\displaystyle 2\pi$ radians so the conversion factor is $\displaystyle \frac{2\pi}{360}= \frac{\pi}{180}$ radians per degree. $\displaystyle \frac{\pi}{180}(380)= \frac{380}{180}\pi$. Just do the fraction part.
Unfortunately, I cannot see the attachments.
to 4 sig.figs
1.018
Solve the equation for all nonnegative values of less than . Do by calculator, if needed, and give the answers to three significant digits in the order of increasing.
(not sure what to do)
A satellite is in a circular orbit 225 km above the equator of the earth. How many kilometres must it travel for its longitude to change by 86.3°?
Assume the radius of the earth equals 6400 kilometres.(round to the nearest whole number)
i got 9639.80 =9600
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Chem Differential Eq HW Solutions Fall 2011 75
# Chem Differential - Section 4.8 Bessel Series Expansions 75 13 Use(6 with p = 4 Then = 8 J4 x J3 x x = J5 x 86 J3(x J2(x J3(x(by(6 with p = 3 xx =
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Section 4.8 Bessel Series Expansions 75 13. Use (6) with p = 4. Then J 5 ( x )= 8 x J 4 ( x ) - J 3 ( x ) = 8 x ± 6 x J 3 ( x ) - J 2 ( x ) ² - J 3 ( x ) (by (6) with p =3) = ³ 48 x 2 - 1 ´ J 3 ( x ) - 8 x J 2 ( x ) = ³ 48 x 2 - 1 ´³ 4 x J 2 ( x ) - J 1 ( x ) ´ - 8 x J 2 ( x ) (by (6) with p =2) = ³ 192 x 3 - 12 x ´ J 2 ( x ) - ³ 48 x 2 - 1 ´ J 1 ( x ) = 12 x ³ 16 x 2 - 1 ´± 2 x J 1 ( x ) - J 0 ( x ) ² - ³ 48 x 2 - 1 ´ J 1 ( x ) (by (6) with p =1) = - 12 x ³ 16 x 2 - 1 ´ J 0 ( x )+ ³ 384 x 4 - 72 x 2 +1 ´ J 1 ( x ) . 17. (a) From (17), A j = 2 J 1 ( α j ) 2 µ 1 0 f ( x ) J 0 ( α j x ) xdx = 2 J 1 ( α j ) 2 µ c 0 J 0 ( α j x ) = 2 α 2 j J 1 ( α j ) 2 µ j 0 J 0 ( s ) sds (let α j x = s ) = 2 α 2 j J 1 ( α j ) 2 J 1 ( s ) s j 0 = 2 cJ 1 ( α j ) α j J 1 ( α j ) 2 . Thus, for 0 <x< 1, f ( x · j =1 2 cJ 1 ( α j ) α j J 1 ( α j ) 2 J 0 ( α j x ) . (b) The function f is piecewise smooth, so by Theorem 2 the series in (a) converges to f ( x ) for all 0 1, except at x = c , where the series converges to the average value f ( c +)+ f ( c - ) 2 = 1 2 . 21. (a) Take m =1 / 2 in the series expansion of Exercise 20 and you’ll get
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
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#### Please solve RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 1 maths textbook solution
Answer: $3a^{2}$
Hint:
Use the formula $(x^{3}-a^{3})$ so that $f(x)\neq \frac{0}{0}$ at $x\neq a$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{x^{3}-a^{3}}{x-a} & , x \neq a \\ b & , x=a \end{array}\right.$ is continuous at $x=a$
Solution:
If $f(x)$ is continuous at $x=a$ , then RHL= LHL
LHL,
\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a} \\ &=\lim _{x \rightarrow a^{-}} \frac{(x-a)\left(x^{2}+a x+a^{2}\right)}{(x-a)} \\ &=\lim _{x \rightarrow a^{-}} x^{2}+a x+a^{2} \\ &=\lim _{\mathbb{\Xi} \rightarrow 0} a^{2}+a^{2}+a^{2} \\ &=3 a^{2} \end{aligned}$\left [ \because x=a \right ]$
RHL,
\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}} b \\ &=b \end{aligned}
As $f(x)$ is continuous at $x=a$, RHL = LHL
$b=3a^{2}$
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## The side lengths of triangle ABC are written in terms of the variable p, where p ≥ 3. Triangle A B C is shown. Side A B has a length of 4
Question
The side lengths of triangle ABC are written in terms of the variable p, where p ≥ 3. Triangle A B C is shown. Side A B has a length of 4 p minus 1, Side B C has a length of 3 p, and side A C has a length of p + 4. Which is correct regarding the angles of the triangle? mAngleA > mAngleC > mAngleB mAngleB > mAngleA > mAngleC mAngleC > mAngleA > mAngleB mAngleC > mAngleB > mAngleA
in progress 0
1 week 2021-09-15T01:10:06+00:00 2 Answers 0
Angle C > Angle A > Angle B
Step-by-step explanation:
In Δ ABC, the sides AB = 4p – 1, BC = 3p and CA = p + 4, where p ≥ 3.
Assume, p = 3, then AB = 11 units, BC = 9 units and CA = 7 units.
We know, the angle of a triangle is larger for the larger length of the opposite side.
So, Angle C > Angle A > Angle B
Therefore, option c is correct. (Answer)
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# 20. In a potato race, a bucket is placed at the starting point, which is from the in and the other potatoes are placed apart in a straight line. There are ten potat bu) built line (see Fig. 5.6). Fig. 5.6 A competitor starts from the bucket, picks up the nearest potato, runs back with it, d it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, she continues in the same way until all the potatoes are in the bucket. What is the competitor has to run? [Hint: To pick up the first potato and the second potato, the total distance (in metre run by a competitor is
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Question Text 20. In a potato race, a bucket is placed at the starting point, which is from the in and the other potatoes are placed apart in a straight line. There are ten potat bu) built line (see Fig. 5.6). Fig. 5.6 A competitor starts from the bucket, picks up the nearest potato, runs back with it, d it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, she continues in the same way until all the potatoes are in the bucket. What is the competitor has to run? [Hint: To pick up the first potato and the second potato, the total distance (in metre run by a competitor is Updated On Mar 19, 2023 Topic All topics Subject Mathematics Class Class 7 Answer Type Video solution: 1 Upvotes 107 Avg. Video Duration 4 min
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Forces and Moments Part 1 University of Sydney To be able to calculate the moment of a force about a specified axis. When the moment of a force is computed about a point, the moment and its axis are always
Moment about an Axis University of Memphis
Module 9 Solve a 2D Moment Problem coursera.org. Moment of Inertia of Beam Sections. Now we have all the information we need to use the "Parallel Axis Theorem" and find the total moment of inertia of the I-beam, Learning Goal: To be able to calculate the moment of a force about a specified axis. When the moment of a force is computed about a point, the moment and its axis are.
Chapter 20 Torque & Circular Motion 124 (a force times a moment arm is a torque, but a force by The magnitude of the torque about the specified axis of Engr210 – Spring 2005 Instructor: Ahmed Abdel-Rahim Lesson # 10: Moment about a line /Couples Page 2 of 2 Example Given: A force of 80 lb acts
Application of Moment of Force in Structural Engineering. point or axis. A moment develops whenever a force example, a common center of moments would along the member’s axis. Bending Moment (+) Shear Force Internal Loads Internal Loadings at a Specified Point (-) Shear Force Internal Loads
Example 1: Moment of Inertia of a Disk About its Central Axis specified axis is joining the point at which the force is applied and the axis of rotation. Example 3: Moment vs The Moment vs. Curvature diagram for bending about Y axis is The Moments vs. Curvature diagrams for the three specified axial force
WMU discontinued personal web page services on homepages.wmich.edu for students, staff and alumni effective January 9, 2018. These pages are now disabled. How to find the x- and y-components of a force vector. For example, imagine the surface The force vector is white, the x-axis is red,
Moment is the measure of the capacity or ability of the force to produce twisting or turning effect about an axis. This axis is perpendicular to the plane containing MOMENT OF A FORCE Section 4'1. When determining the moment of a force about a specified axis, EXAMPLE. Given A force is applied to the tool to open a
Chapter 5 : Moment of forces an axis through a specified determine the desired moment by, for example, aligning the x-axis with the axis AB Water pressure is a function of both water specific weight Area Moment of Inertia Example 3 care to note the location of the x-y coordinate axis. Moments
With the appropriate balance of force, Moment of Inertia Examples. Moment of inertia is defined with respect to a specific rotation axis. The moment of inertia –Vector calculation of a moment •Example Problems What is a moment? What is the load (force) on the hook along moment axis giving
When determining the moment of a force about a specified axis, a 22-year-old african american upper class female. their marriage is an example of: In this lecture you will learn how to find moment about specified axis using scalar as well as vector analysis.
When determining the moment of a force about a specified axis, the axis must be along _____. In this section, students will learn the definition of a moment. Students will calculate the moment of a force about a point, line or axis, and moment due to a couple.
MOMENT OF A FORCE Section 4'1. When determining the moment of a force about a specified axis, EXAMPLE. Given A force is applied to the tool to open a Answer to To be able to calculate the moment of a force about a specified axis. When the moment of a force is computed about a poi...
Answer to ± Moment of a Force About a Specified Axis Learning Goal: To be able to calculate the moment of a force about a specifi... Bedford, Fowler: Statics. Chapter 4: System of Forces and Moments, Examples via TK Solver Take moments about the z‐axis at point A
Moment Formula Physics Definition Formula And Calculation
6161103 4.5 moment of a force about a specified axis. How to find the x- and y-components of a force vector. For example, imagine the surface The force vector is white, the x-axis is red,, This tool will calculate the torque generated around an axis by a force This is the moment of force or turning force generated by a force applied at a specified.
Moment of a force University of Nebraska–Lincoln. 5 Moment of a Force about a Specified Axis To determine the moment created by from CE 2450 at Louisiana State University, Bedford, Fowler: Statics. Chapter 4: System of Forces and Moments, Examples via TK Solver Take moments about the z‐axis at point A.
STATIC-principles of momen Cartesian Coordinate System
X- and Y-Components of a Force Vector Zona Land Education. 2.3 The Statics of Rigid Bodies force about an axis. The moment of a force F about an axis through a point o is Not only must the axis be specified And it's essentially force times the distance to your axis of Let me take a simple example. So essentially, the moment of force created by this force.
• Chapter 5 Moment of forces - Zig" Herzog
• READING QUIZ 1 When determining the moment of a force
• Ch 4 Force System Resultant Torque Force
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With the appropriate balance of force, Moment of Inertia Examples. Moment of inertia is defined with respect to a specific rotation axis. The moment of inertia Application of Moment of Force in Structural Engineering. point or axis. A moment develops whenever a force example, a common center of moments would
5 Moment of a Force about a Specified Axis 45 Moment of a Force about a from 4.5 Moment of a Force about a Specified Axis Example 4.9 The rod is supported Torque = Force (Fm) x Moment Arm; and joint axis. A common example of the effect for a more specific intervention. Moment Arm of a force system is the
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How to find the x- and y-components of a force vector. For example, imagine the surface The force vector is white, the x-axis is red, Forces and Moments: Part 1. Moment of Force F around point O: M O. The moment of a force about a point or an axis provides a measure of the tendency of the force to
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When determining the moment of a force about a specified axis, a 22-year-old african american upper class female. their marriage is an example of: When determining the moment of a force about a specified axis, a 22-year-old african american upper class female. their marriage is an example of:
Moment of a Force about a Specified Axis A moment and its axis are always perpendicular to the plane defined by the force and its moment arm Example 1: Moment of Inertia of a Disk About its Central Axis specified axis is joining the point at which the force is applied and the axis of rotation.
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Module 9 Solve a 2D Moment Problem coursera.org. And it's essentially force times the distance to your axis of Let me take a simple example. So essentially, the moment of force created by this force, For example, the moment of force acting of 'importance' or 'consequence,' and the moment of a force about an axis meant the importance of the force with.
How to find the x- and y-components of a force vector. For example, imagine the surface The force vector is white, the x-axis is red, Chapter 5: Distributed Forces; Centroids and Centers of Gravity They are not discrete forces that act at specific points. What are distributed forces? Examples:
Ch 4 Force System Resultant To provide a method for finding the moment of a force about a specified axis. Example 4. Determine the moment of the force about When determining the moment of a force about a specified axis, the axis must be along _____.
The moment arm is defined as A. the distance from the point of application of a force to a specified axis of rotation B. the distance from the line of action of a Moment about an Axis: Consider the force passing through point A as An example of this case is shown in the figure where the line of action of the force is
5 Moment of a Force about a Specified Axis 45 Moment of a Force about a from 4.5 Moment of a Force about a Specified Axis Example 4.9 The rod is supported 30/12/2007 · [SOLVED] Seriously? Moment of Force about a specified Axis So I really thought I understood this. I have attached the problem and attempt at solution. It should be
Forces and Moments: Let us consider the following example: Force F causes a moment MO If the line of action of a force F is perpendicular to any specific axis READING QUIZ 1 When determining the moment of a force about a specified axis from PCS 211 at Ryerson University. Find Study Resources. EXAMPLE (continued) A B A B.
MOMENT OF A FORCE Section 4'1. When determining the moment of a force about a specified axis, EXAMPLE. Given A force is applied to the tool to open a In this lecture you will learn how to find moment about specified axis using scalar as well as vector analysis.
Chapter 5 : Moment of forces an axis through a specified determine the desired moment by, for example, aligning the x-axis with the axis AB When determining the moment of a force about a specified axis, the axis must be along _____.
How to Draw Bending Moment Diagrams. In our simple example: there will a +10kNm moment added from the first force and -20kNm from the second. Moment of a Force About an Axis A hinge such as the one shown above will only allow for rotation along the axis of the hinge. Since this corresponds to moments
And it's essentially force times the distance to your axis of Let me take a simple example. So essentially, the moment of force created by this force And it's essentially force times the distance to your axis of Let me take a simple example. So essentially, the moment of force created by this force
Example 1: Moment of Inertia of a Disk About its Central Axis specified axis is joining the point at which the force is applied and the axis of rotation. Chapter 20 Torque & Circular Motion 124 (a force times a moment arm is a torque, but a force by The magnitude of the torque about the specified axis of
Chapter 5 Moment of forces - Zig" Herzog. 5. MOMENTS, COUPLES, FORCES SYSTEMS axis or point is called the MOMENT OF A FORCE M=Fxd Objective: An example to illustrate the definition of Moment in Statics., The moment arm is defined as A. the distance from the point of application of a force to a specified axis of rotation B. the distance from the line of action of a.
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ME 200 Test 2 Flashcards Quizlet. Chapter 2 Review of Forces and Moments example forces exerted by a damper or dashpot, (specified in Newtons,, For example, the moment of force acting of 'importance' or 'consequence,' and the moment of a force about an axis meant the importance of the force with.
Seriously? Moment of Force about a specified Axis. In physics, moment of force (often just moment) is a measure of its tendency to cause a body to rotate about a specific point or axis. In this concept the moment arm, –Vector calculation of a moment •Example Problems What is a moment? What is the load (force) on the hook along moment axis giving.
Moment of Inertia statics.marcks.cc
Ch 4 Force System Resultant Torque Force. STATIC-principles of momen 4.5 Moment of a Force about a Specified Axis 4.5 Moment of a Force about a Specified Axis Example 4.9 The rod is supported by To be able to calculate the moment of a force about a specified axis. When the moment of a force is computed about a point, the moment and its axis are always.
To obtain a report (for selected wall zones) of forces along a specified vector, moments about a specified center and along a specified axis, or the center of Moment about an Axis: Consider the force passing through point A as An example of this case is shown in the figure where the line of action of the force is
Water pressure is a function of both water specific weight Area Moment of Inertia Example 3 care to note the location of the x-y coordinate axis. Moments To be able to calculate the moment of a force about a specified axis. When the moment of a force is computed about a point, the moment and its axis are always
Example 3: Moment vs The Moment vs. Curvature diagram for bending about Y axis is The Moments vs. Curvature diagrams for the three specified axial force 12/08/2013 · Moment of Force about specified axis Hi, Finding moments around [tex] Oa[/tex] For I've been throught the worked examples on this topic again,
Engr210 – Spring 2005 Instructor: Ahmed Abdel-Rahim Lesson # 10: Moment about a line /Couples Page 2 of 2 Example Given: A force of 80 lb acts In this lecture you will learn how to find moment about specified axis using scalar as well as vector analysis.
5 Moment of a Force about a Specified Axis 45 Moment of a Force about a from 4.5 Moment of a Force about a Specified Axis Example 4.9 The rod is supported 2.3 The Statics of Rigid Bodies force about an axis. The moment of a force F about an axis through a point o is Not only must the axis be specified
Bedford, Fowler: Statics. Chapter 4: System of Forces and Moments, Examples via TK Solver Take moments about the z‐axis at point A 5 Moment of a Force about a Specified Axis To determine the moment created by from CE 2450 at Louisiana State University
Moment of a force: Part 2. The cross Moment of a force about a specified axis a-a: M a. O: The moment of a couple does not depend on the point one takes the READING QUIZ 1 When determining the moment of a force about a specified axis from PCS 211 at Ryerson University. Find Study Resources. EXAMPLE (continued) A B A B.
You should be able to demonstrate your understanding by analysing simple examples of equilibrium. a force F about a specified axis is moment of a force. Moment of a Force About an Axis A hinge such as the one shown above will only allow for rotation along the axis of the hinge. Since this corresponds to moments
Learning Goal: To be able to calculate the moment of a force about a specified axis. When the moment of a force is computed about a point, the moment and its axis are The moment arm is defined as A. the distance from the point of application of a force to a specified axis of rotation B. the distance from the line of action of a
• It must be specified with respect to a chosen axis of known as the centroidal axis. The equation of the moment Example Find Moment of Inertia of this Moment about an Axis: Consider the force passing through point A as An example of this case is shown in the figure where the line of action of the force is
Moment Components about a Specific Axis moment of force about axis OB Determine the position vector from a point on the specified axis to a point on the Forces and moments Hint: If you are not forces, moments). This moment tends to turn the body about an axis perpendicular to its axis of symmetry,
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# Difference between revisions of "2015 AMC 10B Problems/Problem 22"
## Problem
In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? $[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("A",A,N); label("B",B,E); label("C",C,SE); label("D",D,SW); label("E",E1,W); label("F",F,NW); label("G",G,NE); label("H",H,E); label("I",I,S); label("J",J,W); [/asy]$ $\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10}$
## Solution 1
Triangle $AFG$ is isosceles, so $AG=AF=1$. $FJ = FG$ since $\triangle FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$, notice that $\triangle JHG \cong \triangle AFG$. Therefore, $JH=AF=1$.
Since $\triangle AJH \sim \triangle AFG$, $$\frac{JH}{AF+FJ}=\frac{FG}{FA}$$. $$\frac{1}{1+FG} = \frac{FG}1$$ $$1 = FG^2 + FG$$ $$FG^2+FG-1 = 0$$ $$FG = \frac{-1 \pm \sqrt{5} }{2}$$
So, $FG=\frac{-1 + \sqrt{5}}{2}$ since $FG$ must be greater than 0.
Notice that $CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}$.
Therefore, $FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$
## Solution 2 (Trigonometry)
Note that since $ABCDE$ is a regular pentagon, all of its interior angles are $108^\circ$. We can say that pentagon $FGHIJ$ is also regular by symmetry. So, all of the interior angles of $FGHIJ$ are $108^\circ$. Now, we can angle chase and use trigonometry to get that $FG=2\sin18^\circ$, $JH=2\sin18^\circ*(2\sin18^\circ+1)$, and $DC=2\sin18^\circ*(2\sin18^\circ+2)$. Adding these together, we get that $FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)$. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to $8\sin18^\circ*(1+\sin18^\circ)$, but we can find that this is closest to $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$.
## Solution 3
When you first see this problem you can't help but see similar triangles. But this shape is filled with $36 - 72 - 72$ triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of $FG$ so we can apply similar triangles easily. To simplify the process lets write $FG$ as $x$.
First what is $JH$ in terms of $x$, also remember $AJ = 1+x$: $$\frac{JH}{1+x}=\frac{x}{1}$$$JH$ = ${x}^2+x$
Next, find $DC$ in terms of $x$, also remember $AD = 2+x$: $$\frac{DC}{2+x}=\frac{x}{1}$$$DC$ = ${x}^2+2x$
So adding all the $FG + JH + CD$ we get $2{x}^2+4x$. Now we have to find out what x is. For this, we break out a bit of trig. Let's look at $\triangle AFG$ By the law of sines: $$\frac{x}{sin(36)}=\frac{1}{sin(72)}$$ $$x=\frac{sin(36)}{sin(72)}$$
Now by the double angle identities in trig. $sin(72) = 2sin(36)cos(36)$ substituting in $$x=\frac{1}{2cos(36)}$$
A good thing to memorize for AMC and AIME is the exact values for all the nice sines and cosines. You would then know that: $cos(36)$= $$\frac{1 + \sqrt{5}}{4}$$
so now we know: $$x = \frac{2}{1+\sqrt{5}} = \frac{-1+\sqrt{5}}{2}$$
Substituting back into $2{x}^2+4x$ we get $FG+JH+CD=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$
Notice that $A$ is trisected, meaning that $$AG=BH=EJ=JH=1$$. Since $JH=1$, and the other lines we are supposed to solve for do not look like they can contribute to the whole number value, it is likely that $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$ is our answer.
Note: Only do this if low on time since there could potentially be a weird figure affecting the $1$.
## Solution 5
Because basically everything in pentagon stuff is in ratio of big Phi or small Phi, we use logic to find that the answer is big Phi $+ 1 +$ small Phi = $(\sqrt5+1)/2 + 1 + (\sqrt5-1)/2$ = $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$
2015 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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## Write a factor that you can use to rationalize the denominator of 1/(sqrt(13z))
Question
Write a factor that you can use to rationalize the denominator of 1/(sqrt(13z))
in progress 0
3 months 2022-02-14T01:35:03+00:00 1 Answer 0 views 0
$$\dfrac{1}{\sqrt{13z}}=\dfrac{1}{\sqrt{13z}}\cdot\dfrac{\sqrt{13z}}{\sqrt{13z}}=\dfrac{\sqrt{13z}}{13z}$$
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# ACT Math : How to graph complex numbers
## Example Questions
### Example Question #1 : How To Graph Complex Numbers
The graph of passes through in the standard coordinate plane. What is the value of ?
Explanation:
To answer this question, we need to correctly identify where to plug in our given values and solve for .
Points on a graph are written in coordinate pairs. These pairs show the value first and the value second. So, for this data:
means that is the value and is the value.
We must now plug in our and values into the original equation and solve. Therefore:
We can now begin to solve for by adding up the right side and dividing the entire equation by .
Therefore, the value of is .
### Example Question #1 : Graphing
Point A represents a complex number. Its position is given by which of the following expressions?
Explanation:
Complex numbers can be represented on the coordinate plane by mapping the real part to the x-axis and the imaginary part to the y-axis. For example, the expression can be represented graphically by the point .
Here, we are given the graph and asked to write the corresponding expression.
not only correctly identifies the x-coordinate with the real part and the y-coordinate with the imaginary part of the complex number, it also includes the necessary
correctly identifies the x-coordinate with the real part and the y-coordinate with the imaginary part of the complex number, but fails to include the necessary .
misidentifies the y-coordinate with the real part and the x-coordinate with the imaginary part of the complex number.
misidentifies the y-coordinate with the real part and the x-coordinate with the imaginary part of the complex number. It also fails to include the necessary .
### Example Question #1 : How To Graph Complex Numbers
Which of the following graphs represents the expression ?
Complex numbers cannot be represented on a coordinate plane.
Explanation:
Complex numbers can be represented on the coordinate plane by mapping the real part to the x-axis and the imaginary part to the y-axis. For example, the expression can be represented graphically by the point .
Here, we are given the complex number and asked to graph it. We will represent the real part, , on the x-axis, and the imaginary part, , on the y-axis. Note that the coefficient of is ; this is what we will graph on the y-axis. The correct coordinates are .
### Example Question #2 : How To Graph Complex Numbers
The point is on the graph of . What is the value of ?
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# Integrate (x^3+1)/(x^3+x)
Asked on by svjr
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You should use partial fraction decomposition such that:
`(x^3+1)/(x(x^2+1)) = A/x + (Bx + C)/(x^2+1)`
You should bring both sides to a common denominator;
`x^3 + 1 = A(x^2 + 1) + x*(Bx + C)`
`x^3 + 1 = Ax^2 + A + Bx^2 + Cx`
`` `x^3 + 1 = (A+B)x^2 + Cx + A`
Equate the coefficients that involve the same powers of x:
A+B = 0 => A = -B
C = 0
A = 1 => B = -1
`(x^3+1)/(x(x^2+1)) = 1/x + (-x)/(x^2+1)`
Integrating both sides yields:
`int ((x^3+1)dx)/(x(x^2+1)) = int dx/x + int ((-x)dx)/(x^2+1)`
Use substitution procedure to solve `int ((-x)dx)/(x^2+1)`
Put `x^2+1 = t` .
Differentiating both sides yields dx: 2xdx = dt
Keep xdx to the left side: `xdx = dt/2`
Write the new integral: `int -dt/2t = -(ln|t|)/2 + c`
Replace t by`x^2+1` and remove the absolute value since `x^2+1gt0` .
`int ((-x)dx)/(x^2+1) = -ln sqrt(x^2 + 1) + c`
`int ((x^3+1)dx)/(x(x^2+1)) = ln |x| - ln sqrt(x^2 + 1) + c`
Use the quotient rule of logarithms:
`int ((x^3+1)dx)/(x(x^2+1)) = ln (|x|/sqrt(x^2 + 1)) + c`
Integrating the fraction `(x^3+1)/(x^3+x)` yields `int ((x^3+1)dx)/(x(x^2+1)) = ln (|x|/sqrt(x^2 + 1)) + c` .
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# Difference between revisions of "Power of a Point Theorem/Introductory Problem 4"
## Problem
(ARML) Chords $AB$ and $CD$ of a given circle are perpendicular to each other and intersect at a right angle at $E$ . Given that $BE = 16, DE = 4,$ and $AD = 5$, find $CE$.
## Solution
$ADE$ is a right triangle with hypotenuse 5 and leg 4. Thus, by the Pythagorean Theorem, $AE = 3$ (or by just knowing your Pythagorean Triples). Applying the Power of a Point Theorem gives $AE\cdot BE = CE\cdot DE$, or $3\cdot 16 = x\cdot 4$. Solving gives $x = 12$.
Back to the Power of a Point Theorem.
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# Inverse of set operations
$$A \cup B \equiv C$$ then what is $$A$$ in terms of $$B,C$$?
I tried to use $$A\cup B \equiv (A-B)\cup(A \cap B)\cup(B-A)$$ to find a similar expression for $$A \cap B$$ but got nowhere.
From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.
Complement is easy as it is it's own inverse $$(A^C)^C=A$$
not sure if inverse of difference operator is unique, for example, one can use $$(A-B)\cup A \equiv A$$ to construct one inverse of difference $$-X$$.
when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?
• What does $\equiv$ mean in this context? Why aren't you saying they are equal? – fleablood Jan 20 '19 at 0:38
Inverse operators for union, intersection, or set difference are impossible in general.
Knowing $$B$$ and $$C$$ is not enough to determine what $$A$$ is. Consider for example $$? \cup \{1,2\} = \{1,2,3\}$$ Then either $$\{1,3\}$$ or $$\{2,3\}$$ would be possible solutions (and there are two more), so there's no operator that given just $$\{1,2\}$$ and $$\{1,2,3\}$$ can tell you "which of them $$A$$ really was".
On the other hand symmetric difference $$A \mathbin{\triangle} B = (A\setminus B)\cup(B\setminus A)$$ has an inverse operation, namely itself: $$(A\mathop\triangle B)\mathop\triangle B = A$$.
If you consider the algebra of subsets of some universe $$U$$ under the operations $$\triangle$$ and $$\cap$$, you get a Boolean ring which satisfies the usual ring properties with $$\triangle$$ as addition and $$\cap$$ as multiplication. The ring's $$0$$ is $$\varnothing$$ and $$1$$ is $$U$$ itself.
This gives an opportunity to use more of the usual algebraic rules on sets. And you can express the remaining set operations in this vocabulary too: $$A^\complement = 1 \mathop\triangle A \qquad\qquad A\cup B = A\mathop\triangle B \mathop\triangle(A\cap B)$$
What you lose by doing things this way is the nice duality between $$\cup$$ and $$\cap$$ and De Morgan's laws for sets.
(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $$\mathbb Z$$).
• Nice counter example! – fleablood Jan 20 '19 at 1:31
• So if $C=A\triangle B$ then $C\triangle B=(A\triangle B)\setminus B \cup B\setminus(A\triangle B)=[(A\setminus B \cup B\setminus A)\setminus B ]\cup [B\setminus(A\setminus B \cup B\setminus A]=(A\setminus B)\cup B\setminus(B\setminus A) = (A\setminus B)\cup (B\cap A) = A$. .... nifty! – fleablood Jan 20 '19 at 2:04
It's not possible.
Given sets $$C$$ and $$B$$ there four states of being we can describe. $$x$$ in or not in $$C$$ and $$x$$ in or not in $$B$$ and we can only define sets by some combination of those conditions. Those conditions define the following $$4$$ disjoint basic sets and any set we can possibly be described in terms of $$A$$ and $$B$$ will be a union of these sets.
1) $$\{x \not \in C, x\not \in B\} = C^c$$ (as $$B \subset C$$)
2) $$\{x \not \in C, x \in B\} = \emptyset$$ (as $$B \subset C$$)
3) $$\{x \in C, x \not \in B\} = C\setminus B$$.
4) $$\{x\in C, x \in B\} = C \cap B = B$$ (as $$B \subset C$$).
Although $$A$$ is disjoint from 1) and 2) and 3) $$\subset A$$ set 4) will typically contain elements in $$A$$ and as well as elements not in $$A$$.
Hence in general we can not define $$A$$ solely on the conditions of whether they are or are not in $$C$$ or $$B$$.
====
FWIW
We can list all possible sets possible to describe they are
The four above:
1) and [1 and 2] $$C^c$$, 2)$$C^c\cap B = \emptyset$$, 3) and [2 and 3] $$C\setminus B$$ and 4) and [2 and 4] $$B$$.
[1 and 3] and [1 and 2 and 3] = $$C^c \cup C\setminus B = B^c$$.
[1 and 4] and [1 and 2 and 4] = $$C^c \cup B$$
[3 and 4] and [2 and 3 and 4] = $$(C\setminus B )\cup B = C$$
[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.
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# Square Root of 6
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Last updated date: 05th Aug 2024
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## Root 6 Value
The square root is a topic that many students find difficult to understand. However, once you understand the concept well, finding square roots will no longer be a challenge for you. So, in simple words square root, in mathematics, is a factor of a number that, when multiplied by itself, gives the original number. For example, both 3 and –3 are square roots of 9.
The square root of a number a is the number b such that b² = a. The square root of any number is represented by the symbol and is also often known as radical. The number or expression given under the square root symbol is known as radicand. The square root is a commonly used function in Mathematics. It is widely used in subjects like Mathematics and Physics. Sometimes it is tedious to find the square root of a number, especially the numbers which are not the perfect square of the number. In this article, we will discuss the square root of 6, and how to calculate the root 6 value using the simplifying square root method.
What is the root 6 value?
The root 6 value is 2.449
### Square Root of 6 Definition
The square root of a number 6 is a number y such that y² = 6. The square root of 6 in radical form is written as √6. The square root of 6 in radical form is expressed as√6
### How to Calculate the Under Root 6 Value?
We can calculate the under root 6 value using different methods of square roots. These methods can be the long division methods, prime factorization method, or simplifying square root method. Let us discuss how to calculate the under root 6 value using the simplifying square root method.
To simplify a square root, make the number under the square root as small as possible while keeping it as a whole number. Mathematically, it can be expressed as: √x.y=√x×√y
To express the square root of 6 in the simplest form, we will make the number 6 as small as possible, ensuring to keep it as a whole number. Hence, the square root of 6 in simplest form is represented as√6=√2×√3. This can be further simplified by substituting the value of √2 and √3
$\sqrt{6}=\sqrt{2}\times \sqrt{3}$
$\sqrt{6}=1.414\times 1.732$
$\sqrt{6}=2.449$
Hence, the square root of 6 in simplest form is 2.449. Similarly, we can also calculate the square root of any other whole numbers and their factors. Hence, simplifying the square root method is the simplest method of calculating the square root. We can also calculate the value of under root 6 using the calculator as this will give us the exact value. The exact value of the square root will always be given in a decimal number as it is impossible to determine a positive whole integer as a root for non-rational numbers.
### Simplifying the Square Root Using Perfect Square Method
Following are the steps to simplify the square root using the perfect square method:
1. Find the perfect square that divides the number in the radicand.
2. Express the numbers as a factor of a perfect square.
### Solved Example
1. Simplify $\mathbf{\sqrt{300}}$
Solution:
$\sqrt{300}=\sqrt{100\times 3}$
$\sqrt{300}=\sqrt{10\times 10\times 3}$
$\sqrt{300}=10\sqrt{3}$
Hence,$\sqrt{300}$ can be simplified as $10\sqrt{3}$
2. Simplify the following radical expressions :
1. $\sqrt{48}$
2. $\sqrt{75}$
Solutions:
i.$\mathbf{\sqrt{48}}$
Step 1: The perfect square 16 will divide the number 48.
Step 2: Express 48 as a factor of 16
48=16×3
Step 3: Reduce the square root of 16 as shown below:
$\sqrt{48}=\sqrt{16\times 3}$
$\sqrt{48}=\sqrt{4\times 4\times 3}$
$\sqrt{48}=4\sqrt{3}$
Hence,$\sqrt{48}$ can be simplified as$4\sqrt{3}$
ii.$\mathbf{\sqrt{75}}$
Step 1: The perfect square 25 will divide the number 75.
Step 2: Express 75 as a factor of 25.
75=25×3
Step 3: Simplify the radicals as shown below:
$\sqrt{75}=\sqrt{25\times 3}$
$\sqrt{75}=\sqrt{5\times 5\times 3}$
$\sqrt{75}=5\sqrt{3}$
Hence,$\sqrt{75}$ can be simplified as$5\sqrt{3}$
## FAQs on Square Root of 6
1. What is the Square Root of 6 in the Simplest Radical Form?
The square root of any number can only be simplified if the given number is divisible by a perfect square (other than number 1).√12 can be simplified further because 12 can be divisible by a perfect square i.e. 4. Similarly, √250 can also be simplified further because 250 can be easily divided by perfect square 25. But, √6 cannot be simplified further as 6 cannot be divided by a perfect square. Hence, the square root of 6 in the simplest radical form is expressed as √6.
2. Is the Value of Under Root 6 Rational or Irrational?
The value of under root 6 is considered as an irrational number because the number 6 is not a perfect square. This implies that the square root of 6 will have an infinite number when expressed in decimals. So, if we calculate the value of 6, we get the numbers that will go on into infinity, and will never repeat or terminate. It cannot be written even in the form of x/y, where y is not equal to 0. Hence, the value of 6 that we received is non-terminating.
3. How to Express the Value Under the Square Root 6 in a Fraction Form?
As we know, the value under the square root 6 is an irrational number. Therefore, the root 6 value cannot be expressed in an exact fraction form. However, we can make the value under the square root 6 into an approximate fraction by rounding it to the nearest hundredth. The square root of 6 to the nearest thousand is 2.45. Accordingly,
√6 = 2.45/1
= $\frac{245}{100}$
=$\frac{2\times 9}{20}$Hence, the approximate value of square root 6 in the fraction is 2 (9/20).
4. What is prime factorization?
To understand prime factorization, you need to know what is a factor and what are prime numbers. Take any number and try to find out the factors of that number for say 12 can be written as 1 × 12, 2 × 6, 3 × 4. So the factors are 1, 2, 3, 4, 6, and 12. And any number which is divisible by the number itself and 1. So, 2 and 3 are the prime factors of 12. And 12 can be written as 2×2×3.
5. How to find the square root of a number by the long division method?
Follow the steps given below to find the square root of a number by division method -
• Draw a line on the top of the number and a vertical line on the side of the number.
• Mark pairs of digits from the unit side till we cover the number completely
• Look at the first pair from the left. For eg, if 8281 is your number then your first pair is 82
• Find a perfect square that is equal to or lower than the first pair
• Write the perfect square number on the left of the number
• Then apply the multiply above and add the below method
• Subtract the perfect square from the first pair and write the result below
• Bring down the next pair i,e 81
• Use the trial and error method to find out the perfect multiple of the remainder.
• In simple terms, you divide, multiply, subtract, bring down and repeat the steps.
You can watch the video lecture on Vedantu’s website where we explain the concept using examples for easy understanding.
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Question
Question 1 (essay worth 10 points) (01.01 mc) part a: parallel lines are two lines that never meet. find an example that contradicts this definition. how would you change the definition to make it more accurate
1. ngockhue
The parallel lines will never intersect.
According to the statement
we have to prove that the parallel lines are two lines that never meet.
So, for this purpose, we know that the
Two lines are said to be parallel when their slopes are equal.
Every line can be written in a generalized form as
y= mx+c
Where x, y are two coordinates in axis system.
C is intercept of line on y axis ( coordinate where the line intersects X axis)
M is slope of line, it indirectly tells us inclination (Theta ‘θ’ : it is the angle that the line makes from positive x axis) .
value of m can be determined as
m=tan θ
Thus if lines are parallel both line equations will have same slope that is same angle (inclination) with x axis.
If you want to find if these lines intersect, the intersection point will satisfy both equations of line.
If you try to solve both equations of line for value of x and y using method of simultaneous equations, you will notice that there is no solution to these equations.
This means that these two lines don’t have any point in common.
Thus parallel lines will never intersect.
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# Thinksoft: Numerical Aptitude Questions (Paper 2 of 12)
Examrace Placement Series prepares you for the toughest placement exams to top companies.
1. A person covered some distance in 12 hours. He covered half the distance by rail @ 75 km per hour and the rest by car @ 45 km/hr. The total distance covered by him was
1. 450 km
2. 675 km
3. 337.5 km
4. 1350 km
2. A sum of Rs. 427 is to be divided among A, B and C in such a way that 3 times A's share, 4 times B's share and 7 times C's share are all equal. The share of C is
1. Rs. 84
2. Rs. 147
3. Rs. 196
4. Rs. 240
3. A and B entered into a partnership investing Rs. 12000 and Rs. 9000 respectively. After 3 months C also joined them with a capital of Rs. 15000. The share of C in the half yearly profit of Rs. 9500 is
1. Rs. 3500
2. Rs. 3000
3. Rs. 2500
4. Rs. 4000
4. The ratio of income of A and B is 5: 4 and their expenditure is as 3: 2. If at the end of the year, each saves Rs. 800, then the income of A is
1. Rs. 1700
2. Rs. 1800
3. Rs. 2000
4. Rs. 2200
5. A and B can together finish a work in 30 days. They worked at it for 10 days together and then B left. The remaining work was done by A alone in 30 more days. B alone can finish the work in
1. 48 days
2. 60 days
3. 75 days
4. 90 days
6. The ratio between the curved surface area and the total surface area of a right circular cylinder is 1: 2. If the total surface is 616 sq. Cm, the volume of the cylinder is
1. 1848 cm3
2. 1232 cm3
3. 1078 cm3
4. 980 cm3
7. A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6: 5. The smaller side of the rectangle is
1. 30 cm
2. 60 cm
3. 72 cm
4. 108 cm
8. A man walking at the rate of 6km per hour crosses a square field diagonally in 9 seconds. The area of the field is-
1. 125 sq. Cm
2. 112.5 sq. Cm
3. 110 sq. Cm
4. 100? 2 sq. m
9. A rectangular carpet has an area of 240 sq. Cm. If its diagonal and the longer side are together equal to five times the shorter side, the length of the carpet is-
1. 10 cm
2. 24 cm
3. 26 cm
4. 27.5 cm
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Straight Lines - 2 - JEE MCQ
# Straight Lines - 2 - JEE MCQ
Test Description
## 30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Straight Lines - 2
Straight Lines - 2 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Straight Lines - 2 questions and answers have been prepared according to the JEE exam syllabus.The Straight Lines - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Straight Lines - 2 below.
Solutions of Straight Lines - 2 questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Straight Lines - 2 solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Straight Lines - 2 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Straight Lines - 2 - Question 1
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Straight Lines - 2 - Question 2
Detailed Solution for Straight Lines - 2 - Question 2
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Straight Lines - 2 - Question 3
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Straight Lines - 2 - Question 4
Detailed Solution for Straight Lines - 2 - Question 4
Straight Lines - 2 - Question 5
Triangle formed by the lines x + y = 0 , x – y = 0 and lx + my = 1. If l and m vary subject to the condition l2 + m2 = 1 then the locus of its circumcentre is
Detailed Solution for Straight Lines - 2 - Question 5
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Straight Lines - 2 - Question 7
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Straight Lines - 2 - Question 11
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Straight Lines - 2 - Question 12
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Straight Lines - 2 - Question 13
Through a point A on the x-axis a straight line is drawn parallel to y-axis so as to meet the pair of straight lines ax2 + 2hxy + by2 = 0 in B and C. If AB = BC then
Detailed Solution for Straight Lines - 2 - Question 13
Straight Lines - 2 - Question 14
Triangle formed by the lines x + y = 0 , x – y = 0 and lx + my = 1. If l and m vary subject to the condition l2 + m2 = 1, then the locus of its circumcentre is
Detailed Solution for Straight Lines - 2 - Question 14
Straight Lines - 2 - Question 15
Let a, b, c ∈ R and satisfying (a + c)2 + 4b2 – 4ab – 4bc = 0 then the variable line ax + by + c = 0 passes through a fixed point whose co-ordinates are
Detailed Solution for Straight Lines - 2 - Question 15
Straight Lines - 2 - Question 16
If all lines given by the equation (3 sin θ + 5 cos θ)x + (7 sin θ – 3 cos θ) y + 11(sin θ – cos θ) = 0 pass through a fixed point P for all θ ∈ R, then find the distance of P from Q(7, – 10).
Detailed Solution for Straight Lines - 2 - Question 16
Straight Lines - 2 - Question 17
If a, b, c form an A.P. with common difference d (≠ 0) and x, y, z form a G.P. with common ratio r ( ≠1), then the area of the triangle with vertices (a, x), (b, y) and (c, z) is indepedent of
Detailed Solution for Straight Lines - 2 - Question 17
Straight Lines - 2 - Question 18
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Straight Lines - 2 - Question 19
The line x + 3y - 2 = 0 bisects the angle between a pair of straight lines of which one has equation x - 7y + 5 = 0 . The equation of the other line is :
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Straight Lines - 2 - Question 20
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Straight Lines - 2 - Question 21
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Straight Lines - 2 - Question 22
Vertices of a parallelogram ABCD are A(3, 1), B(13, 6), C(13, 21) and D(3, 16). If a line passing through the origin divides the parallelogram into two congruent parts then the slope of the line is
Detailed Solution for Straight Lines - 2 - Question 22
Straight Lines - 2 - Question 23
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Straight Lines - 2 - Question 24
The lines y = mx + b and y = bx + m intersect at the point (m – b, 9), where m ≠ b. The sum of the x-intercepts of the lines is
Detailed Solution for Straight Lines - 2 - Question 24
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Straight Lines - 2 - Question 26
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Straight Lines - 2 - Question 27
Points A & B are in the first quadrant ; point 'O' is the origin . If the slope of OA is 1, slope of OB is 7 and OA = OB, then the slope of AB is :
Detailed Solution for Straight Lines - 2 - Question 27
Straight Lines - 2 - Question 28
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Straight Lines - 2 - Question 29
A circle of constant radius ' a ' passes through origin ' O ' and cuts the axes of co-ordinates in points P and Q, then the equation of the locus of the foot of perpendicular from O to PQ is :
Detailed Solution for Straight Lines - 2 - Question 29
Straight Lines - 2 - Question 30
Detailed Solution for Straight Lines - 2 - Question 30
## Mathematics (Maths) for JEE Main & Advanced
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## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
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# Chapter 1 - Whole Numbers - 1.5 Dividing Whole Numbers - 1.5 Exercises - Page 54: 85
Check 2, 3, 5 and 10.
#### Work Step by Step
60 is even so is divisible by 2. The sum of the digits is 6, which is divisible by three so 60 is divisible by 3. 60 ends in zero so is divisible by 5. 60 ends in zero so is divisible by 10.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Select Page
I found it helpful to look at a variety of covariance ellipses for tame distributions so that I would have a better feel for those I come across in the wild. Following is a quick tour that I hope will show the effect of changing the parameters of the distributions of the series in question on the covariance ellipses:
1. For uncorrelated series, what is the effect of changing the standard deviations of the series?
2. For series with identical standard deviations, what is the effect of changing the correlation coefficients?
3. For series with different standard deviations, what is the effect of changing the correlation coefficients?
For an introduction to covariance ellipses, refer to my previous post: Covariance Ellipses
## Varying Standard Deviation of Uncorrelated Series
Here is the covariance ellipse for two uncorrelated series each with standard deviation of 20%. Theoretically we should see a circle (a special case of an ellipse) with semi-axes parallel to the x and y axes. I say “theoretically” because for these randomly generated series, the variances of the distributions are not quite equal and the covariance is not exactly zero. So the ellipse is stretched just slightly in the direction of the distribution with the higher variance, and rotated slightly.
When we generate series with markedly different variances, even though the covariance still is not exactly zero the small effect that rotates the axes of the ellipse away from the axes of the plot is overwhelmed. So we see the ellipse stretched in the direction of the distribution with the largest standard deviation. The axes of the ellipse are theoretically parallel to the axes of the plots: The maximum variance portfolio of two uncorrelated series is made up only of the series with the largest variance. The minimum variance portfolio contains only the series with the smaller variance.In these cases, the minor effect of the non-zero covariance makes itself felt in the orientation of the ellipse semi-axes. The eigen vectors of the covariance matrix will be flipped one way or the other depending on the slight deviation from zero in the measured covariance. Notice how, in the second chart the measured correlation is just slightly negative and the semi-axes are -ve and + ve respectively. In the third chart where the correlation comes out just +ve, both the semi-axes are negative.
A final point: if I had plotted correlation ellipse rather than covariance ellipses, all the charts would be the same aside from random variations.
## Varying Correlation
This is another of those “aha” situations. Intuitively, I always expect to see varying correlation cause the ellipse to rotate. Of course, it doesn’t rotate at all it causes the ellipse to change shape. Notice that the axes of all the three ellipses in the following chart have the same orientation of 45 degrees. The eigen vectors have coefficients of $\displaystyle \pm \frac {1} {\sqrt{2}} = \pm sin(45) = \pm cos(45)$ in all positions.
As the magnitude of the correlation coefficient increases the ellipse changes from a perfect circle (ρ = 0) elongating in the direction dictated by the sign of ρ and getting narrower perpendicular to this direction until it ultimately ends up as a straight line.
## Different Variance, Different Correlation
Finally we see some rotation in the covariance ellipses. IF the distributions have different variances (which, in the wild, they mostly will) then when the magnitude of the correlation coefficient increases, the ellipse BOTH stretches AND rotates away from the axis of the series with the larger variance. The final resting angle of the ellipse (compressed to a line) that represents the fully correlated series is dictated by the relationship of the variances of the two series.
This brief synopsis gives me enough understanding to be able to look at changes in covariance ellipses over time, or as a result of partitioning data either by time frame or some other criteria, and make a quick qualitative interpretation of what has happened to the correlation relationships.
Edit: LaTeX issues fixed
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# Expressions - How to Multiply and DivideAlgebraic Expressions
## Expressions - Multiplying and Dividing - How it Works - Video
### Example 1
Example 1:
Here we are simplifying expressions when only multiplication is involved. The expression is (4ab)(3a). The first step is drop to the parenthesis since none of the parenthesis are being raised to a power. The next step is to rearrange the like terms in order to give a better perspective on what is happening. Now we have 4 *3*a*a*b. We can multiply the 4 and 3 to make 12. Now we have 12*a*a*b. There are hidden exponents for a and it is 1 for each letter. If we add the hidden exponents or 1 for each a, we get 2 because 1 + 1 = 2. We can do this because of the Multiplication Rule of Exponents. So our answer is 12*a^2*b.
### Example 2
Example 2:
Here we are simplifying expressions when only multiplication is involved. The expression is (3ab)(2a^2)(5b)^2. The fist step is rewrite (5b)^2 so our exponent is 1. The exponent tells us that we can rewrite it twice. So now we have (3ab)(2a^2)(5b)(5b). Now we can drop the parenthesis since none of the parenthesis are being raised to a power. The next step is rearrange so the like terms are next to each other. So now we have 3*2*5*5*a*a^2*b*b*b. We multiply the numbers, 3*2*5*5, to get 150. Now we add the exponents of like terms, because of the Multiplication Rule of Exponents. For a, we have 1 + 2 = 3 and for b, we have 1 + 1 + 1 = 3. So our answer is 150*a^3*b^3.
### Example 3
Example 3:
Here we are simplifying expressions when only multiplication is involved. The expression is (8xy^2)(2y^2)(3x^2y)^4. Instead of writing the term (3x^2y) four times, we are going to use the Product Rule of Exponents. Now we can write the expression as 3^4*x^8*y^4. We multiply the exponents to get the answer so the hidden 1 for the 3 times the 4 (1*4) to get 3^4; 2 for the x^2 times the 4 (2*4) to get x^8; the hidden 1 for the y times the 4 (1*4) to get y^4. Now we can drop the parenthesis since none of the parenthesis are being raised to a power. The next step is rearrange so the like terms are next to each other. So now we have 8*2*81*x*x^8*y^2*y^2*y^4. Now multiply the numbers, 8*2*81, to get 1296. Now we add the exponents of like terms, because of the Multiplication Rule of Exponents. For x, we have the hidden 1 + 8 = 9 and for y, we have 2 + 2 + 4 = 8. So our answer is 1296*x^9*y^8.
### Example 4
Example 4:
Here we are simplifying expressions when only multiplication is involved. The expression is (5x^4y^3)/(4x^2y). In the picture we have written out all the variables based on their exponents to cancel the terms. Since there are 4 x's on top and 2 x's on bottom, we can cancel 2 of them. Instead of writing the variables each time, we can use the Division Rule of Exponents, where we can subtract the exponents if one like term is one top and one is bottom. So we have x^4 and x^2, we can subtract 4 - 2 to get 2 so we get x^2 on the top. We can do the same to the variable, y, because we have y^3 on top and y^1 on bottom. We can subtract 3 - 1 to get 2 and we get y^2 on the top. So our final answer is (5*x^2*y^2)/4.
### Example 5
Example 5:
Here we are simplifying expressions when only multiplication is involved. The expression is (4x^1y^3)/(8x^5y^2). In the picture we have written out all the variables based on their exponents to cancel the terms. Since there are 1 x on top and 5 x's on bottom, we can cancel 1 of them. Instead of writing the variables each time, we can use the Division Rule of Exponents, where we can subtract the exponents if one like term is one top and one is bottom. So we have x^1 and x^5, we can subtract 5 - 1 to get 4 so we get x^4 on the bottom. We can do the same to the variable, y, because we have y^3 on top and y^2 on bottom. We can subtract 3 - 2 to get 1 and we get y^1. Now we have to divide the 4 and 8 by 4 since both of those numbers are multiples of 4. 4 divided by 4 is 1 and 8 divided by 4 is 2. So our final answer is y/(2x^4).
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Paul's Online Notes
Home / Algebra / Preliminaries / Rational Exponents
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### Section 1-2 : Rational Exponents
Now that we have looked at integer exponents we need to start looking at more complicated exponents. In this section we are going to be looking at rational exponents. That is exponents in the form
${b^{\frac{m}{n}}}$
where both $$m$$ and $$n$$ are integers.
We will start simple by looking at the following special case,
${b^{\frac{1}{n}}}$
where $$n$$ is an integer. Once we have this figured out the more general case given above will actually be pretty easy to deal with.
Let’s first define just what we mean by exponents of this form.
$a = {b^{\frac{1}{n}}}\hspace{0.25in}\hspace{0.25in}{\mbox{is equivalent to }}\hspace{0.25in}{a^n} = b$
In other words, when evaluating $${b^{\frac{1}{n}}}$$ we are really asking what number (in this case $$a$$) did we raise to the $$n$$ to get $$b$$. Often $${b^{\frac{1}{n}}}$$ is called the $$n$$th root of b.
Let’s do a couple of evaluations.
Example 1 Evaluate each of the following.
1. $${25^{\frac{1}{2}}}$$
2. $${32^{\frac{1}{5}}}$$
3. $${81^{\frac{1}{4}}}$$
4. $${\left( { - 8} \right)^{\frac{1}{3}}}$$
5. $${\left( { - 16} \right)^{\frac{1}{4}}}$$
6. $$- {16^{\frac{1}{4}}}$$
Show All Solutions Hide All Solutions
Show Discussion
When doing these evaluations, we will not actually do them directly. When first confronted with these kinds of evaluations doing them directly is often very difficult. In order to evaluate these we will remember the equivalence given in the definition and use that instead.
We will work the first one in detail and then not put as much detail into the rest of the problems.
a $${25^{\frac{1}{2}}}$$ Show Solution
So, here is what we are asking in this problem.
${25^{\frac{1}{2}}} = ?$
Using the equivalence from the definition we can rewrite this as,
${?^{\,2}} = 25$
So, all that we are really asking here is what number did we square to get 25. In this case that is (hopefully) easy to get. We square 5 to get 25. Therefore,
${25^{\frac{1}{2}}} = 5$
b $${32^{\frac{1}{5}}}$$ Show Solution
So what we are asking here is what number did we raise to the 5th power to get 32?
${32^{\frac{1}{5}}} = 2\hspace{0.25in}\hspace{0.25in}{\mbox{because}}\hspace{0.25in}{2^5} = 32$
c $${81^{\frac{1}{4}}}$$ Show Solution
What number did we raise to the 4th power to get 81?
${81^{\frac{1}{4}}} = 3\hspace{0.25in}\hspace{0.25in}{\mbox{because}}\hspace{0.25in}{3^4} = 81$
d $${\left( { - 8} \right)^{\frac{1}{3}}}$$ Show Solution
We need to be a little careful with minus signs here, but other than that it works the same way as the previous parts. What number did we raise to the 3rd power (i.e. cube) to get -8?
${\left( { - 8} \right)^{\frac{1}{3}}} = - 2\hspace{0.25in}\,\,\,{\mbox{because}}\hspace{0.25in}{\left( { - 2} \right)^3} = - 8$
e $${\left( { - 16} \right)^{\frac{1}{4}}}$$ Show Solution
This part does not have an answer. It is here to make a point. In this case we are asking what number do we raise to the 4th power to get -16. However, we also know that raising any number (positive or negative) to an even power will be positive. In other words, there is no real number that we can raise to the 4th power to get -16.
Note that this is different from the previous part. If we raise a negative number to an odd power we will get a negative number so we could do the evaluation in the previous part.
As this part has shown, we can’t always do these evaluations.
f $$- {16^{\frac{1}{4}}}$$ Show Solution
Again, this part is here to make a point more than anything. Unlike the previous part this one has an answer. Recall from the previous section that if there aren’t any parentheses then only the part immediately to the left of the exponent gets the exponent. So, this part is really asking us to evaluate the following term.
$- {16^{\frac{1}{4}}} = - \left( {{{16}^{\frac{1}{4}}}} \right)$
So, we need to determine what number raised to the 4th power will give us 16. This is 2 and so in this case the answer is,
$- {16^{\frac{1}{4}}} = - \left( {{{16}^{\frac{1}{4}}}} \right) = - \left( 2 \right) = - 2$
As the last two parts of the previous example has once again shown, we really need to be careful with parenthesis. In this case parenthesis makes the difference between being able to get an answer or not.
Also, don’t be worried if you didn’t know some of these powers off the top of your head. They are usually fairly simple to determine if you don’t know them right away. For instance, in the part b we needed to determine what number raised to the 5 will give 32. If you can’t see the power right off the top of your head simply start taking powers until you find the correct one. In other words compute $${2^5}$$, $${3^5}$$, $${4^5}$$ until you reach the correct value. Of course, in this case we wouldn’t need to go past the first computation.
The next thing that we should acknowledge is that all of the properties for exponents that we gave in the previous section are still valid for all rational exponents. This includes the more general rational exponent that we haven’t looked at yet.
Now that we know that the properties are still valid we can see how to deal with the more general rational exponent. There are in fact two different ways of dealing with them as we’ll see. Both methods involve using property 2 from the previous section. For reference purposes this property is,
${\left( {{a^n}} \right)^m} = {a^{nm}}$
So, let’s see how to deal with a general rational exponent. We will first rewrite the exponent as follows.
${b^{\frac{m}{n}}} = {b^{\left( {\frac{1}{n}} \right)\left( m \right)}}$
In other words, we can think of the exponent as a product of two numbers. Now we will use the exponent property shown above. However, we will be using it in the opposite direction than what we did in the previous section. Also, there are two ways to do it. Here they are,
${b^{\frac{m}{n}}} = {\left( {{b^{\frac{1}{n}}}} \right)^m}\hspace{0.25in}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\hspace{0.25in}{b^{\frac{m}{n}}} = {\left( {{b^m}} \right)^{\frac{1}{n}}}$
Using either of these forms we can now evaluate some more complicated expressions
Example 2 Evaluate each of the following.
1. $${8^{\frac{2}{3}}}$$
2. $${625^{\frac{3}{4}}}$$
3. $${\left( {\displaystyle \frac{{243}}{{32}}} \right)^{\frac{4}{5}}}$$
Show All Solutions Hide All Solutions
Show Discussion
We can use either form to do the evaluations. However, it is usually more convenient to use the first form as we will see.
a $${8^{\frac{2}{3}}}$$ Show Solution
Let’s use both forms here since neither one is too bad in this case. Let’s take a look at the first form.
${8^{\frac{2}{3}}} = {\left( {{8^{\frac{1}{3}}}} \right)^2} = {\left( 2 \right)^2} = 4\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{8^{\frac{1}{3}}} = 2\,\,{\mbox{because }}{2^3} = 8$
Now, let’s take a look at the second form.
${8^{\frac{2}{3}}} = {\left( {{8^2}} \right)^{\frac{1}{3}}} = {\left( {64} \right)^{\frac{1}{3}}} = 4\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{64^{\frac{1}{3}}} = 4\,\,{\mbox{because }}{{\mbox{4}}^3} = 64$
So, we get the same answer regardless of the form. Notice however that when we used the second form we ended up taking the 3rd root of a much larger number which can cause problems on occasion.
b $${625^{\frac{3}{4}}}$$ Show Solution
Again, let’s use both forms to compute this one.
${625^{\frac{3}{4}}} = {\left( {{{625}^{\frac{1}{4}}}} \right)^3} = {\left( 5 \right)^3} = 125\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{625^{\frac{1}{4}}} = 5{\mbox{ because }}{5^4} = 625$ ${625^{\frac{3}{4}}} = {\left( {{{625}^3}} \right)^{\frac{1}{4}}} = {\left( {244140625} \right)^{\frac{1}{4}}} = 125\hspace{0.25in}{\mbox{because}}\;\;{125^4} = 244140625$
As this part has shown the second form can be quite difficult to use in computations. The root in this case was not an obvious root and not particularly easy to get if you didn’t know it right off the top of your head.
c $${\left( {\displaystyle \frac{{243}}{{32}}} \right)^{\frac{4}{5}}}$$ Show Solution
In this case we’ll only use the first form. However, before doing that we’ll need to first use property 5 of our exponent properties to get the exponent onto the numerator and denominator.
${\left( {\frac{{243}}{{32}}} \right)^{\frac{4}{5}}} = \frac{{{{243}^{\frac{4}{5}}}}}{{{{32}^{\frac{4}{5}}}}} = \frac{{{{\left( {{{243}^{\frac{1}{5}}}} \right)}^4}}}{{{{\left( {{{32}^{\frac{1}{5}}}} \right)}^4}}} = \frac{{{{\left( 3 \right)}^4}}}{{{{\left( 2 \right)}^4}}} = \frac{{81}}{{16}}$
We can also do some of the simplification type problems with rational exponents that we saw in the previous section.
Example 3 Simplify each of the following and write the answers with only positive exponents.
1. $${\left( {\displaystyle \frac{{{w^{ - 2}}}}{{16{v^{\frac{1}{2}}}}}} \right)^{\frac{1}{4}}}$$
2. $${\left( {\displaystyle \frac{{{x^2}{y^{ - \frac{2}{3}}}}}{{{x^{ - \frac{1}{2}}}{y^{ - 3}}}}} \right)^{ - \frac{1}{7}}}$$
Show All Solutions Hide All Solutions
a $${\left( {\displaystyle \frac{{{w^{ - 2}}}}{{16{v^{\frac{1}{2}}}}}} \right)^{\frac{1}{4}}}$$ Show Solution
For this problem we will first move the exponent into the parenthesis then we will eliminate the negative exponent as we did in the previous section. We will then move the term to the denominator and drop the minus sign.
$\frac{{{w^{ - 2\left( {\frac{1}{4}} \right)}}}}{{{{16}^{\frac{1}{4}}}{v^{\frac{1}{2}\left( {\frac{1}{4}} \right)}}}} = \frac{{{w^{ - \frac{1}{2}}}}}{{2{v^{\frac{1}{8}}}}} = \frac{1}{{2{v^{\frac{1}{8}}}{w^{\frac{1}{2}}}}}$
b $${\left( {\displaystyle \frac{{{x^2}{y^{ - \frac{2}{3}}}}}{{{x^{ - \frac{1}{2}}}{y^{ - 3}}}}} \right)^{ - \frac{1}{7}}}$$ Show Solution
In this case we will first simplify the expression inside the parenthesis.
${\left( {\frac{{{x^2}{y^{ - \frac{2}{3}}}}}{{{x^{ - \frac{1}{2}}}{y^{ - 3}}}}} \right)^{ - \frac{1}{7}}} = {\left( {\frac{{{x^2}{x^{\frac{1}{2}}}{y^3}}}{{{y^{\frac{2}{3}}}}}} \right)^{ - \frac{1}{7}}} = {\left( {\frac{{{x^{2 + \frac{1}{2}}}{y^{3 - \frac{2}{3}}}}}{1}} \right)^{ - \frac{1}{7}}} = {\left( {{x^{\frac{5}{2}}}{y^{\frac{7}{3}}}} \right)^{ - \frac{1}{7}}}$
Don’t worry if, after simplification, we don’t have a fraction anymore. That will happen on occasion. Now we will eliminate the negative in the exponent using property 7 and then we’ll use property 4 to finish the problem up.
${\left( {\frac{{{x^2}{y^{ - \frac{2}{3}}}}}{{{x^{ - \frac{1}{2}}}{y^{ - 3}}}}} \right)^{ - \frac{1}{7}}} = \frac{1}{{{{\left( {{x^{\frac{5}{2}}}{y^{\frac{7}{3}}}} \right)}^{\frac{1}{7}}}}} = \frac{1}{{{x^{\frac{5}{{14}}}}{y^{\frac{1}{3}}}}}$
We will leave this section with a warning about a common mistake that students make in regard to negative exponents and rational exponents. Be careful not to confuse the two as they are totally separate topics.
In other words,
${b^{ - n}} = \frac{1}{{{b^n}}}$
and NOT
${b^{ - n}} \ne {b^{\frac{1}{n}}}$
This is a very common mistake when students first learn exponent rules.
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## Vector product of two Vectors – Class 11 | Chapter – 7 | Physics Short Notes Series PDF for NEET & JEE
Vector product of two Vectors: The vector product has many important applications in physics and engineering, such as in the calculation of torque and angular momentum in rotational motion, in electromagnetism for the calculation of magnetic fields, and in fluid mechanics for the calculation of vorticity. It is also used in 3D computer graphics to calculate lighting and shading effects.
## Vector product of two Vectors
The vector product, also known as the cross product, is an operation between two vectors that results in a third vector that is perpendicular to both of the original vectors. The cross product of two vectors a and b is denoted by a × b and is defined as:
a × b = |a| |b| sin(θ) n
where |a| and |b| are the magnitudes of the vectors a and b, θ is the angle between the two vectors, and n is a unit vector perpendicular to both a and b in the direction given by the right-hand rule.
The right-hand rule is used to determine the direction of the resulting vector. To apply the right-hand rule, place the fingers of your right hand in the direction of the first vector (a) and curl them towards the second vector (b). Your thumb will then point in the direction of the resulting vector (a × b).
The magnitude of the resulting vector is given by:
|a × b| = |a| |b| sin(θ)
The cross product has many important applications in physics and engineering, such as in the calculation of torque and angular momentum in rotational motion, in electromagnetism for the calculation of magnetic fields, and in fluid mechanics for the calculation of vorticity. It is also used in 3D computer graphics to calculate lighting and shading effects.
## Applications of Vector product of two Vectors
The vector product, also known as the cross product, has many important applications in physics and engineering. Some of the most common applications of the vector product include:
• Calculation of torque: The cross product is used to calculate torque, which is a measure of the force that causes an object to rotate. The torque produced by a force F acting at a distance r from an axis of rotation is given by the cross product τ = r × F.
• Calculation of angular momentum: The cross product is used to calculate angular momentum, which is a measure of the rotational motion of an object. The angular momentum of an object with mass m and velocity v rotating around an axis is given by the cross product L = r × p, where r is the position vector of the object with respect to the axis of rotation, and p is its linear momentum.
• Calculation of magnetic fields: The cross product is used to calculate magnetic fields, which are produced by moving electric charges. The magnetic field produced by a current I flowing through a wire in a magnetic field B is given by the cross product B = I × r, where r is the position vector of a point in space with respect to the wire.
• Calculation of vorticity: The cross product is used to calculate vorticity, which is a measure of the local rotation of a fluid. The vorticity of a fluid element with velocity v is given by the cross product ω = ∇ × v, where ∇ is the gradient operator.
• 3D Computer graphics: The cross product is used extensively in 3D computer graphics to calculate lighting and shading effects, such as surface normals and specular highlights.
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# Diaspora
## Chapter 17: Partition of Unity
### Mathematical details
The surface of a hypersphere in 5 dimensions can be described by the equation:
x2 + y2 + z2 + u2 + w2 = R2 (1)
where x, y, z, u, w are the 5 spatial coordinates, and the origin of the coordinate system lies at the centre of the hypersphere.
Suppose this hypersphere is rotating as a rigid body. In general (in any number of dimensions) the velocity v of any point of a rotating body is given by the product of the body’s angular velocity matrix, Ω, with the vector for the point in question, r.
v = Ω r (2)
The angular velocity matrix must be antisymmetric: Ωij = –Ωji. To see this, note that Ω is the derivative with respect to time of the linear transformation that takes any point from its original position at t = 0 to its rotated position; let’s call the matrix for this transformation M(t). Since the motion is rigid, if we take two basis vectors, ei and ej, and rotate them with M(t), the dot product of the resulting vectors (which measures the angle between them) must remain constant, and so the rate of change of this with respect to time must be zero.
d/dt [(M(t) ei) · (M(t) ej)] = 0 (3a) d/dt [(M(t)kiek) · (M(t)rjer)] = 0 (3b) d/dt [M(t)kiM(t)kj] = 0 (3c) M(t)kiΩ(t)kj + Ω(t)kiM(t)kj = 0 (3d) Ω(0)ij + Ω(0)ji = 0 (3e) Ωij = –Ωji (3f)
Here we’ve used the Einstein summation convention of summing over all values of repeated indices, such as k and r. To get from (3d) to (3e), note that M(0) is just the identity matrix. In writing (3f) we’ve dropped any dependence on time from the angular velocity matrix, since we’re assuming that the body isn’t subject to external forces, leaving Ω constant.
In 5 dimensions, a completely general 5×5 antisymmetric matrix Ω will have 10 independent parameters, but it’s always possible to choose a basis that reduces it to the “canonical form”:
Ω =
0 ω1 0 0 0 –ω1 0 0 0 0 0 0 0 ω2 0 0 0 –ω2 0 0 0 0 0 0 0
(4)
Here, the x and y coordinates have been chosen to lie in one plane of rotation, the z and u coordinates in the other, and the w coordinate lies perpendicular to both planes. To see why it’s always possible to choose a basis that puts the matrix in this form, first note that the determinant of any antisymmetric N × N matrix where N is odd must be zero. This is because det Ω = det ΩT, and the determinant is a sum of products of N terms which all change sign in the transpose for an antisymmetric matrix, yielding det Ω = –(det Ω) for N odd. This means that there must be at least one non-zero vector in the null space of Ω. Choosing this as the direction for the w coordinate, the axis of rotation, makes the last row and the last column of Ω all zeros, and reduces the problem to 4 dimensions. The 4-dimensional case is described in detail below.
Multiplying the vector for a general point, r = (x,y,z,u,w), by this canonical matrix for Ω yields a velocity vector of:
v = (ω1y, –ω1x, ω2u, –ω2z, 0) (5)
So, for any point with x = y = z = u = 0, the velocity is zero. The set of points that meet this condition is just the w axis, the body’s axis of rotation. This intersects the surface of the hypersphere at w = ±R, giving two poles.
In the physically possible (but cosmologically unlikely) case that ω2 = 0, merely setting x = y = 0 is enough to make the velocity zero, and since the other three coordinates are free to take any values, they trace out a 3-dimensional volume. This volume intersects the hypersphere to form a single connected “pole”, the 2-sphere z2 + u2 + w2 = R2.
The two equatorial circles are {z = u = w = 0; x2 + y2 = R2}, where the speed at which the surface is moving is ω1R, and {x = y = w = 0; z2 + u2 = R2}, where the speed is ω2R.
### Mathematical details
In 4 dimensions, a completely general angular velocity matrix is described by 6 parameters:
A =
0 a b c –a 0 d e –b –d 0 f –c –e –f 0
(6)
It’s always possible to re-orient the coordinate system in such a way that A is converted to the canonical form. One method of doing this is to find the eigenvectors of AA, the matrix product of A with itself. This is a real symmetric matrix, and hence it must have 4 orthogonal eigenvectors; it turns out that they come in two pairs, with eigenvalues –ω12 and –ω22. This makes sense geometrically: applying A to any vector that lies in one of the planes of rotation simply rotates that vector by 90° and multiples it by the appropriate ω, so applying A twice reverses the vector’s direction and multiplies it by ω2. So each of these pairs of eigenvectors spans one of the planes of rotation.
Another way to find the planes of rotation involves a linear operator called the Hodge dual; the Hodge dual of a matrix M is usually written as *M. In general, this operator takes an algebraic description of a geometrical object, such as a plane, and produces the corresponding description of the perpendicular object; in the context of 4-dimensional Euclidean geometry, it maps planes to other planes. For example, if the 4 coordinates we’re using are called x, y, z and u, the Hodge dual of the xy plane is the zu plane, and similarly the dual of the plane spanned by any two coordinates is the plane spanned by the other two. With the small added complication that you need to stick to a consistent orientation scheme (to decide between the two possible directions of rotation within each plane) this is enough for us to write the Hodge dual of the matrix A. We just treat A as a sum of rotations in all 6 coordinate planes, and take the duals of each of them; this amounts to swapping the xy coordinate of A with the zu coordinate, etc., and changing a few signs to keep the orientation consistent.
*A =
0 f –e d –f 0 c –b e –c 0 a –d b –a 0
(7)
Now, what we want to do is write A as a sum of rotations in just two planes: one plane whose matrix we’ll call S, and the plane perpendicular to it, whose matrix will be *S. In other words, we want to find S, ω1, and ω2 such that:
A = ω1S + ω2*S (8)
Taking the dual of this, and noting that applying the dual twice to anything just gives you back the original matrix, yields:
*A = ω1* S + ω2S (9)
Multiplying Eqn (8) by ω1 and Eqn (9) by ω2 and taking the difference allows us to find S in terms of A, ω1 and ω2:
S = (ω1A – ω2*A) / (ω12 – ω22) (10)
To find the values of ω1 and ω2, note that if S is applied to a vector perpendicular to its plane, the result must be zero, and this is only possible if the determinant of S is zero. Writing out S in full, and then computing its determinant, gives:
S =
1/(ω12 – ω22)
0 ω1a–ω2f ω1b+ω2e ω1c–ω2d –ω1a+ω2f 0 ω1d–ω2c ω1e+ω2b –ω1b–ω2e –ω1d+ω2c 0 ω1f–ω2a –ω1c+ω2d –ω1e–ω2b –ω1f+ω2a 0
(11)
det S = [ω2/(ω12 – ω22)]4 [D(ω1/ω2)2 + |A|2(ω1/ω2) + D]2 (12)
where |A|2 = (a2+b2+c2+d2+e2+f2), the square of the magnitude of A, and D = (b ec da f) is the square root of the determinant of A. Eqn (12) can be solved to find the value of ω12 that makes det S equal to zero; call this value r.
r = [±√(|A|4 – 4 D2) – |A|2] / 2 D (13a) 1+r2 = |A|2 [|A|2 – √(|A|4 – 4 D2)] / 2 D2 (13b)
Then if S is normalised by requiring that |S|2 = |*S|2 = 1, ω1 and ω2 can be found individually. There’s a nice Pythagorean relationship between the squares of the magnitudes of the matrices involved, once A is split into dual parts, which makes it easy to find the individual rates of rotation.
|A|2 = ω12|S|2 + ω22|*S|2 = ω12 + ω22 (14a) ω12 = |A|2 r2 / (1+r2) = [|A|2 – √(|A|4 – 4 D2)] / 2 (14b) ω22 = |A|2 / (1+r2) = 2 D2 / [|A|2 – √(|A|4 – 4 D2)] = [|A|2 + √(|A|4 – 4 D2)] / 2 (14c)
Choosing new coordinates that put A into canonical form then requires picking one pair of orthogonal vectors spanning the plane defined by S, and then another pair spanning the plane defined by *S. There are standard linear algebra techniques for doing this, starting with a pair of row or column vectors from each matrix.
Diaspora / Chapter 17: Partition of Unity (detailed) / created Monday, 3 August 1998
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# Factorize the following:
Question:
Factorize the following:
20a12b2 − 15a8b4
Solution:
The greatest common factor of the terms $20 a^{12} b^{2}$ and $-15 a^{8} b^{4}$ of the expression $20 a^{12} b^{2}-15 a^{8} b^{4}$ is $5 a^{8} b^{2}$.
$20 a^{12} b^{2}=5 \times 4 \times a^{8} \times a^{4} \times b^{2}=5 a^{8} \times b^{2} \times 4 a^{4}$ and $-15 a^{8} b^{4}=5 \times-3 \times a^{8} \times b^{2} \times b^{2}=5 a^{8} b^{2} \times-3 b^{2}$
Hence, the expression 20a12b2 - 15a8b4 can be factorised as 5a8b2(4a4-3b2)
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Day-3-filled
# Day-3-filled - Math 1314 Day 3 Some Applications of the...
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Unformatted text preview: Math 1314 Day 3 Some Applications of the Derivative Section 11.4 Marginal Cost Suppose a business owner is operating a plant that manufactures a certain product at a known level. Sometimes the business owner will want to know how much it costs to produce one more unit of this product. The cost of producing this additional item is called the marginal cost . Suppose the total cost in dollars per week by ABC Corporation for producing its best- selling product is given by . 4 . 3000 000 , 10 ) ( 2 x x x C- + = We can find the actual cost of producing the 101 st item. The cost of producing the 101 st item can be found by computing the average rate of change, that is by computing 100 101 ) 100 ( ) 101 (-- C C . Note that x h x x C h x C C C- +- + =-- ) ( ) ( ) ( 100 101 ) 100 ( ) 101 ( where x = 100 and h = 1. Suppose the total cost in dollars per week by ABC Corporation for producing its best- selling product is given by . 4 . 3000 000 , 10 ) ( 2 x x x C- + = Find ) 100 ( ' C and interpret the results. Example 1 : A company produces noise-canceling headphones. Management of the company has determined that the total daily cost of producing x headsets can be modeled by the function 000 , 15 135 03 . 0001 . ) ( 2 3 + +- = x x x x C . Find the marginal cost function. Use the marginal cost function to approximate the actual cost of producing the 21 st , and 181 st headsets. Average Cost and Marginal Average Cost Suppose ) ( x C is the total cost function for producing x units of a certain product. If we divide this function by the number of units produced, x , we get the average cost function . We denote this function by ) ( x C . Then we can express the average cost function as x x C x C ) ( ) ( = . The derivative of the average cost function is called the marginal average cost....
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# Linear Equations Worksheet
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# College Chemistry
posted by Anonymous
a mixture contains two crystalline solid X and Y. Decide if pure X can be obtained by a proper recrystallization using solvent A. Include the exact volumes of solvent used and the amount of pure X obtained.
GIVEN:
Solubility of X in boiling A :200mg/ml
solubility of X at room temp.: 20mg/ml
solubility of Y at boiling A : 200mg/ml
solubility of Y at room temp.: 20mg/ml
amount of X in mixture: 400mg
amount of Y in mixture: 30mg
1. DrBob222
Suppose you take 2 mL of boiling A.
X = 2 mL x 200 mg/mL = 400 mg so all of X will dissolve.
Y = 2 mL x 200 mg/mL = 400 mg so all of Y will dissolve.
Now cool the solution to room temperature. How much X and Y will recrystallize?
2 mL x 20 mg/mL = 40 mg X will stay in solution; 400 - 40 = 360 mg X will crystallize.
For Y,
2 mL x 20 mg/mL = 40 mg Y will stay in solution; you had only 30 so none will crystallize.
2. Anonymous
I'm guessing that 2mL of boiling A came from (400mg)(1mL/200mg)= 2mL
So we don't have worry about the Y at boiling A? which will be 0.15mL if we do the calculation.
3. DrBob222
You add enough boiling A to dissolve all of X (of course it will dissolver all of Y, too). Using 0.15 mL of boiling A will dissolve all of Y and some of X. In fact I suppose you could get pure X this way, too, although it isn't the usual way we think of recrystallization.
200 mg/mL x 0.15 mL = 30 mg X will dissolve leaving 400-30 = 370 mg X behind, BUT
1. 0.15 mL is a small volume to be working with, and you would have trouble controlling the volume and losing volume as it sticks to the walls of the vessel,
2. I would want to add just a little more than 0.15 because this JUST barely gets all of Y in solution.
3. I doubt this is the way your prof expects you to answer.
4. Anonymous
I was also thinking so we can't add the two volume of A together? So like for X in boiling A there is 2mL and in Y there is 0.15ml. So all together solvent A would be 2.15mL.
all of Y would still be gone but for X instead of 360mg i would get 357mg
5. DrBob222
In general, we assume that we need only one volume. In this case, however, an extra 0.15 mL couldn't hurt anything. As you point, however, it affects the yield.
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# Absurdity?
zamba
Posers and Puzzles 17 Jun '03 00:33
1. 17 Jun '03 00:33
Given:
a = b
Multiply with a on both sides:
a^2 = ab
a^2 + (a^2 - 2ab) = ab + (a^2 - 2ab)
2a^2 - 2ab = a^2 - ab
2(a^2 - ab) = 1(a^2 - ab)
2 = 1
How is this possible? Anything wrong somewhere?
2. royalchicken
CHAOS GHOST!!!
17 Jun '03 00:38
Since a=b, a^2=ab, so a^2-ab=0. Thus in the final step you divided by 0.
3. 17 Jun '03 00:42
I'm impressed of your quick problem-solving. 🙂
I really like that problem. Still, I was hoping that at least _one_ person would answer wrong... *sob*
😛
4. royalchicken
CHAOS GHOST!!!
17 Jun '03 00:441 edit
Thanks. No problem then. WHen you post a problem, I'll give it at least half an hour before I post 😛.
5. 17 Jun '03 00:54
Hehehe. Don't worry about it. From now on, I'm only posting new problems when you are asleep. 🙂
However, if you've got a problem, please feel free to make a new thread ... 😛
*sits down, waiting for a new problem to pop up*
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Generating non-consecutive combinations
I am trying to create a generator (iterator which supports doing a next, perhaps using yield in python) which gives all combinations of r elements from {1,2,...n} (n and r are parameters) such that in the selected r elements, no two are consecutive.
For example, for r = 2 and n= 4
The generated combinations are `{1,3}, {1,4}, {2, 4}`.
I could generate all combinations(as an iterator) and filter those which don't satisfy the criteria, but we will be doing unnecessary work.
Is there some generation algorithm such that the `next` is O(1) (and if that is not possible, O(r) or O(n)).
The order in which the sets are returned is not relevant (and hopefully will allow an O(1) algorithm).
Note: I have tagged it python, but a language-agnostic algorithm will help too.
Update:
I have found a way to map it to generating pure combinations! A web search reveals that O(1) is possible for combinations (though it seems complicated).
Here is the mapping.
Suppose we have a combination x_1, x_2, ... , x_r with x_1 + 1 < x_2, x_2 + 1 < x_3, ...
We map to y_1, y_2, ..., y_r as follows
y_1 = x_1
y_2 = x_2 - 1
y_3 = x_3 - 2
...
y_r = x_r - (r-1)
This way we have that y_1 < y_2 < y_3 ... without the non-consecutive constraint!
This basically amounts to choosing r elements out of n-r+1. Thus all I need to do is run the generation for (n-r+1 choose r).
For our purposes, using the mapping after things are generated is good enough.
Here are some reasons why
1) It is effectively stateless (or "Markovian" to be more precise). The next permutation can be generated from the previous one. It is in a way almost optimal: O(r) space and time.
2) It is predictable. We know exactly the order(lexicographic) in which the combinations are generated.
These two properties make it easy to parallelise the generation (split at predictable points and delegate), with fault tolerance thrown in (can pick off from the last generated combination if a CPU/machine fails)!
Sorry, parallelisation was not mentioned earlier, because it didn't occur to me when I wrote the question and I got that idea only later.
-
Isn't generating and filtering going to be O(n)? Or, actually, O(r)? There's only one illegal value at each slot from 1 to r, so at most (r-1) combinations to skip. – abarnert Mar 14 '13 at 23:09
PS, "generating all combinations(as an iterator)" is a one-liner with `itertools`. – abarnert Mar 14 '13 at 23:09
@abarnert: It could potentially be Omega(nr) (or worse) can't it? Thanks for tip about itertools. – Knoothe Mar 14 '13 at 23:11
How could it be nr? One of us needs to think this through in detail. But keep in mind that you can assume the combinations arrive in sorted order. – abarnert Mar 14 '13 at 23:13
@abarnert: The underlying generator could be Omega(n), and so if you skip r calls to next, you have Omega(nr). You are right, though, more thought needs to be given (by me). – Knoothe Mar 14 '13 at 23:15
``````def nonconsecutive_combinations(r, n):
# first combination, startng at 1, step-size 2
combination = range(1, r*2, 2)
# as long as all items are less than or equal to n
while combination[r-1] <= n:
yield tuple(combination)
p = r-1 # pointer to the element we will increment
a = 1 # number that will be added to the last element
# find the rightmost element we can increment without
# making the last element bigger than n
while p > 0 and combination[p] + a > n:
p -= 1
a += 2
# increment the item and
# fill the tail of the list with increments of two
combination[p:] = range(combination[p]+1, combination[p] + 2*(r-p), 2)
``````
Each `next()` call should have an O(r) .. I got the idea while thinking about how this would translate to natural numbers, but it took quite some time to get the details right.
``````> list(nonconsecutive_combinations(2, 4))
[(1, 3), (1, 4), (2, 4)]
> list(nonconsecutive_combinations(3, 6))
[(1, 3, 5), (1, 3, 6), (1, 4, 6), (2, 4, 6)]
``````
Let me try to explain how this works.
Conditions for a tuple `c` with `r` elements to be part of the result set:
1. Any element of the tuple is at least as large as the preceding element plus 2. (`c[x] >= c[x-1] + 2`)
2. All elements are less than, or equal to `n`. Because of 1. it is sufficient to say that the last element is less than or equal to `n`. (`c[r-1] <= n`)
The smallest tuple that may be part of the result set is `(1, 3, 5, ..., 2*r-1)`. When I say a tuple is "smaller" than another, I am assuming the lexicographical order.
As Blckknght is pointing out, even the smallest possible tuple may be to large to satisfy condition 2.
The function above contains two while loops:
• The outer loop steps through the results and assumes they appear in lexicographical order and satisfy condition one. As soon as the tuple in question violates condition two, we know that we have exhausted the result set and are done:
``````combination = range(1, r*2, 2)
while combination[r-1] <= n:
``````
The first line initializes the result-tuple with the first possible result according to condition one. Line two squarely translates to condition two.
• The inner loop finds the next possible tuple satisfying condition one.
``````yield tuple(combination)
``````
Since the `while` condition (2) is true and we made sure the result satisfies condition one we can yield the current result-tuple.
Next, to find the lexicographically next tuple, we would add "1" to the last element.
``````# we don't actually do this:
combination[r-1] += 1
``````
However, that may break condition 2 too early. So, if that operation would break condition 2, we increment preceding element and adjust the last element accordingly. This is a little like counting integers base 10: "If the last digit is larger than 9, increment the previous digit and make the last digit a 0." But instead of adding zeros, we fill the tuple so that condition 1 is true.
``````# if above does not work
combination[r-2] += 1
combination[r-1] = combination[r-2] + 2
``````
Problem is, the second line may break condition two yet again. So what we actually do is, we keep track of the last element and that is what is done with the `a`. Also we use the `p` variable to refer to the index current element we are looking at.
``````p = r-1
a = 1
while p > 0 and combination[p] + a > n:
p -= 1
a += 2
``````
We are iterating right-to-left (p = r-1, p -= 1) through the items of the result tuple. Initially we want to add one to the last item (a = 1) but when stepping through the tuple we actually want to replace the last item with the value of a preceding item plus `2*x`, where `x` is the distance between the two items. (a += 2, combination[p] + a)
Finally, we have found the item we want to increment, and fill the rest of the tuple with a sequence starting at the incremented item, with a step size of 2:
``````combination[p:] = range(combination[p]+1, combination[p] + 2*(r-p), 2)
``````
And that's it. It seemed so easy when I first thought of it, but all the arithmetic throughout the function make a great place for off-by-one errors and describing it is harder than it should be. I should have known I'm in trouble when I added that inner loop :)
On performance ..
Unfortunately while loops filled with arithmetic are not the most efficient thing to write in Python. The other solutions accept that reality and use list comprehensions or filtering to push the heavy lifting down into the Python runtime. This seems to me to be the right thing to do.
On the other hand, I'm quite certain that my solution would perform a lot better than most if this were C. The inner while loop is O(log r) and it mutates the result in-place and the same O(log r). It does not consume additional stack frames and does not consume any memory besides the result and two variables. But obviously this is not C, so none of this really matters.
-
I did my best; hope this is of some help! – svckr Mar 15 '13 at 7:25
Thanks for the explanation. You can modify it to get combinations too :-)! – Knoothe Mar 15 '13 at 9:25
here is my recursive generator(it just skips `n+1`th item if `n`th item is selected):
``````def non_consecutive_combinator(rnge, r, prev=[]):
if r == 0:
yield prev
else:
for i, item in enumerate(rnge):
for next_comb in non_consecutive_combinator(rnge[i+2:], r-1, prev+[item]):
yield next_comb
print list(non_consecutive_combinator([1,2,3,4], 2))
#[[1, 3], [1, 4], [2, 4]]
print list(non_consecutive_combinator([1,2,3,4,5], 2))
#[[1, 3], [1, 4], [1, 5], [2, 4], [2, 5], [3, 5]]
print list(non_consecutive_combinator(range(1, 10), 3))
#[[1, 3, 5], [1, 3, 6], [1, 3, 7], [1, 3, 8], [1, 3, 9], [1, 4, 6], [1, 4, 7], [1, 4, 8], [1, 4, 9], [1, 5, 7], [1, 5, 8], [1, 5, 9], [1, 6, 8], [1, 6, 9], [1, 7, 9], [2, 4, 6], [2, 4, 7], [2, 4, 8], [2, 4, 9], [2, 5, 7], [2, 5, 8], [2, 5, 9], [2, 6, 8], [2, 6, 9], [2, 7, 9], [3, 5, 7], [3, 5, 8], [3, 5, 9], [3, 6, 8], [3, 6, 9], [3, 7, 9], [4, 6, 8], [4, 6, 9], [4, 7, 9], [5, 7, 9]]
``````
on efficiency :
this code can't be O(1) because of traversing stack and building a new collection upon each iteration won't be O(1). also, a recursive generator means you have to use `r` maximum stack depth to get `r`-item combination. this means with low `r`, the cost of calling stacks can be more expensive then non-recursive generation. with enough `n` and `r`, it is potentially much more efficient than itertools-based solution.
i've tested two uploaded codes in this question:
``````from itertools import ifilter, combinations
import timeit
def filtered_combi(n, r):
def good(combo):
return not any(combo[i]+1 == combo[i+1] for i in range(len(combo)-1))
return ifilter(good, combinations(range(1, n+1), r))
def non_consecutive_combinator(rnge, r, prev=[]):
if r == 0:
yield prev
else:
for i, item in enumerate(rnge):
for next_comb in non_consecutive_combinator(rnge[i+2:], r-1, prev+[item]):
yield next_comb
def wrapper(n, r):
return non_consecutive_combinator(range(1, n+1), r)
def consume(f, *args, **kwargs):
deque(f(*args, **kwargs))
t = timeit.timeit(lambda : consume(wrapper, 30, 4), number=100)
f = timeit.timeit(lambda : consume(filtered_combi, 30, 4), number=100)
``````
results and more results(edit) (on windows7, python 64bit 2.7.3, core i5 ivy bridge with 8gb ram) :
``````(n, r) recursive itertools
----------------------------------------
(30, 4) 1.6728046 4.0149797 100 times 17550 combinations
(20, 4) 2.6734657 7.0835997 1000 times 2380 combinations
(10, 4) 0.1253318 0.3157737 1000 times 35 combinations
(4, 2) 0.0091073 0.0120918 1000 times 3 combinations
(20, 5) 0.6275073 2.4236898 100 times 4368 combinations
(20, 6) 1.0542227 6.1903468 100 times 5005 combinations
(20, 7) 1.3339530 12.4065561 100 times 3432 combinations
(20, 8) 1.4118724 19.9793801 100 times 1287 combinations
(20, 9) 1.4116702 26.1977839 100 times 220 combinations
``````
as you can see, the gap between recursive solution and itertools.combination based solution gets wider as `n` goes up.
in fact, with the gap between both solutions is heavily dependent on `r` - bigger `r` means you have to throw away more combinations generated from `itertools.combinations`. for example, in case of `n=20, r=9` : we filter and take only 220 combinations among 167960(20C9) combinations. if `n` and `r` are small, using `itertools.combinations` can be faster because it is more efficient at less r and does not use stack as I explained. (as you can see, itertools are very optimized(if write your logic with `for`, `if`, `while` and bunch of generator and list comprehensions, it won't be as fast as the abstracted one with itertools), and this is an one of reasons why people love python - you bring your code to higher level, and you get rewarded. not many languages to that.
-
Thank you for your answer. – Knoothe Mar 14 '13 at 23:15
This is clearly not O(1) on the `next` call. Calculating exactly what it is (in terms of `n` and `r`) is a bit harder… but the good news is that it seems like it has the same complexity as the filtered-itertools solution, so if we figure out how to calculate one, we know the answer for the other. – abarnert Mar 14 '13 at 23:56
PS, why `[each for each in foo]` instead of just `list(foo)`? – abarnert Mar 14 '13 at 23:58
@abarnert I rewrote them from `for each in foo: .. `; just adding some brackets meant less typing. – thkang Mar 14 '13 at 23:59
@thkang: OK, but still, how is `list(foo)` more typing than a list comprehension that doesn't do anything? – abarnert Mar 15 '13 at 0:35
If there were a way to generate all combinations in O(1), you could do this in O(r) just by generating and filtering. Assuming `itertools.combinations` has an O(1) `next`, there are at most r-1 values to skip over, so the worst case is r-1 times O(1), right?
Jumping ahead a bit to avoid confusion, I don't think there is an O(1) implementation of `combinations`, and therefore this is not O(r). But is there anything possible that is? I'm not sure. Anyway…
So:
``````def nonconsecutive_combinations(p, r):
def good(combo):
return not any(combo[i]+1 == combo[i+1] for i in range(len(combo)-1))
return filter(good, itertools.combinations(p, r))
r, n = 2, 4
print(list(nonconsecutive_combinations(range(1, n+1), r))
``````
This prints:
``````[(1, 3), (1, 4), (2, 4)]
``````
The `itertools` documentation doesn't guarantee that `combinations` has an O(1) `next`. But it seems to me that if there is a possible O(1) algorithm, they'd use it, and if there isn't, you're not going to find one.
You can read the source code, or we could time it… but we're going to do that, let's time the whole thing.
http://pastebin.com/ydk1TMbD has my code, and thkang's code, and a test driver. The times it's printing are the cost of iterating the entire sequence, divided by the length of the sequence.
With `n` ranging from 4 to 20 and `r` fixed at 2, we can see that the times for both go down. (Remember, the total time to iterate the sequence is of course going up. It's just sublinear in `the total length`) With `n` ranging from 7 to 20 and `r` fixed at 4, the same is true.
With `n` fixed at 12 and `r` ranging from 2 to 5, the times for both go up linearly from 2 through 5, but they're much higher for 1 and 6 than you'd expect.
On reflection, that makes sense—there are only 6 good values out of 924, right? And that's why the time per `next` was going down as `n` went up. The total time is going up, but the number of values yielded is going up even faster.
So, `combinations` doesn't have an O(1) `next`; what it does have is something complicated. And my algorithm does not have an O(r) `next`; it's also something complicated. I think the performance guarantees would be a lot easier to specify over the whole iteration than per `next` (and then it's easy to divide by the number of values if you know how to calculate that).
At any rate, the performance characteristics are exactly the same for the two algorithms I tested. (Oddly, switching the wrapper `return` to `yield from` made the recursive one faster and the filtering one slower… but it's a small constant factor anyway, so who cares?)
-
Thanks, I need to try it out and think more about your claim. Actually now that I see your good, it is potentially Omega(r^2) (and so Omega(n^2)) even if underlying generator is O(1), isn't it? Still, more thought(from me) is required. – Knoothe Mar 14 '13 at 23:18
@Knoothe: Would it help if I used `timeit` to demonstrate how the time grows with n and with r? – abarnert Mar 14 '13 at 23:26
I will try it out on my workloads, but I do need this to be as fast as I can make it. My current solution is based on the filtering, and that does not seem sufficient, based on the perf measurements I have done. – Knoothe Mar 14 '13 at 23:56
I think this can't be `O(r)` per each `next()` because of identifying consecutive combinations in total combinations from `itertools`. – thkang Mar 15 '13 at 0:05
@thkang: I'm not sure what that explanation means, but if you read through the whole thing, I made it pretty clear that it's not actually O(r), and in fact it's apparently exactly the same complexity as your code (but I'm not sure what that is). – abarnert Mar 15 '13 at 0:36
Here's my attempt at a recursive generator:
``````def combinations_nonseq(r, n):
if r == 0:
yield ()
return
for v in range(2*r-1, n+1):
for c in combinations_nonseq(r-1, v-2):
yield c + (v,)
``````
This is approximately the same algorithm as thkang's recursive generator, but it has better performance. If `n` is close to `r*2-1` the improvement is very large, while for smaller `r` values (relative to `n`) it is a small improvement. It's also a bit better than svckr's code, without a clear connection to `n` or `r` values.
The key insight I had was that when `n` is less than `2*r-1`, there can be no combinations that do not have adjacent values. This lets my generator do less work than thkang's.
Here's some timings, run using a modified version of thkang's `test` code. It uses the `timeit` module to find out how much time it takes to consume the whole contents of the generator ten times. The `#` column shows the number of values are yielded by my code (I'm pretty sure all the others are the same).
``````( n, r) # |abarnert | thkang | svckr |BlckKnght| Knoothe |JFSebastian
===============+=========+=========+=========+=========+=========+========
(16, 2) 105 | 0.0037 | 0.0026 | 0.0064 | 0.0017 | 0.0047 | 0.0020
(16, 4) 715 | 0.0479 | 0.0181 | 0.0281 | 0.0117 | 0.0215 | 0.0143
(16, 6) 462 | 0.2034 | 0.0400 | 0.0191 | 0.0125 | 0.0153 | 0.0361
(16, 8) 9 | 0.3158 | 0.0410 | 0.0005 | 0.0006 | 0.0004 | 0.0401
===============+=========+=========+=========+=========+=========+========
(24, 2) 253 | 0.0054 | 0.0037 | 0.0097 | 0.0022 | 0.0069 | 0.0026
(24, 4) 5985 | 0.2703 | 0.1131 | 0.2337 | 0.0835 | 0.1772 | 0.0811
(24, 6) 27132 | 3.6876 | 0.8047 | 1.0896 | 0.5517 | 0.8852 | 0.6374
(24, 8) 24310 | 19.7518 | 1.7545 | 1.0015 | 0.7019 | 0.8387 | 1.5343
``````
For larger values of `n`, abarnert's code was taking too long, so I've dropped it from the next tests:
``````( n, r) # | thkang | svckr |BlckKnght| Knoothe |JFSebastian
===============+=========+=========+=========+=========+========
(32, 2) 465 | 0.0069 | 0.0178 | 0.0040 | 0.0127 | 0.0064
(32, 4) 23751 | 0.4156 | 0.9330 | 0.3176 | 0.7068 | 0.2766
(32, 6) 296010 | 7.1074 | 11.8937 | 5.6699 | 9.7678 | 4.9028
(32, 8)1081575 | 37.8419 | 44.5834 | 27.6628 | 37.7919 | 28.4133
``````
The code I've been testing with is here.
-
could you run your benchmark for `combinations_stack()` and `combinations_knoothe()` from my answer? – J.F. Sebastian Mar 15 '13 at 7:14
@J.F.Sebastian: I'm playing around with it, but I'm getting different answers depending on whether or not I include the time required to create the input list as part of your function's runtime. Interestingly, on Python 3, your can also run your code on a `range` object (`range`s can be sliced), but performance is worse than with a list. Once I figure out how to account for this, I'll replace my big timing dump above with something new. – Blckknght Mar 15 '13 at 8:00
you could use wrapper: `f = lambda r, n: combinations_stack(list(range(1, n+1)), r)`. Or replace `seq[i]` with `i+1`. There are `(n-r+1)! / r! / (n-2r+1)!` items in the result so unless `n` is small; the initial O(n) setup shouldn't matter. – J.F. Sebastian Mar 15 '13 at 8:23
Thank you for your answer. – Knoothe Mar 15 '13 at 9:22
I have chosen svkcr's answer, and updated the question with the reason. Thanks! – Knoothe Mar 15 '13 at 9:41
Here's a solution similar to @thkang's answer but with explicit stack:
``````def combinations_stack(seq, r):
stack = [(0, r, ())]
while stack:
j, r, prev = stack.pop()
if r == 0:
yield prev
else:
for i in range(len(seq)-1, j-1, -1):
stack.append((i+2, r-1, prev + (seq[i],)))
``````
Example:
``````print(list(combinations_stack(range(1, 4+1), 2)))
# -> [(1, 3), (1, 4), (2, 4)]
``````
For some `(n, r)` values it is the fastest solution according to the benchmark on my machine:
``````name time ratio comment
combinations_knoothe 17.4 usec 1.00 8 4
combinations_blckknght 17.9 usec 1.03 8 4
combinations_svckr 20.1 usec 1.16 8 4
combinations_stack 62.4 usec 3.59 8 4
combinations_thkang 69.6 usec 4.00 8 4
combinations_abarnert 123 usec 7.05 8 4
name time ratio comment
combinations_stack 965 usec 1.00 16 4
combinations_blckknght 1e+03 usec 1.04 16 4
combinations_knoothe 1.62 msec 1.68 16 4
combinations_thkang 1.64 msec 1.70 16 4
combinations_svckr 1.84 msec 1.90 16 4
combinations_abarnert 3.3 msec 3.42 16 4
name time ratio comment
combinations_stack 18 msec 1.00 32 4
combinations_blckknght 28.1 msec 1.56 32 4
combinations_thkang 40.4 msec 2.25 32 4
combinations_knoothe 53.3 msec 2.96 32 4
combinations_svckr 59.8 msec 3.32 32 4
combinations_abarnert 68.3 msec 3.79 32 4
name time ratio comment
combinations_stack 1.84 sec 1.00 32 8
combinations_blckknght 2.27 sec 1.24 32 8
combinations_svckr 2.83 sec 1.54 32 8
combinations_knoothe 3.08 sec 1.68 32 8
combinations_thkang 3.29 sec 1.79 32 8
combinations_abarnert 22 sec 11.99 32 8
``````
where `combinations_knoothe` is an implementation of the algorithm described in the question:
``````import itertools
from itertools import imap as map
def _combinations_knoothe(n, r):
def y2x(y):
"""
y_1 = x_1
y_2 = x_2 - 1
y_3 = x_3 - 2
...
y_r = x_r - (r-1)
"""
return tuple(yy + i for i, yy in enumerate(y))
return map(y2x, itertools.combinations(range(1, n-r+1+1), r))
def combinations_knoothe(seq, r):
assert seq == list(range(1, len(seq)+1))
return _combinations_knoothe(len(seq), r)
``````
and other functions are from corresponding answers (modified to accept input in the unified format).
-
Thank you for your answer. – Knoothe Mar 15 '13 at 9:21
I have chosen svkcr's answer, and updated the question with the reason. Thanks! – Knoothe Mar 15 '13 at 9:40
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### Updated sec. 5.2
1. 1. What is the biggest number that you can divide both 16 and 24 by?
2. 2. Learning Target: Students will learn how to find the greatest common factor (GCF) of two or more numbers and they will learn the different strategies for doing so.
3. 3. A common factor is a whole number that is a factor of two or more nonzero whole numbers. The biggest of the common factors is called the greatest common factor (GCF)
4. 4. There are 3 methods for finding the GCF. 1. List all the factors. (rainbow) 2. Factor Tree and multiply the common numbers to find GCF. 3. CAKE – multiply the numbers on the left side of the cake
5. 5. Find the GCF of 14, 35. 1. List all the factors. (rainbow) 14: 1, 2, 7, 14 35: 1, 5, 7, 35 Common factors: 1, 7 GCF: 7
6. 6. Find the GCF of 12, 54 2. Factor Tree and multiply the common numbers to find GCF. 12 54 3 4 6 9 3 2 2 2 3 3 3 2 x 3 = 6 GCF = 6
7. 7. Find the GCF of 27, 45 3. CAKE – multiply the numbers on the left side of the cake 2, 3, 5, 7, 11, 13 3 27 45 3 9 15 3 5 GCF = 3 x 3 = 9
8. 8. CAKE works the same as the ladder except both numbers must be divisible by the number before you write it down in each step and you stop when one or both of your numbers turn prime. Then to find the GCF you multiply the numbers on the left side of the cake. This works the same when you are given three numbers.
9. 9. Find the GCF using any of the methods. 20, 75 12, 21, 30
10. 10. Answers: 20, 75 GCF: 5 12, 21, 30 GCF: 3
11. 11. Can you find the greatest common factor (GCF) of two or more numbers?
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# Prime Number Iterator and Calculations
This library provides a structure for iterating through prime numbers as well as methods for calculating prime factors and classifying numbers as prime or composite.
# Using this library
Add the following to your `Cargo.toml` file
``````[dependencies]
elr_primes = "0.1.0"
``````
# Examples
Basic Usage:
```use elr_primes::Primes;
// Provides an iterator for all prime numbers less than or equal to 1000
let mut p = Primes::new(1000);```
Once the structure has been initiated, the `primes()` method provides an iterator for the prime numbers.
```let p = Primes::new(10); // Primes less than or equal to 10
let mut prime_iter = p.primes();
let primes: Vec<usize> = prime_iter.copied().collect();
let expected: [usize; 4] = [2, 3, 5, 7];
assert_eq!(primes, expected);```
Since `primes()` returns an iterator, you can also use it to directly find a specific prime number.
```let p = Primes::new(100); // Primes less than or equal to 100
let n = 20;
// Iterators are zero-based, so to get the 20th prime we need to search for the 19th
match p.primes().nth(n - 1) {
Some(x) => println!("The {}th prime is: {}", n, x),
None => println!("The {}th prime is outside the current bounds", n)
};```
Methods are also available to find the prime factors of a number, and whether a number is prime or composite.
```use elr_primes::{Primes, Primality};
let p = Primes::new(100);
let n = 96;
match p.factors(n) {
Ok(factors) => println!("{:?}", factors),
Err(_) => println!("Could not find all prime factors within the bounds"),
};
let n = 23;
match p.primality(n) {
Primality::Prime => println!("{} is prime", n),
Primality::Composite => println!("{} is composite", n),
Primality::Unknown => println!("Primality of {} is undetermined", n),
};```
## Structs
Primes Prime Iterator
## Enums
Primality Primality types for numbers.
## Type Definitions
Factors Vector of prime factor tuples.
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# The velocity of the particle after t seconds .
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 3.4, Problem 69E
To determine
## To find: The velocity of the particle after t seconds.
Expert Solution
The velocity of the particle after t seconds is v(t)=2.5πcos(10πt) cm/s.
### Explanation of Solution
Given:
The equation of vibrating string is s(t)=10+14sin(10πt), where t is in seconds and s is in centimeters.
Recall:
If x(t) is a displacement of a particle and the time t is in seconds, then the velocity of the particle is v(t)=dxdt.
Calculation:
Obtain the velocity of the particle after t seconds.
v(t)=ddt(s(t))=ddt(10+14sin(10πt))=0+14cos(10πt)10π=5π2cos(10πt)
Therefore, the velocity of the particle after t seconds is v(t)=2.5πcos(10πt) cm/s.
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!
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# Integrating functions
A volume form on a discrete surface $M$ is a 2-form $dA$ such that $\textrm{vol}(\varphi):=dA(\varphi)>0$ for all faces $\varphi \in F$. The interpretation is that a volume form assign an area $\textrm{vol}(\varphi)$ to each face $\varphi$. The total area of $M$ is then defined as
$\displaystyle \textrm{vol}(M):=\int_M dA = \sum_{\varphi \in F} \textrm{vol}(\varphi)$
Given a volume form $dA$, we can define the integral of a dual function $f\in \Omega_0(M^*)$:
$\displaystyle \int_M f :=\int_M f\,dA= \sum_{\varphi \in F} f|_\varphi\,\textrm{vol}(\varphi)$.
Here we have used the notation
$f|_\varphi := f(\varphi)$
for the evaluation of dual 0-forms on faces (recall they are to be viewed as functions that are constant on faces).
We can use this to make $\Omega_0(M^*)$ into a euclidean vector-space by defining a scalar product
$\displaystyle \langle \! \langle f,g \rangle \! \rangle =\int_M fg\, \,dA$.
One way to view such a scalar product is that it defines an isomorphism $\star_2^{-1}$ into the dual space:
$\star_2^{-1}: \Omega_0(M^*) \rightarrow \Omega_0(M^*)^*=\Omega_2(M)$
$\displaystyle \langle \star_2^{-1}(f), g \rangle := \langle \! \langle f,g \rangle \! \rangle= \int_M gf \, dA$.
Thus $\star_2^{-1}(f) = f \, \sigma$. Setting $\star_2^{-1}(f)=:\tau$ this implies $\tau(\varphi)= \star_2(\varphi) \sigma(\varphi)$ for all $\varphi \in F$ or
$\displaystyle (\star_2 \tau)|_\varphi = \frac{1}{\textrm{vol}(\varphi)} \int_\varphi \tau$
Similarly, a volume form on $M^*$ allows to integrate functions $f \in \Omega_0(M)$, now seen as functions that are constant on each dual cell. Then we also have a euclidean scalar product on $\Omega_0(M)$. Now it seems unreasonable to choose $\sigma$ and $\sigma^*$ in a completely independent way. For example, a natural demand seems to be that $M$ and $M^*$ have the same area:
$\textrm{vol}(M^*) = \textrm{vol}(M)$.
We can ensure this and other natural properties by deriving both $\sigma$ and $\sigma^*$ from a volume-form on the kite complex of $M$. This is a discrete surface that kind of mediates between $M$ and $M^*$. The kite complex has vertices at the vertices of $M$, the vertices of $M^*$ and the “centers” of the edges of $M$ (understood combinatorially). Its faces (the kites) are all quadrilaterals and correspond to the intersections of faces of $M$ (the so-called primal faces) with faces of $M^*$ (the so-called dual faces). In the picture below, the triangle is a primal face and is the union of three kites. Each of the three dual faces is a union of six kites.
It is a nice exercise to construct the Kite complex as a discrete surface $M^K=(E^K,s^K,\rho^K)$. What is clear is that (up to a canonical bijection) the set of faces of $M^K$ is
$F^K = \{(v,\varphi)\in V \times F \,|\, v \cap \varphi \neq \emptyset\}$.
Here the intersection of $v$ and $\varphi$ is to be understood as subsets of $E$.
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# Parallelogram
In the parallelogram we know one internal angle 67°33`. Calculate the other internal angles.
Correct result:
β = 112.45 °
γ = 67.55 °
δ = 112.45 °
#### Solution:
$\beta =180-\left(67+33\mathrm{/}60\right)={\frac{2249}{20}}^{\circ }=112.4{5}^{\circ }=11{2}^{\circ }2{7}^{\mathrm{\prime }}$
$\gamma =67+33\mathrm{/}60={\frac{1351}{20}}^{\circ }=67.5{5}^{\circ }=6{7}^{\circ }3{3}^{\mathrm{\prime }}$
$\delta =180-\left(67+33\mathrm{/}60\right)={\frac{2249}{20}}^{\circ }=112.4{5}^{\circ }=11{2}^{\circ }2{7}^{\mathrm{\prime }}$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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or
# If (sin 3theta)/(cos 2theta)lt 0, then theta lies in
Question from Class 12 Chapter Trigonometric Equations
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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`((3pi),(8),(23pi)/(48))``((7pi)/(24),(3pi)/(8))``((13pi)/(48),(7pi)/(24))``((2pi)/(4),(7pi)/(12))`
C
Solution :
Let `(sin 3theta)/(cos 2theta)lt 0` <br> Case I. `sin theta lt 0` and `cos 2theta gt 0` <br> `rArr 3theta in (pi, 2pi)` and `2theta in (-(pi)/(2),(-pi)/(2))` <br> `rArr theta in((pi)/(3),(2pi)/(3))` and `theta in (-(pi)/(4), (pi)/(4))` <br> Case II. `sin 3theta gt 0` and `cos 2 theta lt 0` <br> `therefore 3theta in(0, pi)` and `2theta in((pi)/(2),(3pi)/(2))` <br> `rArr theta in(0,(pi)/(3))` and `theta in((pi)/(4),(3pi)/(4))` <br> From case I and II, we have <br> `theta in((pi)/(4),(pi)/(3))` <br> Since `((13pi)/(48),(7pi)/(24))in((pi)/(4),(pi)/(3))` <br> `therefore theta in((13pi)/(48),(7pi)/(24))`
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Eureka Math Algebra 2 Module 1 Lesson 31 Answer Key
Engage NY Eureka Math Algebra 2 Module 1 Lesson 31 Answer Key
Eureka Math Algebra 2 Module 1 Lesson 31 Exercise Answer Key
Exercise 1.
Draw a graph of the circle with equation x2 + y2 = 9.
a. What are the solutions to the system of circle and line when the circle is given by x2 + y2 = 9, and the line is given by y = 2?
Substituting y = 2 in the equation of the circle yields x2 + 4 = 9, so x2 = 5, and x = √5 or x = -√5. The solutions are (-√5, 2) and (√5, 2).
b. What happens when the line is given by y = 3?
Substituting y = 3 in the equation of the circle yields x2 + 9 = 9, so x2 = 0. The line is tangent to the circle, and the solution is (0, 3).
c. What happens when the line is given by y = 4?
Substituting y = 4 in the equation of the circle yields x2 + 16 = 9, so x2 = – 7. Since there are no real numbers that satisfy x2 = – 7, there is no solution to this equation. This indicates that the line and circle do not intersect.
Exercise 2.
By solving the equations as a system, find the points common to the line with equation x – y = 6 and the circle with equation x2 + y2 = 26. Graph the line and the circle to show those points.
(5, – 1) and (1, – 5). See picture to the right.
Exercise 3.
Graph the line given by 5x + 6y = 12 and the circle given by x2 + y2 = 1. Find all solutions to the system of equations.
There is no real solution; the line and circle do not intersect. See picture to the right.
Exercise 4.
Graph the line given by 3x + 4y = 25 and the circle given by x2 + y2 = 25. Find all solutions to the system of equations. Verify your result both algebraically and graphically.
The line is tangent to the circle at (3, 4), which is the only solution. See picture to the right.
Exercise 5.
Graph the line given by 2x + y = 1 and the circle given by x2 + y2 = 10. Find all solutions to the system of equations. Verify your result both algebraically and graphically.
The line and circle intersect at (- 1, 3) and $$\left(\frac{9}{5},-\frac{13}{5}\right)$$, which are the two solutions. See picture to the right.
Exercise 6.
Graph the line given by x + y = – 2 and the quadratic curve given by y = x2 – 4. Find all solutions to the system of equations. Verify your result both algebraically and graphically.
The line and the parabola intersect at (1, – 3) and (- 2, 0), which are the two solutions. See picture to ‘ the right.
Eureka Math Algebra 2 Module 1 Lesson 31 Problem Set Answer Key
Question 1.
Where do the lines given by y = x + b and y = 2x + 1 intersect?
Since we do not know the value of b, we cannot solve this problem by graphing, and we will have to approach it algebraically. Eliminating y gives the equation
x + b = 2x + 1
x = b – 1.
Since x = b – 1, we have y = x + b = (b – 1) + b = 2b – 1. Thus, the lines intersect at the point (b – 1, 2b – 1).
Question 2.
Find all solutions to the following system of equations.
(x – 2)2 + (y + 3)2 = 4
x – y = 3
Illustrate with a graph.
Solve the linear equation for one of the variables: x = y + 3.
Substitute that variable in the quadratic equation:
(y + 3 – 2)2 + (y + 3)2 = 4.
Rewrite the equation in standard form: 2y2 + 8y + 6 = 0.
Solve the quadratic equation: 2(y + 3)(y + 1) = 0, so
y = – 3 or y = -1.
If y = – 3, then x = 0. If y = -1, then x = 2.
As the graph shows, the solution is the two points (0, -3) and (2, -1).
Question 3.
Find all solutions to the following system of equations.
x + 2y = 0
x2 – 2x + y2 – 2y – 3 = 0
Illustrate with a graph.
Solve the linear equation for one of the variables: x = – 2y.
Substitute that variable in the quadratic equation:
(-2y)2 – 2(-2y) + y2 – 2y – 3 = 0.
Rewrite the equation in standard form: 5y2 + 2y – 3 = 0.
Solve the quadratic equation: (5y – 3)(y + 1) = 0, so y = $$\frac{3}{5}$$ or
y = – 1.
If y = $$\frac{3}{5}$$, then x = –$$\frac{6}{5}$$. If y = – 1, then x = 2.
As the graph shows, the solutions are the two points: (-$$\frac{6}{5}$$, $$\frac{3}{5}$$) and (2, -1).
Question 4.
Find all solutions to the following system of equations.
x + y = 4
(x + 3)2 + (y – 2)2 = 10
Illustrate with a graph.
Solve the linear equation for one of the variables: x = 4 – y.
Substitute that variable in the quadratic equation:
(4 – y + 3)2 + (y – 2)2 = 10.
Rewrite the equation in standard form: 2y2 – 18y + 43 = 0.
Solve the equation using the quadratic formula:
y = $$\frac{18+\sqrt{324-344}}{4}$$ or y = $$\frac{18-\sqrt{324-344}}{4}$$
So we have y = $$\frac{1}{2}$$(9 + √-5) or y = $$\frac{1}{2}$$(9 – √-5).
Therefore, there is no real solution to the system.
As the graph shows, the line and circle do not intersect.
Question 5.
Find all solutions to the following system of equations.
y = – 2x + 3
y = x2 – 6x + 3
Illustrate with a graph.
The linear equation is already solved for one of the variables: y = -2x + 3.
Substitute that variable in the quadratic equation: – 2x + 3 = x2 – 6x + 3.
Rewrite the equation in standard form: x2 – 4x = 0.
Solve the quadratic equation: x(x – 4) = 0.
So, x = 0 or x = 4.
If x = 0, then y = 3. If x = 4, then y = – 5.
As the graph shows, the solutions are the two points (0, 3) and (4, – 5).
Question 6.
Find all solutions to the following system of equations.
– y2 + 6y + x – 9 = 0
6y = x + 27
Illustrate with a graph.
Solve the second equation for x: x = 6y – 27.
Substitute in the first equation: – y2 + 6y + 6y – 27 – 9 = 0.
Combine like terms: – y2 + 12y – 36 = 0.
Rewrite the equation in standard form and factor: – (y – 6)2 = 0.
Therefore, y = 6. Then x = 6y – 27, so x = 9.
There is only one solution (9, 6), and as the graph shows, the line is tangent to the parabola.
An alternative solution would be to solve the linear equation for y instead of x, getting the quadratic equation (x – 9)(x – 9) = 0, which gives the repeated root x = 9 and the same point of tangency (9, 6).
Another alternative solution would be to solve the quadratic equation for x, so that x = y2 – 6y + 9. Substituting in the linear equation would yield 6y = y2 – 6y + 9 + 27. Converting that to standard form would give y2 – 12y + 36 = 0, which gives the repeated root y = 6, as in the first solution. Note that in this case, unlike when the graph of the quadratic equation is a circle, the quadratic equation can be solved for x in terms of y without getting a radical expression.
Question 7.
Find all values of k so that the following system has two solutions.
x2 + y2 = 25
y = k
Illustrate with a graph.
The center of the circle is the origin, and the line is parallel to the x-axis. Therefore, as the graph shows, there are two solutions only when – 5 < k < 5.
Question 8.
Find all values of k so that the following system has exactly one solution.
y = 5 – (x – 3)2
y = k
Illustrate with a graph.
The parabola opens down, and its axis of symmetry is the vertical line x = 3. The line y = k is a horizontal line and will intersect the parabola in either two, one, or no points. It intersects the parabola in one point only if it passes through the vertex of the parabola, which is k = 5.
Question 9.
Find all values of k so that the following system has no solutions.
x2 + (y – k)2 = 36
y = 5x + k
Illustrate with a graph.
The circle has radius 6 and center (0, k). The line has slope 5 and crosses the y-axis at (0, k). Since for any value of k the line passes through the center of the circle, the line intersects the circle twice. (In the figure on the left below, k = 2, and in the one on the right below, k = – 3.) There is no value of k for which there is no solution.
Eureka Math Algebra 2 Module 1 Lesson 31 Exit Ticket Answer Key
Make and explain a prediction about the nature of the solution to the following system of equations, and then solve the system.
x2 + y2 = 25
4x + 3y = 0
Illustrate with a graph. Verify your solution, and compare it with your initial prediction.
Prediction: By inspecting the equations, students should conclude that the circle is centered at the origin, and that the line goes through the origin. So, the solution should consist of two points.
Solution: Solve the linear equation for one of the variables: y = – $$\frac{4 x}{3}$$
Substitute that variable in the quadratic equation: x2 + $$\left(-\frac{4 x}{3}\right)^{2}$$ = 25.
Remove parentheses and combine like terms: 25x2 – 25 ∙ 9 = 0,
so x2 – 9 = 0.
Solve the quadratic equation in x: (x + 3)(x – 3) = 0, which gives the roots 3 and – 3.
Substitute into the linear equation: If x = 3, then y = – 4; if x = – 3, then y = 4.
As the graph shows, the solution is the two points of intersection of the circle and the line: (3, – 4) and (- 3, 4).
An alternative solution would be to solve the linear equation for x instead of y, getting the quadratic equation (y + 4)(y – 4) = 0, which gives the roots 4 and – 4 and the same points of intersection.
Eureka Math Algebra 2 Module 1 Lesson 31 Exploratory Challenge Answer Key
Exploratory Challenge 1
a. Sketch the lines given by x + y = 6 and – 3x + y = 2 on the same set of axes to solve the system graphically. Then solve the system of equations algebraically to verify your graphical solution.
The point (1, 5) should be easily identifiable from the sketch. See the graph to the right.
Solving each equation for y gives the system
y = – x + 6
y = 3x + 2.
This leads to the single-variable equation
– x + 6 = 3x + 2
4x = 4
x = 1
y = – 1 + 6
y = 5.
Thus, the solution is the point (1, 5).
b. Suppose the second line is replaced by the line with equation x + y = 2. Plot the two lines on the same set of axes, and solve the pair of equations algebraically to verify your graphical solution.
The lines are parallel, and there is no point in common. See the graph to the right.
If we try to solve the system algebraically, we have
y = – x + 6
y = – x + 2,
which leads to the single-variable equation
– x + 6 = – x + 2
4 = 0.
Since 4 = 0 is not a true number sentence, the system has no solution.
c. Suppose the second line ¡s replaced by the line with equation 2x = 12 – 2y. Plot the lines on the same set of axes, and solve the pair of equations algebraically to verify your graphical solution.
The lines coincide, and they have all points in common. See the graph to the right.
Algebraically, we have the system
y = – x + 6
y = – x + 6,
– x + 6 = – x + 6
0 = 0.
Thus all points (x, – x + 6) are solutions to the system.
d. We have seen that a pair of lines can intersect in 1, 0, or an infinite number of points. Are there any other possibilities?
No. Students should convince themselves and each other that these three options exhaust the possibilities for the intersection of two lines.
Exploratory Challenge 2
a. Suppose that instead of equations for a pair of lines, you were given an equation for a circle and an equation for a line. What possibilities are there for the two figures to intersect? Sketch a graph for each possibility.
They can intersect in 0, 1, or 2 points as shown below.
b. Graph the parabola with equation y = x2. What possibilities are there for a line to intersect the parabola? Sketch each possibility.
The parabola and line can intersect in 0, 1, or 2 points as shown below. Note that, in contrast to the circle, where all the lines intersecting the circle in one point are tangent to it, lines intersecting the parabola in one point are either tangent to it or are parallel to the parabola’s axis of symmetry in this case, the y-axis.
c. Sketch the circle given by x2 + y2 = 1 and the line given by y = 2x + 2 on the same set of axes. One solution to the pair of equations is easily identifiable from the sketch. What is it?
Factoring or using the quadratic formula, students should find that the solutions to x2 + (2x + 2)2 = 1 are – 1 and –$$\frac{3}{5}$$.
There are two intersections of the line and the circle. When x = – 1, then y = 0, as the sketch shows, so (- 1, 0) is a solution. When x = – $$\frac{3}{5}$$, then y = 2 (-$$\frac{3}{5}$$) + 2 = $$\frac{4}{5}$$, so $$\left(-\frac{3}{5}, \frac{4}{5}\right)$$ is another solution.
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TOPICS
# Twin Peaks
For an integer , let denote the least prime factor of . A pair of integers is called a twin peak if
1. ,
2. ,
3. For all , implies .
A broken-line graph of the least prime factor function resembles a jagged terrain of mountains. In terms of this terrain, a twin peak consists of two mountains of equal height with no mountain of equal or greater height between them. Denote the height of twin peak by . By definition of the least prime factor function, must be prime.
Call the distance between two twin peaks
(1)
Then must be an even multiple of ; that is, where is even. A twin peak with is called a -twin peak. Thus we can speak of -twin peaks, -twin peaks, etc. A -twin peak is fully specified by , , and , from which we can easily compute .
The set of -twin peaks is periodic with period , where is the primorial of . That is, if is a -twin peak, then so is . A fundamental -twin peak is a twin peak having in the fundamental period . The set of fundamental -twin peaks is symmetric with respect to the fundamental period; that is, if is a twin peak on , then so is .
The question of the existence of twin peaks was first raised by David Wilson (pers. comm., Feb. 10, 1997). Wilson already had privately showed the existence of twin peaks of height to be unlikely, but was unable to rule them out altogether. Later that same day, John H. Conway, Johan de Jong, Derek Smith, and Manjul Bhargava collaborated to discover the first twin peak. Two hours at the blackboard revealed that admits the -twin peak
(2)
which settled the existence question. Immediately thereafter, Fred Helenius found the smaller -twin peak with and
(3)
The effort now shifted to finding the least prime admitting a -twin peak. On Feb. 12, 1997, Fred Helenius found , which admits 240 fundamental -twin peaks, the least being
(4)
Helenius's results were confirmed by Dan Hoey, who also computed the least -twin peak and number of fundamental -twin peaks for , 79, and 83. His results are summarized in the following table (OEIS A009190).
71 7310131732015251470110369 240 73 2061519317176132799110061 40296 79 3756800873017263196139951 164440 83 6316254452384500173544921 6625240
The -twin peak of height is the smallest known twin peak. Wilson found the smallest known -twin peak with , as well as another very large -twin peak with . Richard Schroeppel noted that the latter twin peak is at the high end of its fundamental period and that its reflection within the fundamental period is smaller.
Many open questions remain concerning twin peaks, e.g.,
1. What is the smallest twin peak (smallest )?
2. What is the least prime admitting a -twin peak?
3. Do -twin peaks exist?
4. Is there, as Conway has argued, an upper bound on the span of twin peaks?
5. Let be prime. If and each admit -twin peaks, does then necessarily admit a -twin peak?
Andrica's Conjecture, Divisor Function, Least Common Multiple, Least Prime Factor
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## References
Sloane, N. J. A. Sequence A009190 in "The On-Line Encyclopedia of Integer Sequences."
Twin Peaks
## Cite this as:
Weisstein, Eric W. "Twin Peaks." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/TwinPeaks.html
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# Powers and roots 5.1. Previous learning. Objectives based on NC levels and (mainly level ) Lessons 1 Working with integer powers of numbers
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1 N 5.1 Powers and roots Previous learning Before they start, pupils should be able to: recognise and use multiples, factors (divisors), common factor, highest common factor, lowest common multiple and primes use squares, positive and negative square roots, cubes and cube roots, and index notation for small positive integer powers. Objectives based on NC levels and (mainly level ) In this unit, pupils learn to: represent problems and synthesise information in different forms use accurate notation calculate accurately, selecting mental methods or a calculator as appropriate estimate, approximate and check working record methods, solutions and conclusions make convincing arguments to justify generalisations or solutions review and refine own findings and approaches on the basis of discussions with others and to: use index notation for integer powers know and use the index laws for multiplication and division of positive integer powers extend mental methods of calculation with factors, powers and roots use the power and root keys of a calculator use ICT to estimate square roots and cube roots use the prime factor decomposition of a number. Lessons 1 Working with integer powers of numbers About this unit Assessment Estimating square roots 3 Prime factor decomposition Sound understanding of powers and roots of numbers helps pupils to generalise the principles in their work in algebra. It also helps pupils to be aware of the relationships between numbers and to know at a glance which properties they possess and which they do not. This unit includes: an optional mental test which could replace part of a lesson (p. 00); a self-assessment section (N5.1 How well are you doing? class book p. 00); a set of questions to replace or supplement questions in the exercises or homework tasks, or to use as an informal test (N5.1 Check up, CD-ROM). Common errors and misconception N5.1 Powers and roots Look out for pupils who: think that n means n, or that n means n ; wrongly apply the index laws, e.g , or ; think that 1 is a prime number; include 1 in the prime factor decomposition of a number; confuse the highest common factor (HCF) and lowest common multiple (LCM); assume that the lowest common multiple of a and b is always a b.
2 Key terms and notation Practical resources Exploring maths Useful websites problem, solution, method, pattern, relationship, expression, solve, explain, systematic, trial and improvement calculate, calculation, calculator, operation, multiply, divide, divisible, product, quotient positive, negative, integer factor, factor pair, prime, prime factor decomposition, power, root, square, cube, square root, cube root, notation n and n, n 3 and 3 n scientific calculators for pupils individual whiteboards Tier 5 teacher s book N5.1 Mental test, p. 00 Answers for Unit N5.1, pp Tier 5 CD-ROM PowerPoint files N5.1 Slides for lessons 1 to 3 Excel file N5.1 SquareRoot Tools and prepared toolsheets Calculator tool Tier 5 programs Multiples and factors quiz HCF and LCM Ladder method computers with spreadsheet software, e.g. Microsoft Excel, or graphics calculators Tier 5 class book N5.1, pp Tier 5 home book N5.1, pp Tier 5 CD-ROM N5.1 Check up Topic B: Indices: simplifying Factor tree nlvm.usu.edu/en/nav/category_g_3_t_1.html Grid game N5.1 Powers and roots
3 1 Working with integer powers of numbers Learning points A number a raised to the power 4 is a 4 or a a a a. The number that expresses the power is its index, so, 5 and 7 are the indices of a, a 5 and a 7. To multiply two numbers in index form, add the indices, so a m a n a m1n. To divide two numbers in index form, subtract the indices, so a m a n a m n. Starter Use slide 1.1 to discuss the objectives for the unit. This lesson is about finding positive and negative integer powers of numbers. Remind pupils that when a number is multiplied by itself the product is called a power of that number. So a a, or a squared, is the second power of a, and is written as a, a a a, or a cubed, is the third power of a and is written as a 3, and so on. For a 4 we say a to the power of 4, and similarly with higher powers. The number that expresses the power is its index. So 5 and 7 are the indices of a 5 and a 7. When the index is 1, it is usually omitted: we write a, rather than a 1. Ask pupils to calculate mentally powers of positive and negative integers, e.g. ( 8), 8, ( ) 5, 5, ( 3) 4, 3 4, ( 5) 3, 5 3 Record answers on the board, and ask pupils what they notice. Draw out that for negative numbers even powers are positive, and odd powers are negative. What is ( 1) 13? What is ( 1) 14? Extend to decimals, e.g. ask pupils to calculate mentally: (0.1) 3, (0.7), ( 0.) 4 TO Main activity Use the Calculator tool to show pupils how to use the x y keys of their calculators. Use the key to explore raising a number to the power 0. Explain that this always has the answer 1. Discuss negative indices. Ask pupils to consider this pattern. How is each number found from the one above it? What is the pattern of the indices? What are the next few lines of the pattern? Establish that: Similarly 1 1, 1 4 and Show the table of powers of on slide 1.. Ask pupils in pairs to make up and record some multiplications using the table. Ask questions to help pupils to discover for themselves the rules for calculations with indices. N5.1 Powers and roots
4 What do you notice about the indices in these calculations? What is a quick way of multiplying powers of? Why does it work? What is this calculation in index form? [ ] [ ] [ ] What do you notice about the indices in these calculations? What is a quick way of dividing one power of by another? Why does it work? Repeat with the powers of 4 on slide 1.3. Now generalise. Write on the board m m 3. What will this simplify to? Explain why. [m m 3 (m m) (m m m) m m m m m m 5 ] Stress that the indices have been added, so that: m m 3 m 1 3 m 5 Repeat with m 5 m. Stress that for division the indices are subtracted, so that: m 5 m m 5 m 3 Discuss negative indices, e.g. m 3 m 7 m 3 7 m 4 = 1 m 4 m 5 m 3 m 5 3 m. Select individual work from N5.1 Exercise 1 in the class book (p. 00). Review Homework Show slide 1.4. Point to two different powers of 10. Ask pupils to multiply or divide them and to write the answer on their whiteboards. Stress that the rules for multiplying and dividing numbers in index form apply to both positive and negative indices. Ask pupils to remember the points on slide 1.5. Homework Ask pupils to do N5.1 Task 1 in the home book (p. 00). N5.1 Powers and roots
5 Estimating square roots Learning points n is the square root of n, e.g n is the cube root of n, e.g , Trial and improvement can be used to estimate square roots when a calculator is not available. Starter Tell pupils that in this lesson they will be estimating the value of square roots. Remind them that the square root of a is denoted by a, or more simply as a, and that a square root of a positive number can be positive or negative, e.g. if a 5 9, a Show the grid on slide.1. Write on the board: x 51, z 5 4. Point to an expression on the grid. Ask pupils to work out its value mentally and to write the answer on their whiteboards. Ask someone to explain how they calculated it. After a while, change the values for x and z to: x 5 9 and z 5 5. Main activity Discuss how to estimate the positive square root of a number that is not a perfect square, e.g. 70 must lie between 64 and 81, so Since 70 is closer to 64 than to 81, we expect 70 to be closer to 8 than to 9, perhaps about Show the class how they could find 7 if they had only a basic calculator with no square-root key. Tell them that the process is called trial and improvement. Explain that 7 must lie between and 3, because 7 lies between and 3. Try Try.6 = Try.7 = 7.9. Try.65 = Try.64 = Try.645 = Too low Too low Too high. Very close but a little bit too high Very close but too low Still too low The answer lies between.645 and.65. All numbers between.645 and.65 round up to.65. So correct to two decimal places. Ask pupils to work in pairs and, using only the key on their calculator, to find 1 to two decimal places [answer: 3.46]. Establish first that it must lie between 3 and 4. N5.1 Powers and roots
6 Show the class how they could use a spreadsheet for this activity, without using the square-root function, e.g. use the Excel file N5.1 SquareRoot. Point out that the strategy here is different. We work systematically in tenths from 3 to 4, then in hundredths from 3.4 to 3.5, then in thousandths from 3.46 to You can use this file to estimate other square roots by overtyping 3 and 3.4. If possible, pupils should develop similar spreadsheets, using either a computer or a graphics calculator. XL Select individual work from N5.1 Exercise in the class book (p. 00). Review Introduce root notation. Explain that if 79 is the cube of 9, then 9 is the cube root of 79, which is written as 3 _ The cube root, fourth root, fifth root, of a are denoted by 3 a, 4 a, 5 a, Use the Calculator tool to demonstrate how to find a cube root. You may need to explain that some calculators have a cube root key 3. Others have a key like x, or other variations. For example, to find the value of 3 _ 16, key in 1 6 x 3. [Answer: 6] Ask pupils to work out 3 64 and Explain that the cube root of a positive number is positive, and the cube root of a negative number is negative. TO Sum up the lesson by reminding pupils of the learning points. Homework Ask pupils to do N5.1 Task in the home book (p. 00). N5.1 Powers and roots
7 3 Prime factor decomposition Learning points Writing a number as the product of its prime factors is called the prime factor decomposition of the number. You can use a tree method or a ladder method to find a number s prime factors. To find the highest common factor (HCF) of a pair of numbers, find the product of all the prime factors common to both numbers. To find the lowest common multiple (LCM) of a pair of numbers, find the smallest number that is a multiple of each of the numbers. QZ Starter Tell pupils that in this lesson they will be finding the prime factors of numbers and using them to find common factors and multiples of a pair of numbers. Remind them of the definitions of multiple, factor, factor pair and prime number. Launch Multiples and factors quiz. Ask pupils to answer on their whiteboards. Use Next and Back to move through the questions at a suitable pace. Main activity Write on the board three products such as: What do you notice about the numbers in these products? Establish that they are all prime numbers. Explain that when a number is expressed as the product of its prime factors it is called the prime factor decomposition of a number. Stress that because 1 is not a prime number it is not included in the decomposition. How can we find the prime factor decomposition of 80? First explain the tree method, i.e. split 80 into a product such as 0 3 4, then continue factorising any number in the product that is not a prime. Repeat with SIM Launch Ladder method. Use it to show the alternative method, where the number is repeatedly divided by any prime that will divide into it exactly. Demonstrate with 63, dragging numbers from the grid to the relevant positions. Continue to divide by prime numbers until the answer is 1. Express the answer as Repeat with SIM Show how to find the highest common factor (HCF) and lowest common multiple (LCM) of a pair of numbers. Launch HCF and LCM. N5.1 Powers and roots
8 Choose lowest common multiple. Select 8 and 6 using the arrows by the numbers. Drag multiples of 8 and multiples of 6 from the 100-square to the answer boxes. (Numbers snap back to the 100-square if dragged from answer box.) Numbers common to both boxes change colour to blue. Which numbers are both multiples of 8 and multiples of 6? Which is the lowest number that is both a multiple of 8 and 6? Drag the HCF into the box below the 100-square. Repeat several times with different numbers, then change to highest common factor, which works similarly. Select 4 and 18, then drag factors to the answer boxes. Which numbers are both factors of 4 and factors of 18? Which is the highest number that is both a factor of 4 and 18? Repeat with different numbers. Show how to use a Venn diagram to find the HCF and LCM of a pair of numbers such as 36 and 30. Explain that: the overlapping prime factors give the HCF ( ); all the prime factors give the LCM ( ). Repeat with 18 and Select individual work from N5.1 Exercise 3 in the class book (p. 00). Review Sum up the lesson by stressing the points on slide 3.1. Round off the unit by referring again to the objectives. Suggest that pupils find time to try the self-assessment problems in N5.1 How well are you doing? in the class book (p. 00). Homework Ask pupils to do N5.1 Task 3 in the home book (p. 00). N5.1 Powers and roots
9 N5.1 Mental test Read each question aloud twice. Allow a suitable pause for pupils to write answers. 1 What number is five to the power three? Write all the prime factors of forty-two. 3 Write down a factor of thirty-six that is greater than ten and less than twenty. 005 KS3 4 What is the next number in the sequence of square numbers? 004 KS3 One, four, nine, sixteen,... 5 Look at the numbers. Write down each number that is a factor of one hundred Y7 [Write on board ] 6 Write two factors of twenty-four which add to make eleven. 005 KS 7 What is the square root of eighty-one? 001 KS3 8 What number is five cubed? 003 KS3 9 The volume of a cube is sixty-four centimetres cubed. 00 KS3 What is the length of an edge of the cube? 10 What is the square of three thousand? 001 KS3 11 To the nearest whole number, what is the square root of 004 KS3 eighty-three point nine? 1 I think of a number. I square my number and get the answer 007 KS3 one thousand six hundred. What could my number be? Key: KS3 Key Stage 3 test KS Key Stage test Y7 Year 7 optional test (1999) Questions 3 to 7 are at level 5; 8 to 11 are at level 6; 1 is at level 7 Answers 1 15, 3, or , 0, 5, and cm N5.1 Powers and roots
10 N5.1 Check up and resource sheets Check up Answer these questions by writing in your book. Powers and roots (no calculator) N level 6 Which two of the numbers below are not square numbers? David says that What is 10? 3 To the nearest whole number, what is the square root of 93.7? _ 4 If 81 n 144, then n could be which of the following numbers? 5 Year 8 optional test level 6 Terry has 4 centimetre cubes. He uses them to make a cuboid that is one cube high. Tina has 4 centimetre cubes. She uses them to make a solid cuboid that is two cubes high What could the dimensions of her cuboid be? 1 cm high 4 cm wide 6 cm long 6 What is the biggest number that is a factor of both 105 and 135? 7 What is the smallest number that is a multiple of both 1 and 7? Powers and roots (calculator allowed) level 6 Mary thinks of a number. First I subtract 3.76 Then I find the square root of what I get My answer is 6.80 Which number did Mary think of? Pearson Education 008 Tier 5 resource sheets N5.1 Powers and roots N5.1 N5.1 Powers and roots 11
11 N5.1 Answers Class book Exercise 1 1 a 3 b 4 5 c 3 8 d ( 1) 4 e 5 f 6 1 a 64 b 43 c 56 d 18 e 1 f 1 g 1 h a 401 b c 1331 d e 104 f g h a 8 b 3 5 c d a 8 e 5 4 f 1 4 g 8 0 h b a b c d e f Rachel and Hannah are 14 and 11 years old. Extension problem 8 The smallest whole numbers are 6 and 10: Exercise 1 a x 3 b x 7 c x 1 d x 1 a 1.41 to d.p. b.15 to d.p. c 4 d 0. e 5 f 1. to d.p. g 1.73 to d.p. h 1 3 a b 7 c 11 d 8 4 Each answer is correct to 1 d.p. a.4 b 6.7 c 10.7 d Between 700 and 750 slabs will be used. 6 6 is 678, which is too few, and 8 8 is 784, which is too many. So the exact number of slabs is a a 9.7 to 1 d.p. b a 1.3 to 1 d.p. c a 0.4 to 1 d.p metres to d.p. Extension problem 8 98 = is not a perfect square. There is no whole number between the square root of 8880 and the square root of 8889 but 98 lies between the and Exercise 3 1 a 3 b 3 5 c 3 7 d 3 3 e 3 3 f 3 3 a 1,, 5, 10, 5, 50 b 5 3 a 1, 3, 5, 9, 15, 45 b e.g. 63 (with factors 1, 3, 7, 9, 1, 63) 5 a 7 and 30: HCF 6, LCM 360 b 50 and 80: HCF 10, LCM 400 c 48 and 84: HCF 1, LCM HCF 15, LCM HCF 10, LCM N5.1 Powers and roots
12 8 a and 3 b 6 c a 8 and 40: HCF 4, LCM 80 b 00 and 175: HCF 5, LCM 1400 c 36 and 64: HCF 4, LCM Extension problem 11 a 4 (with factors 1, and 4) b 4 16 (with factors 1,, 4, 8, 16) c 7 factors: factors: factors: factors: days from now, since 40 is the lowest common multiple of 1,, 3, 4, 5, 6 and 7. N5.1 How well are you doing? 1 a 3, 4, 5, 3 3 or 9, 16, 5, 7 b a is the largest b 5 and 7 are not square numbers. 3 a 3 9 b 7 18 c 3 = 18 4 a a 3 b b 5 a HCF is 1 b LCM is Suzy s number is Home book TASK 1 1 a 3 11 b 6 c 11 1 d x 6 e 4 3 f 10 4 g 8 10 h z a b c a (15) 5 (5) 65 b (11) 11 (19) 361 (1) 441 (9) 841 (31) 961 c (6) 3 16 d (13) TASK 1 a 19 b 9 c 9 d 5 e 4 f g 8.67 to d.p. h 4.4 to d.p cm to 1 d.p. 3 a 3 b 4 c 6 d 10 TASK 3 1 a b 3 5 a 3 5 b , 13 and 17 CD-ROM CHECK UP 1 5 and 7 are not square numbers B: 11 5 cm high by 1 cm wide by 1 cm long cm high by cm wide by 6 cm long cm high by 3 cm wide by 4 cm long N5.1 Powers and roots 13
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# Why do we usually not want the eigenvalues of non-symmetric matrices?
I came across this line in a class note I am reading where it discusses finding eigenvalues of matrices.
In reality we don't go all the way with Arnoldi. We stop at a decent value of π. Then the π eigenvalues of π» are (usually) good approximations to π extreme eigenvalues of π΄. Trefethen and Bau emphasize for non-symmetric π΄ that we may not want eigenvalues of π΄ in the first place! When they are badly conditioned, this led Trefethen and Embree to the theory of pseudospectra.
Why is this the case? I understand that for symmetric matrices, there are many nice properties of eigenvalues. For example the eigenvalues of a real symmetric matrix are real. SVD comes from the eigenvalues of $$A^TA$$ which is symmetric, etc.
But why are we so confident that we usually don't need to find the eigenvalues of non-symmetric matrix? Is it purely because of the nice properties of symmetric matrix that make us tend to formulate our problems that way? If someone can explain or point to where Trefethen and Bau explains it, that would be great. I have that book, but I can't find the explanation based in the relevant chapters I went through.
# Stability under perturbations
Let $$E$$ be a perturbation such that $$\|E\| \leq \varepsilon$$.
If $$A$$ is symmetric, then the eigenvalues of $$A+E$$ are at a distance $$\varepsilon$$ from those of $$A$$. (Bauer-Fike Theorem.)
If $$A$$ is non-symmetric, then the eigenvalues of $$A+E$$ could be just anywhere in the complex plane. No bound can be formulated a priori. (Counterexample to show that no bound exist: start with a Jordan block of size $$n$$, and perturb the $$(1,n)$$ entry to $$\varepsilon$$; then the eigenvalues are the $$k$$th complex roots of $$\varepsilon$$, which have magnitude $$\varepsilon^{1/n}$$).
If $$A$$ is non-symmetric, then the $$\delta$$-pseudospectrum of $$A+E$$ is at a distance $$\varepsilon$$ from that of $$A$$ (follows from the definition).
So, the point is, how sure are you that the numbers in your matrix $$A$$ are correct? What if that measurement is only accurate to the third decimal digit? What about that tiny $$2^{-52}$$ error that you make when you truncate the coefficient 2/3 to a double? Those small perturbations affect your computed eigenvalues. There is little point in computing a number if you don't know how accurate it is in the first place. Or, worse, if you know from the start that it is inaccurate.
Pseudospectra capture nicely this concept that the true location of the eigenvalues is uncertain, for non-symmetric matrices, and provide a concept analogous to eigenvalues that is stable under perturbations.
• This is a good answer. But the asker failed to give context, so it may not be immediately relevant. The excerpt is from Gilbert Strang's Learning from Data book, particularly in the Numerical Linear Algebra section between "Krylov Subspaces and Arnoldi iteration" and "Linear systems by Arnoldi and GMRES" Aug 8, 2021 at 7:53
• Interesting, can you explain a bit on why symmetric matrix is stable under perturbation while the non-symmetric ones can have eigenvalues all over the place? Aug 8, 2021 at 14:46
• @CuriousMind I have added some more detail. Aug 8, 2021 at 14:58
• @FedericoPoloni thank you let me study a bit more Aug 8, 2021 at 15:05
• "If $A$ is non-symmetric, then the eigenvalues of $A+E$ can be just anywhere on the complex plane. No bound can be formulated a priori." This is wrong. Bhatia's Matrix Analysis Theorem VIII.1.5 contains the bound $4(\|A\|+\epsilon)^{1-1/n}\epsilon^{1/n}$ in the optimal matching distance between the eigenvalues of $A$ and $A+E$. No doubt, $\epsilon^{1/n}$ is an unfathomably bad scaling, but the claim that the eigenvalues of $A+E$ can be anywhere in the complex plane and that no a priori bound is possible is overstated. Aug 10, 2021 at 22:01
Amazing question which has a long answer, but I will try to be concise. In the context of Krylov subspace methods for general matrices, the eigenvalues of a non-symmetric matrix mean very little. In βAny nonincreasing convergence curve is possible for GMRESβ, Greenbaum et al. show that any nonincreasing convergence curve is possible for GMRES independent of the eigenvalue distribution of the matrix. To remedy this problem, Trefethen suggests pseudospectra of a matrix and pseudospectra-convergence analysis for various Krylov subspace methods (βGeneralizing eigenvalue theorems to pseudospectra theoremsβ and βPseudospectra and Spectraβ, but really checkout anything Trefethen wrote on Krylov methods and pseudospectra). However, in some cases, determining pseudospectra of a matrix is difficult (without having the matrix). For example, if you are investigating the matrix representation of a linear operator (say, discretization of a PDE), you have a lot of functional analysis tools which don't help with the pseudospectra analysis at all.
Hence, other people are pushing for another analysis technique called numerical range (or, field of values). This technique is amenable to functional analysis techniques and there has been some exciting developments recently. (See βThe Field of Values Bounds on Ideal GMRESβ for example, but the last 10 or so years were amazing in terms of new theoretical results and their applications).
However, all of these methods provide an asymptotical limit, not a practical one. For example, you regularly see optimality (in the context of preconditioners) proofs, but to observe optimality in practice one has to be solving very large problems. Mark Embree (who worked with Trefethen on pseudospectra) has the paper βHow descriptive are GMRES convergence bounds?β, where he demonstrates that all three types of bounds fail to provide informative and meaningful predictions regarding the convergence. They are usually incredibly pessimistic, and the convergence is achieved much earlier than predicted in practice.
I tend to use field of values in general, even for symmetric matrices -though you have to be careful with indefinite matrices, ratio field of values or other variants are better to use in that case, or use eigenvalues-. Most of the time, I interpret the result I obtained qualitatively rather than quantitatively. Which means that if I can prove that the residual reduces by 10% every iteration (miserably slow) independent of a parameter, I don't conclude that the convergence (tol < 1e-8) will be achieved in 175 iterations. I conclude that independent of the parameter of interest, the problem is solvable at most some number $$k$$ iterations which will be much smaller than the matrix size $$n$$.
There will be a lot of typos/grammatical mistakes in this answer, but it is late here. My apologies.
Comment on: "But why are we so confident that we usually don't need to find the eigenvalues of non-symmetric matrix?"
Eigenvalues and eigenspaces are relevant when you have an endomorphism, the same vector space as domain and range, like for instance in the linearization of an ODE system around some (equilibrium or periodic) solution. Then you can ask to decompose the corresponding system into smaller, decoupled systems, which the transformation to an eigen-basis would deliver. Many such tasks are derived from a variational formulation (finite elements, Galerkin) leading to eigenvalue or generalized eigenvalue problems with symmetric matrices.
In a more general situation, domain and range will be different, the matrix components will not encode easily visible structural information. Metric properties of the matrix or effective rank computations then lead to the spectral radius and singular spectrum. Restoring the endomorphism property for instance via an approximative inverse matrix and then considering the defect will also be better controlled via the singular spectrum.
As observed, structurally, the singular spectrum is related to the spectrum of the symmetric matrix $$A^TA$$ or more directly of the block matrix $$\pmatrix {0&A\\A^T&0}$$
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# Find the term n=12, given the arithmetic sequence a_1=5 and d=1/3
Find the term n=12, given the arithmetic sequence ${a}_{1}=5$ and d=1/3
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The ${n}^{\text{th}}$ term, ${a}_{n}$ of an arithmetic sequence, with initial term, ${a}_{1}$ and incremental difference d is given by the formula:
$\text{XXX}{a}_{n}={a}_{1}+\left(n-1\right)\cdot d$
In this case with $n=12,{a}_{1}=5,\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}d=\frac{1}{3}$
$\text{XXX}{a}_{12}=5+\left(12-1\right)\cdot \frac{1}{3}=5+\frac{11}{3}=8\frac{2}{3}$
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# Helix and radius of curvature
#### Jonny_trigonometry
I was wondering how to find the radius of curvature of a helix. If it's circling around the z axis, the radius of it's projection onto the xy axis is a circle of radius r. Let one full cycle of the helix around the z-axis cover a distance d along the z-axis, then what is R, the radius of curvature of the helix in terms of d and r? I know it must be larger than d + r... Is there a handy formula for this?
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#### amcavoy
Jonny_trigonometry said:
I was wondering how to find the radius of curvature of a helix. If it's circling around the z axis, the radius of it's projection onto the xy axis is a circle of radius r. Let one full cycle of the helix around the z-axis cover a distance d along the z-axis, then what is R, the radius of curvature of the helix in terms of d and r? I know it must be larger than d + r... Is there a handy formula for this?
Hmm. From what I know about these, the equations are in the form of:
$$\vec{r}=\left<r\cos{t},r\sin{t},\alpha t\right>$$
You know the radius projected onto the x-y plane, and also that d is proportional to the period. Assuming you know the formula for the radius of curvature:
$$R=\frac{1}{\left|\kappa\right|}$$
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#### Jonny_trigonometry
hmm, ya. The parametric curve looks good, but what is kappa?
forget the "radius of curvature", what I mean is radius...
I guess what I really want to know is what is the radius R of the circle that is made from the length of a string that is wound around a cyninder with radius r as it spans a distance d (along the longitudinal axis of the cylinder) to make one cycle around the cylinder.
If i have to integrate the parametric curve to find the length, then I guess thats what I have to do... I just don't like the complexity involved in doing so, and I figured someone has already done that and found a relationship between the variables R, d and r.
#### amcavoy
Jonny_trigonometry said:
hmm, ya. The parametric curve looks good, but what is kappa?
forget the "radius of curvature", what I mean is radius...
I guess what I really want to know is what is the radius R of the circle that is made from the length of a string that is wound around a cyninder with radius r as it spans a distance d (along the longitudinal axis of the cylinder) to make one cycle around the cylinder.
If i have to integrate the parametric curve to find the length, then I guess thats what I have to do... I just don't like the complexity involved in doing so, and I figured someone has already done that and found a relationship between the variables R, d and r.
I might be doing this wrong, but this is what it looks like:
$$2\pi R=\int_{0}^{2\pi}\sqrt{r^{2}+\alpha^{2}}\,dt=2\pi\sqrt{r^{2}+\alpha^{2}}=2\pi\sqrt{r^{2}+\frac{d^{2}}{4\pi^{2}}}$$
Which would represent the length of the helix (I calculated that by the definition of arc length). Now you know that the length above (circumference) is really 2piR where R is the radius of the circle you want. Is this what you were getting at or did I misinterpret your question?
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#### Jonny_trigonometry
thanks! this is exactly what i was looking for. I reviewed arc length in 3d and checked your solution. It must be correct. I didn't think it would be that easy, I thought there would be a triple integral for some reason. Eh, I got a c in calc 3, so I'm not proficient enough in doing problems like this. Now that I think of it, triple integrals really don't show up unless you're calculating volume, and doubles are usually for area, or to simplify a more difficult single integral... thanks a lot
#### bobb513
Helix Radius
I've seen vaiants of formulas such as amcavoy suggests in his second post. They do the job, but it bothered me that a Pathagorean approach was used when trig should offer a streamlined version. This is what I formulated:
R = r(cos)^2
where the cos is derived from the slope of the curve around the cylinder.
I recognize that amcavoy did in fact introduce trig into his forms, suggested in his first post, but without squaring the cos, the value for t is unattainable.
Regards, Bob
#### DeltaT
Re: Helix Radius
I've seen vaiants of formulas such as amcavoy suggests in his second post. They do the job, but it bothered me that a Pathagorean approach was used when trig should offer a streamlined version. This is what I formulated:
R = r(cos)^2
where the cos is derived from the slope of the curve around the cylinder.
I recognize that amcavoy did in fact introduce trig into his forms, suggested in his first post, but without squaring the cos, the value for t is unattainable.
Regards, Bob
I've seen that result quoted before in a text book, unfortunately the derivation wasn't given, and so far it eludes me. Any chance you could provide a step by step explanation of how the R = r(cos)^2 result was obtained?
DeltaT
#### adriank
Well, the curvature of a curve in R3 is $$\kappa = \frac{\lvert \vec r' \times \vec r'' \rvert}{\lvert \vec r' \rvert^3}$$, and using $$R = \frac{1}{\lvert \kappa \rvert}$$ should give you the radius of curvature.
#### DeltaT
Well, the curvature of a curve in R3 is $$\kappa = \frac{\lvert \vec r' \times \vec r'' \rvert}{\lvert \vec r' \rvert^3}$$, and using $$R = \frac{1}{\lvert \kappa \rvert}$$ should give you the radius of curvature.
Thanks, I don't mean to sound ungrateful, but I was particularly hoping to avoid using vectors, and was hoping for a solution using ordinary algebra and trigonometry. A previous poster, bobb513 appears to be saying he reached his result that way, where the angle involved is the slope of the curve around the cylinder.
I would appreciate any help in reaching the R = r(cos)^2 result just using the trig functions and simple algebra if possible.
I would just add, I don't need this for any specific purpose, other than personal curiosity. It is a result I've seen stated several times, but so far I have never seen it derived in a way I could follow.
Regards
DeltaT
#### DeltaT
Hi
Ok, I've found a website that has allowed me to find the solution I wanted.
http://ca.geocities.com/web_sketches/calculators/baluster_radius/baluster_radius.html [Broken]
From the result given on that site for R, and using the fact that cos(pitch) can be found from the geometry given, it is easy to show that:
r = R cos^2 (pitch)
which was the result I wanted to be able to find.
However, there is still a slight catch. I can follow the math on that page, and I was even able to extend it to reach the trigonometric result. However, I can't see why the opening statement is true:
Helix_Length = C * c/Helix_Length
I can't think of a justification for that statement, can anyone here see what I'm missing?
DeltaT
Last edited by a moderator:
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Question
# A solution contains 0.010 M Ba2+ and 0.010 M Ag+. If we add chromate (CrO42-) to...
A solution contains 0.010 M Ba2+ and 0.010 M Ag+. If we add chromate (CrO42-) to the solution:
Which ion starts to precipitate first?
Can 99.90% of either ion be precipitated by chromate without precipitating the other metal ion?
pKsp BaCrO4 = 9.67, pKsp Ag2CrO4=11.92
Ksp of BaCrO4 is 2.14 x 10^-10
Ksp of Ag2CrO4 is 1.20 x 10^-12
As the Ksp of Ag2CrO4 is lower than the Ksp of BaCrO4, Ag+ would precipitate first from the solution
[CrO4^2-] needed for Ba2+ precipitation = 2.14 x 10^-10/0.01 = 2.14 x 10^-8 M
at this chromate concentration, [Ag+] = sq.rt.(1.20 x 10^-12/2.14 x 10^-8) = 7.5 x 10^-3 M
percentage of Ag+ precipitated = 7.5 x 10^-3 x 100/0.01 = 75%
So, precipitation of 99.9% Ag+ is not possible without precipitating the other metal ion
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Question
# I keep getting the z score wrong on this question? Please advise According to a recent...
I keep getting the z score wrong on this question? Please advise
According to a recent census,17% of the people in the United States are of Hispanic origin. One county supervisor believes her county has a different proportion of Hispanic people than the nation as a whole. She looks at their most recent survey data, which was a random sample of 493 county residents, and found that 50 of those surveyed are of Hispanic origin.
Solution :
This is the two tailed test .
The null and alternative hypothesis is
H0 : p = 0.17
Ha : p 0.17
n = 493
x = 50
= x / n = 50 / 493 = 0.1014
P0 = 0.17
1 - P0 = 1 - 0.17 = 0.83
z = - P0 / [P0 * (1 - P0 ) / n]
= 0.1014 - 0.17 / [(0.17 * 0.83) / 493]
= -4.05
Test statistic =
P(z < -4.05) = 0
P-value = 2 * 0 = 0
= 0.05
P-value <
Reject the null hypothesis .
There is sufficient evidence to that the county supervisor believes her county has a different proportion of
Hispanic people than the nation as a whole .
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# Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 8
• 00000 to 99999 and 100000 is what is required. As sum of digits has been asked, we do not need place values. So, 100000/10 will be the occurrence of each digit at each place. So, 45 * 10000 * 5 = 2250000. Also, we have to add 1 that comes because of 100000. So, total of 2250001.
• Q26) P is a prime number greater than 30. When P is divided by 30, the remainder is x. How many different values of x are possible?
a. 8
b. 9
c. 10
d. 11
• P = 30k + x
x cannot be a multiple of 2, 3 or 5. Also, x has to be less than 30. So, the possible remainders are 1, 7, 11, 13, 17, 19, 23, and 29. Total of 8 values.
• Q27) Raman took classes for 20 consecutive days. On nth day, there were (n + 1) students present in his class, where n is a natural number. Each day Raman distributed Rs.1100 equally among the students present in the class. Student ‘x’, who attended all the 20 classes of Raman, donated 1/(m + 1) th part of the amount received by him (from Raman on that particular day) to a charity at the end of each day, where ‘m’ is the number of students attending the class on that day. What is the total amount donated by student ‘x’ to the charity?
a. Rs. 300
b. Rs. 1000
c. Rs. 666
d. Rs. 500
• On the first day, the student got Rs. 1100/2 and gave away 1/3 of what he got. So, amount donated = 1100/6
On the second day, the student got Rs. 1100/3 and gave away ¼ of what he got. So, amount donated = 1100/12
On the third day, he would have donated 1100/20, on the fourth day, 1100/30 and so on.
So, total amount donated will be
1100 (1/6 + 1/12 + 1/20 + … + 1/462)
1100 (1/2 – 1/3 + 1/3 – ¼ + ¼ - 1/5 + … + 1/21 – 1/22)
1100 * 10/22
Rs. 500.
• Q28) Let M be a three-digit number denoted by ‘ABC’ where A, B and C are numerals from 0 to 9. Let N be a number formed by reversing the digits of M. It is known that M – N + (396 × C) is equal to 990. How many possible values of M are there which are greater than 300?
a. 10
b. 18
c. 30
d. 20
• 99A – 99C + 396C = 990
99A + 297C = 990
A + 3C = 10
C = 1, 2, 3 and A = 7, 4, 1. For each value of A and C, B can take 10 values. Also, A has to be greater than or equal to 3. So, 4B2 and 7B1 will be the numbers. B can take any value from 0 to 9. So, 20 values in total.
• Q29) The product of three positive integers is 6 times their sum. One of these integers is the sum of the other two integers. If the product of these three numbers is denoted by P, then find the sum of all distinct possible values of P.
a. 432
b. 252
c. 144
d. 336
• Let a = b + c
(b + c)bc = 6(2b + 2c)
bc = 12.
The values of a, b and c could be 13, 12, 1 or 8, 6, 2 or 7, 4, 3. So, the distinct products will be 156 + 96 + 84 = 336.
• Q30) P1, P2 and P3 are three consecutive prime numbers and P1 × P2 × P3 = 190747. What is the value of P1 + P2 + P3?
• Can start by understanding that the numbers will be close to the cube root of 190740 which should lie between 50 and 60 (between 125000 and 216000). Can start with the smallest prime numbers and check. It will be satisfied for 53 * 59 * 61. Total of 53 + 59 + 61 = 173.
63
61
71
61
61
61
51
63
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← 18 19 20 →
Cardinalnineteen
Ordinal19th
(nineteenth)
Factorizationprime
Prime8th
Divisors1, 19
Greek numeralΙΘ´
Roman numeralXIX
Binary100112
Ternary2013
Senary316
Octal238
Duodecimal1712
Hebrew numeralי"ט
Babylonian numeral𒌋𒐝
19 (nineteen) is the natural number following 18 and preceding 20. It is a prime number.
## Mathematics
${\displaystyle 19}$ is the eighth prime number, and forms a sexy prime with 13,[1] a twin prime with 17,[2] and a cousin prime with 23.[3] It is the third full reptend prime in decimal,[4] the fifth central trinomial coefficient,[5] and the seventh Mersenne prime exponent.[6] 19 is the second Keith number, and more specifically the first Keith prime.[7] It is also the second octahedral number, after 6.[8]
### Number theory
19 is the maximum number of fourth powers needed to sum up to any natural number, and in the context of Waring's problem, 19 is the fourth value of g(k).[9]
The Collatz sequence for nine requires nineteen steps to return back to one, more than any other number below it.[10] On the other hand, nineteen requires twenty steps, like eighteen. Less than ten thousand, only thirty-one other numbers require nineteen steps to return back to one:
{56, 58, 60, 61, 352, 360, 362, 368, 369, 372, 373, 401, 402, 403, 2176, ..., and 2421}.[11]
19 is the sixth Heegner number.[12] 67 and 163, respectively the 19th and 38th prime numbers, are the two largest Heegner numbers, of nine total.
#### Prime properties
The sum of the squares of the first 19 primes is divisible by 19.[13]
19 is the first prime number that is not a permutable prime in decimal, as its reverse (91) is composite; where 91 is also the fourth centered nonagonal number.[14]
1729 is also the nineteenth dodecagonal number.[17]
19, alongside 109, 1009, and 10009, are all prime (with 109 also full reptend), and form part of a sequence of numbers where inserting a digit inside the previous term produces the next smallest prime possible, up to scale, with the composite number 9 as root.[18] 100019 is the next such smallest prime number, by the insertion of a 1.
• Numbers of the form 10n9 equivalent to 10x + 9 with x = n + 1, where n is the number of zeros in the term, are prime for n = {0, 1, 2, 3, 8, 17, 21, 44, 48, 55, 68, 145, 201, 271, 2731, 4563}, and probably prime for n = {31811, 43187, 48109, 92691}.[19]
R19 is the second base-10 repunit prime, short for the number 1111111111111111111.[20]
#### Figurate numbers and magic figures
19 is the third centered triangular number as well as the third centered hexagonal number.[21][22]
19 is the first number in an infinite sequence of numbers in decimal whose digits start with 1 and have trailing 9's, that form triangular numbers containing trailing zeroes in proportion to 9s present in the original number; i.e. 19900 is the 199th triangular number, and 1999000 is the 1999th.[24]
• Like 19, 199 and 1999 are also both prime, as are 199999 and 19999999. In fact, a number of the form 19n, where n is the number of nines that terminate in the number, is prime for:
n = {1, 2, 3, 5, 7, 26, 27, 53, 147, 236, 248, 386, 401}.[25]
The number of nodes in regular hexagon with all diagonals drawn is nineteen.[26]
• Distinguishably, the only nontrivial normal magic hexagon is composed of nineteen cells, where every diagonal of consecutive hexagons has sums equal to 38, or twice 19.[27]
• A hexaflexagon is a strip of nineteen alternating triangular faces that can flex into a regular hexagon, such that any two of six colorings on triangles can be oriented to align on opposite sides of the folded figure.[28]
• Nineteen is also the number of one-sided hexiamonds, meaning there are nineteen ways of arranging six equiangular triangular polyforms edge-to-edge on the plane without turn-overs (and where holes are allowed).[29]
${\displaystyle {\tfrac {1}{19))}$ can be used to generate the first full, non-normal prime reciprocal magic square in decimal whose rows, columns and diagonals — in a 18 x 18 array — all generate a magic constant of 81 = 92.[30]
• The next prime number to generate a like-magic square in base-ten is 383,[31] the seventy-sixth prime number (where 19 × 4 = 76).[32] A regular 19 x 19 magic square, on the other hand, has a magic constant ${\displaystyle M_{19))$ of 3439 = 19 × 181.[33]
### In abstract algebra
The projective special linear group ${\displaystyle \mathrm {L(19)} }$ represents the abstract structure of the 57-cell: a universal 4-polytope with a total of one hundred and seventy-one (171 = 9 × 19) edges and vertices, and fifty-seven (57 = 3 × 19) hemi-icosahedral cells that are self-dual.[34]
In total, there are nineteen Coxeter groups of non-prismatic uniform honeycombs in the fourth dimension: five Coxeter honeycomb groups exist in Euclidean space, while the other fourteen Coxeter groups are compact and paracompact hyperbolic honeycomb groups.
• There are also specifically nineteen uniform honeycombs inside the Euclidean ${\displaystyle {\tilde {C))_{4))$ tesseractic honeycomb group in 4-space. In 5-space, there are nineteen uniform polytopes with ${\displaystyle \mathrm {A} _{5))$ simplex symmetry.
There are infinitely many finite-volume Vinberg polytopes up through dimension nineteen, which generate hyperbolic tilings with degenerate simplex quadrilateral pyramidal domains, as well as prismatic domains and otherwise.[35]
On the other hand, a cubic surface is the zero set in ${\displaystyle \mathbb {P^{3)) }$ of a homogeneous cubic polynomial in four variables ${\displaystyle f=c_{3}000x_{1}^{3}+c_{2}100x_{1}^{2}x_{2}+c_{1}200x^{1}x_{2}^{2}+c_{0}300x_{2}^{3}+\cdots +c_{0}003x_{4}^{3},}$ a polynomial with a total of twenty coefficients, which specifies a space for cubic surfaces that is 19-dimensional.[37]
#### Finite simple groups
19 is the eighth consecutive supersingular prime. It is the middle indexed member in the sequence of fifteen such primes that divide the order of the Friendly Giant ${\displaystyle \mathrm {F_{1)) }$, the largest sporadic group: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 47, 59, 71}.[38]
• Janko groups ${\displaystyle \mathrm {J_{1)) }$ and ${\displaystyle \mathrm {J_{3)) }$ are the two-smallest of six pariah groups that are not subquotients of ${\displaystyle \mathrm {F_{1)) }$, which contain 19 as the largest prime number that divides their orders.[39]
${\displaystyle \mathrm {J_{1)) }$ holds (2,3,7) as standard generators (a,b,ab) that yield a semi-presentation where o(abab2) = 19, while ${\displaystyle \mathrm {J_{3)) }$ holds as standard generators (2A, 3A, 19), where o([a, b]) = 9.[40][41]
• ${\displaystyle 10944}$ is the dimensionality of the minimal faithful complex representation of O'Nan group ${\displaystyle O'N}$ — the second-largest after ${\displaystyle \mathrm {F_{1)) }$ of like-representation in ${\displaystyle {\text{dim)){\text{ ))196,883}$ and largest amongst the six pariahs[42] — whose value lies midway between primes (10939, 10949), the latter with a prime index of ${\displaystyle 1330}$,[43] which is the nineteenth tetrahedral number.[44]
• On the other hand, the Tits group ${\displaystyle \mathrm {T} }$, as the only non-strict group of Lie type that can loosely categorize as sporadic, has group order 211 · 33 · 52 · 13, whose prime factors (inclusive of powers) generate a sum equal to 54, which is the smallest non-trivial 19-gonal number.[45]
In the Happy Family of sporadic groups, nineteen of twenty-six such groups are subquotients of the Friendly Giant, which is also its own subquotient.[46] If the Tits group is indeed included as a group of Lie type,[47] then there are nineteen classes of finite simple groups that are not sporadic groups.
Worth noting, 26 is the only number to lie between a perfect square (52) and a cube (33); if all primes in the prime factorizations of 25 and 27 are added together, a sum of 19 is obtained.
## Religion
### Islam
• The number of angels guarding Hell ("Hellfire") ("Saqar") according to the Qur'an: "Over it are nineteen" (74:30), after which the Qur'an describes this number as being "a trial for those who disbelieve" (74:31), a sign for people of the scripture to be "convinced" (74:31) and that believers "will increase in faith" (74:31) due to it.
• The Number of Verse and Sura together in the Qur'an which announces Jesus son of Maryam's (Mary's) birth (Qur'an 19:19).
• A group called United Submitter International claim the Quran has mathematical structure based on the number 19. The gematrical value of WAHD = 6+1+8+4=19, Wahd means 'One' (God) to the first verse (1:1), known as Bas-malah, consists of 19 Arabic letters or the Quran consists of 114 (19x6) surat etc.
### Baháʼí faith
In the Bábí and Baháʼí Faiths, a group of 19 is called a Váhid, a Unity (Arabic: واحد, romanizedwāhid, lit.'one'). The numerical value of this word in the Abjad numeral system is 19.
### Celtic paganism
19 is a sacred number of the goddess Brigid because it is said to represent the 19-year cycle of the Great Celtic Year and the amount of time it takes the Moon to coincide with the winter solstice.[48]
## Music
• "19" is a 1985 song by Paul Hardcastle, including sampled soundbites taken from a documentary about the Vietnam War in which 19 is claimed to have been the average age of United States soldiers killed in the conflict.[49] The song was parodied by British satirist Rory Bremner under the pseudonym 'The Commentators,' as N-n-nineteen, Not Out, the title referring to the batting average of David Gower, the England cricket captain, during his side's risible performance against the West Indies in 1984 when they lost 5–0.
• "I Was Only Nineteen" by the Australian group Redgum reached number one on the Australian charts in 1983. In 2005 a hip hop version of the song was produced by The Herd.
• 19 is the name of Adele's 2008 debut album, so named since she was 19 years old at the time.
• "Hey Nineteen" is a song by American jazz rock band Steely Dan, on the 1980 album Gaucho.
• Nineteen has been used as an alternative to twelve for a division of the octave into equal parts. This idea goes back to Salinas in the sixteenth century, and is interesting in part because it gives a system of meantone tuning, being close to 1/3 comma meantone. See 19 equal temperament.
• Some organs use the 19th harmonic to approximate a minor third.
## Games
• The game of Go is played on a grid of 19×19 lines (though variants can be played on grids of other sizes).
• Though the maximum score for a cribbage hand is 29, there is no combination of cards that adds up to 19 points. Many cribbage players, therefore, jokingly refer to a zero-point hand as "a 19 hand".
• In the base version of Settlers of Catan there are 19 hexagonal pieces that can be randomly or intentionally placed to form the board.
## In sports
• In golf, the '19th hole' is the clubhouse bar and in match play, if there is a tie after 18 holes, an extra hole(s) is played. In miniature golf it is an extra hole on which the winner earns an instant prize.
## References
1. ^ Sloane, N. J. A. (ed.). "Sequence A046117 (Primes p such that p-6 is also prime. (Upper of a pair of sexy primes.))". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
2. ^ Sloane, N. J. A. (ed.). "Sequence A006512 (Greater of twin primes.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
3. ^ Sloane, N. J. A. (ed.). "Sequence A088762 (Numbers n such that (2n-1, 2n+3) is a cousin prime pair.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
4. ^ Sloane, N. J. A. (ed.). "Sequence A001913 (Full reptend primes: primes with primitive root 10.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
5. ^ Sloane, N. J. A. (ed.). "Sequence A002426 (Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
6. ^ "Sloane's A000043 : Mersenne exponents". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-31.
7. ^ Sloane, N. J. A. (ed.). "Sequence A007629 (Repfigit (REPetitive FIbonacci-like diGIT) numbers (or Keith numbers).)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
8. ^ Sloane, N. J. A. (ed.). "Sequence A005900 (Octahedral numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-08-17.
9. ^ Sloane, N. J. A. (ed.). "Sequence A002804 ((Presumed) solution to Waring's problem.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-08-05.
10. ^ Sloane, N. J. A. "3x+1 problem". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-01-24.
11. ^ Sloane, N. J. A. (ed.). "Sequence A006577 (Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-01-24.
"Table of n, a(n) for n = 1..10000".
12. ^ "Sloane's A003173 : Heegner numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-31.
13. ^ Sloane, N. J. A. (ed.). "Sequence A111441 (Numbers k such that the sum of the squares of the first k primes is divisible by k)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-06-02.
14. ^ a b Sloane, N. J. A. (ed.). "Sequence A060544 (Centered 9-gonal (also known as nonagonal or enneagonal) numbers. Every third triangular number, starting with 1)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-11-30.
15. ^ "19". Prime Curios!. Retrieved 2022-08-05.
16. ^ Sloane, N. J. A. (ed.). "Sequence A005349 (Niven (or Harshad, or harshad) numbers: numbers that are divisible by the sum of their digits.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-10-11.
17. ^ Sloane, N. J. A. (ed.). "Sequence A051624 (12-gonal (or dodecagonal) numbers: a(n) equal to n*(5*n-4).)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-12-21.
18. ^
19. ^ Sloane, N. J. A. (ed.). "Sequence A088275 (Numbers n such that 10^n + 9 is prime)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-07-28.
20. ^ Guy, Richard; Unsolved Problems in Number Theory, p. 7 ISBN 1475717385
21. ^ "Sloane's A125602 : Centered triangular numbers that are prime". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-31.
22. ^ "Sloane's A003215 : Hex (or centered hexagonal) numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-31.
23. ^ Sloane, N. J. A. (ed.). "Sequence A000217 (Triangular numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-07-13.
24. ^ Sloane, N. J. A. "Sequence A186076". The On-line Encyclopedia of Integer Sequences. Retrieved 2022-07-13. Note that terms A186074(4) and A186074(10) have trailing 0's, i.e. 19900 = Sum_{k=0..199} k and 1999000 = Sum_{k=0..1999} k...". "This pattern continues indefinitely: 199990000, 19999900000, etc.
25. ^ Sloane, N. J. A. (ed.). "Sequence A055558 (Primes of the form 1999...999)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-07-26.
26. ^ Sloane, N. J. A. (ed.). "Sequence A007569 (Number of nodes in regular n-gon with all diagonals drawn.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-04-04.
27. ^ Trigg, C. W. (February 1964). "A Unique Magic Hexagon". Recreational Mathematics Magazine. Retrieved 2022-07-14.
28. ^ Gardner, Martin (January 2012). "Hexaflexagons". The College Mathematics Journal. 43 (1). Taylor & Francis: 2–5. doi:10.4169/college.math.j.43.1.002. JSTOR 10.4169/college.math.j.43.1.002. S2CID 218544330.
29. ^ Sloane, N. J. A. (ed.). "Sequence A006534 (Number of one-sided triangular polyominoes (n-iamonds) with n cells; turning over not allowed, holes are allowed.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-12-08.
30. ^ Andrews, William Symes (1917). Magic Squares and Cubes (PDF). Chicago, IL: Open Court Publishing Company. pp. 176, 177. ISBN 9780486206585. MR 0114763. OCLC 1136401. Zbl 1003.05500.
31. ^
32. ^ Sloane, N. J. A. (ed.). "Sequence A000040 (The prime numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-09-06.
33. ^ Sloane, N. J. A. (ed.). "Sequence A006003 (a(n) equal to n*(n^2 + 1)/2.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-09-04.
34. ^ Coxeter, H. S. M. (1982). "Ten toroids and fifty-seven hemidodecahedra". Geometriae Dedicata. 13 (1): 87–99. doi:10.1007/BF00149428. MR 0679218. S2CID 120672023.
35. ^ Allcock, Daniel (11 July 2006). "Infinitely many hyperbolic Coxeter groups through dimension 19". Geometry & Topology. 10 (2): 737–758. arXiv:0903.0138. doi:10.2140/gt.2006.10.737. S2CID 14378861.
36. ^ Tumarkin, P. (2004). "Hyperbolic Coxeter n-polytopes with n + 2 facets". Mathematical Notes. 75 (5/6). Springer: 848–854. arXiv:math/0301133v2. doi:10.1023/B:MATN.0000030993.74338.dd. MR 2086616. S2CID 15156852. Zbl 1062.52012.
37. ^ Seigal, Anna (2020). "Ranks and symmetric ranks of cubic surfaces". Journal of Symbolic Computation. 101. Amsterdam: Elsevier: 304–306. arXiv:1801.05377. Bibcode:2018arXiv180105377S. doi:10.1016/j.jsc.2019.10.001. S2CID 55542435. Zbl 1444.14091.
38. ^ Sloane, N. J. A. (ed.). "Sequence A002267 (The 15 supersingular primes.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-12-11.
39. ^ Ronan, Mark (2006). Symmetry and the Monster: One of the Greatest Quests of Mathematics. New York: Oxford University Press. pp. 244–246. doi:10.1007/s00283-008-9007-9. ISBN 978-0-19-280722-9. MR 2215662. OCLC 180766312. Zbl 1113.00002.
40. ^ Wilson, R.A (1998). "Chapter: An Atlas of Sporadic Group Representations" (PDF). The Atlas of Finite Groups - Ten Years On (LMS Lecture Note Series 249). Cambridge, U.K: Cambridge University Press. p. 267. doi:10.1017/CBO9780511565830.024. ISBN 9780511565830. OCLC 726827806. S2CID 59394831. Zbl 0914.20016.
List of standard generators of all sporadic groups.
41. ^ Nickerson, S.J.; Wilson, R.A. (2011). "Semi-Presentations for the Sporadic Simple Groups". Experimental Mathematics. 14 (3). Oxfordshire: Taylor & Francis: 365. CiteSeerX 10.1.1.218.8035. doi:10.1080/10586458.2005.10128927. MR 2172713. S2CID 13100616. Zbl 1087.20025.
42. ^
43. ^ Sloane, N. J. A. (ed.). "Sequence A000040 (The prime numbers.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2024-02-28.
44. ^ Sloane, N. J. A. (ed.). "Sequence A000292". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2024-02-28.
45. ^ Sloane, N. J. A. (ed.). "Sequence A051871 (19-gonal (or enneadecagonal) numbers: n(17n-15)/2.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2023-12-09.
46. ^ John F.R. Duncan; Michael H. Mertens; Ken Ono (2017). "Pariah moonshine". Nature Communications. 8 (1): 2 (Article 670). arXiv:1709.08867. Bibcode:2017NatCo...8..670D. doi:10.1038/s41467-017-00660-y. PMC 5608900. PMID 28935903. ...so [sic] moonshine illuminates a physical origin for the monster, and for the 19 other sporadic groups that are involved in the monster.
47. ^ R. B. Howlett; L. J. Rylands; D. E. Taylor (2001). "Matrix generators for exceptional groups of Lie type". Journal of Symbolic Computation. 31 (4): 429. doi:10.1006/jsco.2000.0431. ...for all groups of Lie type, including the twisted groups of Steinberg, Suzuki and Ree (and the Tits group).
48. ^ Brigid: Triple Goddess of the Flame (Health, Hearth, & Forge)
49. ^ Roush, Gary (2008-06-02). "Statistics about the Vietnam War". Vietnam Helicopter Flight Crew Network. Archived from the original on 2010-01-06. Retrieved 2009-12-06. Assuming KIAs accurately represented age groups serving in Vietnam, the average age of an infantryman (MOS 11B) serving in Vietnam to be 19 years old is a myth, it is actually 22. None of the enlisted grades have an average age of less than 20.
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# Pedometer Steps vs. Calories Burned
If you're already using a pedometer, you're probably doing it because you want to lose weight or get in better shape. By walking on a regular basis, you're already on your way toward your goals -- but if you want to get even more detailed, you may also want to translate the number of steps you're taking into a calorie count.
## Data to Consider
You need to create a deficit of 3,500 calories in order to lose 1 pound -- but a lot of other factors come into play when it comes to calculating how many you're burning during any given activity. Your weight is one of them. Typically, heavier people burn more calories than lighter people. For example, a 185-pound person will burn 178 calories walking at a 3.5 mph pace for 30 minutes, while a 125-pound person will only burn 120 calories in that same amount of time. Intensity matters too; that 185-pound person will burn 222 calories by speeding up to a 4.5 mph pace. The 125-pound person will burn 150 calories in that same amount of time.
## The Intensity Issue
Here's where you might see the first problem in calculating the number of calories by the number of steps you've taken: A simple pedometer is not going to calculate your speed -- and thus won't take intensity into consideration. With that in mind, if you're calculating the number of calories you've burned based on the steps you've taken, use it as an estimate and not absolute gospel. Estimating your stride length -- which is a marker of your intensity -- can help. To gauge your stride length, make a chalk line on the sidewalk. Place both heels against that line and then walk 10 steps using a stride similar to your exercise walking stride. On the 10th step, make another chalk line behind that heel. Then measure the distance between the two chalk lines in inches and divide it by 10; that's your average stride length in inches.
## Using a Formula
After your pedometer has calculated the number of steps you've taken for the day, use an online calculator to determine how that number -- and those other factors -- translate into calories burned. The Geared to be Fit website has a pedometer calculator that factors in your weight, stride length and the number of steps you've taken. For example, if you weigh about 185 pounds, have an average stride length of 27 inches and you walked 10,000 steps in a day, the calculator estimates that you've burned about 370 calories.
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# 12.3 The most general applications of bernoulli’s equation
Page 1 / 2
• Calculate using Torricelli’s theorem.
• Calculate power in fluid flow.
## Torricelli’s theorem
[link] shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $h$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is
${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{.}$
Both ${P}_{1}$ and ${P}_{2}$ equal atmospheric pressure ( ${P}_{1}$ is atmospheric pressure because it is the pressure at the top of the reservoir. ${P}_{2}$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving
$\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}=\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{.}$
Solving this equation for ${v}_{2}^{2}$ , noting that the density $\rho$ cancels (because the fluid is incompressible), yields
${v}_{2}^{2}={v}_{1}^{2}+2g\left({h}_{1}-{h}_{2}\right)\text{.}$
We let $h={h}_{1}-{h}_{2}$ ; the equation then becomes
${v}_{2}^{2}={v}_{1}^{2}+2\text{gh}$
where $h$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $h$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem . Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.
All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See [link] .)
## Calculating pressure: a fire hose nozzle
Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?
Strategy
Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
Solution
Bernoulli’s equation states
${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{,}$
where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds ${v}_{1}$ and ${v}_{2}$ . Since $Q={A}_{1}{v}_{1}$ , we get
${v}_{1}=\frac{Q}{{A}_{1}}=\frac{\text{40}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}}{\pi \left(3\text{.}\text{20}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}}=\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$
Similarly, we find
${v}_{2}=\text{56.6 m/s}\text{.}$
(This rather large speed is helpful in reaching the fire.) Now, taking ${h}_{1}$ to be zero, we solve Bernoulli’s equation for ${P}_{2}$ :
${P}_{2}={P}_{1}+\frac{1}{2}\rho \left({v}_{1}^{2}-{v}_{2}^{2}\right)-\rho {\text{gh}}_{2}\text{.}$
Substituting known values yields
${P}_{2}=1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}+\frac{1}{2}\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left[\left(\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}-\left(\text{56}\text{.}6\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}\right]-\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left(9\text{.}80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(\text{10}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m}\right)=0\text{.}$
Discussion
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.
the meaning of phrase in physics
is the meaning of phrase in physics
Chovwe
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
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Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
Besmellah
Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?
what's electromagnetic induction
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
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mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
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Gabriel
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
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Ajayi
What is atomic number?
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oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm
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# Meera borrowed a sum of Rs 1000 from Sita for two years.
Question:
Meera borrowed a sum of Rs 1000 from Sita for two years. If the rate of interest is 10% compounded annually, find the amount that Meera has to pay back.
Solution:
Given:
$\mathrm{P}=\mathrm{Rs} 1,000$
$\mathrm{R}=10 \%$ p. $\mathrm{a} .$
$\mathrm{n}=2$ years
We know that amount $\mathrm{A}$ at the end of n years at the rate $\mathrm{R} \%$ per annum when the interest is compounded annually is given by $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)$.
$\therefore \mathrm{A}=1,000\left(1+\frac{10}{100}\right)^{2}$
$=1,000(1.1)^{2}$
$=1,210$
Thus, the required amount is Rs 1,210 .
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# Pythagorean Theorem 1
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Taught by YourMathGal
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Part of video series
Meets NCTM Standards:
Lesson Summary:
In this lesson, we learn about the Pythagorean theorem, one of the most famous theorems in math, and how to use it to find the length of a missing side of a right triangle. The theorem states that if we have a right triangle and we have two legs, A and B, then the sum of the squares of the legs is the same as the square of the hypotenuse. We are shown how to use this theorem to find the exact and approximate lengths of the missing sides of two right triangles. Finally, we learn how to simplify the square root of a number to make it easier to work with.
Lesson Description:
Find side of a right triangle using the Pythagorean Theorem.
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
Questions answered by this video:
• What is the Pythagorean Theorem?
• If a right triangle has one leg of length 5 inches and a hypotenuse of 8 inches, what is the length of the other side?
• If two legs of a right triangle are 3 feet and 5 feet, how can you use the Pythagorean Theorem to find the length of the hypotenuse?
• How can you use the Pythagorean Theorem to find the length of a missing side of a right triangle?
• What does a^2 + b^2 = c^2 mean?
• How do you solve x^2 + 5^2 = 8^2?
• How do you solve 3^2 + 5^2 = h^2?
• #### Staff Review
• Currently 4.0/5 Stars.
This lesson explains the Pythagorean Theorem and how it can be used to find missing lengths of right triangles. A couple of example problems are shown in this lesson and all steps involved are explained. This is a great introduction to the Pythagorean Theorem.
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TA: Tomoyuki Nakayama Monday, January 25 th , 2010 PHY 2048: Physic 1, Discussion Section 3081 Quiz 2 (Homework Set #3) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ There are two forces on the 4.00-kg box in the overhead view of the figure below, but only one is shown, where F 1 = 50.0 N, a = 8.00 m/s and θ = 30.0º. a) Express a and F 1 in unit-vector notation. The angle between a and the +x axis is 30º + 180º = 210º Thus a is expressed as a = acos210º i + asin210º j = -6.92 m/s 2 i - 4.00 m/s 2 j F 1
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Unformatted text preview: is along the y axis, thus it has y component only. F 1 = 50.0 N j b) Find the second force in unit-vector notation. We apply the Newton’s 2 nd law to the block and solve it for F 2 . We get m a = F 1 + F 2 ⇒ F 2 = m a-F 1 = 4 × (-6.92 i - 4.00 j ) -50.0 j = -27.7 N i-66.0 N j c) What are the magnitude and direction (relative to the + x axis) of the second force? Using Pythagorean Theorem, the magnitude is F 2 = √(F 1x 2 + F 1y 2 ) = √((-27.7) 2 + (-66) 2 ) = 71.6 N The angle is given by the inverse tangent function. Since F 2 is in the 3 rd quadrant, we add 180º to the answer your calculator displays. θ 2 = tan-1 (F 1y /F 1x ) +180º = 247º...
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# If $f\left(x\right)=\left\{\frac{{x}^{2}}{2},\phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}0\le x<1,2{x}^{2}-3x+\frac{3}{2},\phantom{\rule{1ex}{0ex}}if\phantom{\rule{1ex}{0ex}}1\le x\le 2$ .Show that $f$is continuous at $x=1$ .
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## To show that the function f(x) is continuous at x=1, we need to verify that the following three conditions are satisfied:1. The left-hand limit as x approaches 1 exists.2. The right-hand limit as x approaches 1 exists.3. The value of the function at x=1 equals both limits.The function is defined as follows:f(x)={x22if 0≤x<12x2−3x+32if 1≤x≤2Step 1: Calculate the left-hand limit as x approaches 1The left-hand limit is given by:limx→1−f(x)=limx→1−x22Substituting x=1:limx→1−f(x)=122=12Step 2: Calculate the right-hand limit as x approaches 1The right-hand limit is given by:limx→1+f(x)=limx→1+(2x2−3x+32)Substituting x=1:limx→1+f(x)=2(12)−3(1)+32=2−3+32=−1+32=12Step 3: Calculate the value of the function at x=1Since x=1 falls in the second piece of the function:f(1)=2(12)−3(1)+32=2−3+32=−1+32=12ConclusionNow we have:- Left-hand limit: limx→1−f(x)=12- Right-hand limit: limx→1+f(x)=12- Value of the function: f(1)=12Since all three values are equal, we conclude that f(x) is continuous at x=1.
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# Fundamental Theorems of Calculus
## Overview
In simple terms these are the fundamental theorems of calculus:
1
Derivatives and Integrals are the inverse (opposite) of each other.
2
When we know the indefinite integral:
F =
f(x) dx
We can then calculate a definite integral between a and b by the difference between the values of the indefinite integrals at b and a:
b
a
f(x) dx = F(b) − F(a)
Let's explore the details:
## First Fundamental Theorem of Calculus
For a continuous function f(x) on an interval [a, b] with the integral:
F(x) =
x
a
f(t) dt
then the derivative of the integral F(x) gets us the original function f(x) back again:
F'(x) = f(x)
This means that the derivative of the integral of f with respect to its upper limit is the function f itself.
### Example: f(x) = 2x
The integral of 2x is x2, and using the second theorem (below):
F(x) =
x
a
2t dt = x2 − a2
Taking the derivative:
F'(x) = ddx(x2 − a2) = 2x − 0 = 2x
So the derivative of the integral of 2x got us 2x back again.
### Example: Constant speed
A car travels at a constant speed of 50 km per hour for exactly one hour:
Speed:
50 km per hour
Integral of 50 km per hour for one hour:
50 km
Derivative of 50 km over one hour:
50 km per hour
## Second Fundamental Theorem of Calculus
When we have a continuous function f(x) on an interval [a, b], and its indefinite integral is F(x), then:
b
a
f(x) dx = F(b) − F(a)
In other words the definite integral of f(x) from a to b equals the difference in the values of F(x) at b and a
This makes calculating a definite integral easy if we can find its indefinite integral.
### Example
Imagine we're filling a tank with water, and the rate at which water flows into the tank is given by f(t), where t is time in minutes.
If F(t) measures the total volume of water in the tank at any time t, then the amount of water added to the tank between times a and b is F(b) - F(a).
f(t):
t2
F(t):
t3/3
So the amount of water added to the tank between 3 and 6 minutes is
6
3
f(t) dt:
F(6) − F(3)
= 63/3 − 33/3
= 63
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Jacob has 5 fishbowls. There are 3 fish in each bowl. How many fish does Jacob have?
02:16
Video Transcript
Jacob has five fishbowls. There are three fish in each bowl. How many fish does Jacob have?
In this question, we need to find out how many fish Jacob has. Now we’re given a picture to help us, so we could just count all the fish: one, two, three, and so on. But counting one by one like this can take a while, and there are quicker ways we could find the answer. Let’s use the information that we’re given to help us. To begin with, we’re told that Jacob has five fishbowls: one, two, three, four, five. We’re then told that there are three fish in each bowl. They’re easy to spot in each bowl because they’re three different colors. Each bowl contains a yellow fish, a red fish, and an orange fish. Every bowl has an equal group of fish. In other words, each bowl contains the same number of fish.
So, really, what this question is asking us is five bowls of three fish is how many fish? Or five groups of three is what? To find the answer, we could count in threes, one for each fish bowl. Otherwise, we need to find the answer to three plus three plus three plus three plus three. We know that three plus another three makes six. And if we add another group of three to six, we have seven, eight, nine. If we add another group of three to nine, we have 10, 11, 12. That’s four groups of three or four fish bowls that we’ve added. We just need to add one more lot of three. 12 add another three is 13, 14, 15.
You know, if we knew how to skip count in threes, we could have found the answer even quicker. Three, six, nine, 12, 15. Five bowls of three fish is 15 fish. Five groups of three is 15. The total number of fish that Jacob has is 15.
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## The output of a NOT gate is HIGH in case
Electrical Engineering XYZ MCQs The output of a NOT gate is HIGH in case: Correct answer: 1. The input is LOW
## Extended Binary Coded Decimal Interchange Code is
Electrical Engineering XYZ MCQs Extended Binary Coded Decimal Interchange Code is: Correct answer: 4. 8 bit code Explanation: The Extended Binary Coded Decimal Interchange Code (EBCDIC) is an 8-bit character encoding widely used in IBM mainframe computers and other systems. Each character in EBCDIC is represented by an 8-bit binary code, allowing for a total … Read more
## The difference between the diagram of a NOR and OR gate is
Electrical Engineering XYZ The difference between the diagram of a NOR and OR gate is: Correct answer: 1. NOR has got a bubble at its output terminal
## Maximum number in decimal that can be represented by 4 bits (binary) is
Electrical Engineering XYZ MCQs Maximum number in decimal that can be represented by 4 bits (binary) is: Correct answer: 1. 15 Explanation: In binary, each bit represents a power of 2. With 4 bits, you can have combinations of 2^0, 2^1, 2^2, and 2^3. Adding up these values: 1 + 2 + 4 + 8 … Read more
## A binary number system is of base
Electrical Engineering XYZ MCQs A binary number system is of base: Correct answer: 3. 2 Explanation:
## Complement of NOR and OR gate is ABC and XYZ respectively, here ABC and XYZ imply
Complement of NOR and OR gate is ABC and XYZ respectively, here ABC and XYZ imply: Correct answer: 1. OR, NOR
## A boolean function can be transformed into logical
Electrical Engineering XYZ MCQs A boolean function can be transformed into logical: Correct answer: 4. Diagram Explanation: Boolean functions can be transformed into logical maps using techniques like Karnaugh maps (K-maps). K-maps provide a systematic way to represent and simplify Boolean expressions, making it easier to understand and optimize logical circuits.
## Stack is also known as
Electrical Engineering XYZ MCQs Stack is also known as: Correct answer: 1. LIFO memory Explanation: In conclusion, a stack is also known as LIFO memory due to its characteristic of Last In, First Out.
## Which number system has a base of 16
Electrical Engineering XYZ MCQs Which number system has a base of 16: Correct answer: 1. Hexadecimal Explanation:
## Which of the following is not a transmission medium
Electrical Engineering XYZ MCQs Which of the following is not a transmission medium: Correct answer: 1. Modem Explanation: A modem (modulator-demodulator) is not a transmission medium but a device that modulates and demodulates digital data to enable communication between digital devices over analog transmission lines. It is used to convert digital signals from a computer … Read more
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# Explain why ( 7 × 5 × 3 × 2 + 3 ) is a composite number?
2
(7 x 5 x 3 x 2 + 3)
= 3 (7 x 5 x 2 + 1)
= 3 (71)
as it has to factors 3 and 71 other than 1 and no. itself so its a composite no.
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### 1: Number Sense, Properties, and Operations
#### 1.1: Understand the structure and properties of our number system. At their most basic level numbers are abstract symbols that represent real-world quantities
1.1: The decimal number system to the hundredths place describes place value patterns and relationships that are repeated in large and small numbers and forms the foundation for efficient algorithms
1.1.a: Students can: Generalize place value understanding for multi-digit whole numbers.
1.1.a.i: Explain that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right.
1.1.a.ii: Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form.
1.1.a.iii: Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.
1.1.a.iv: Use place value understanding to round multi-digit whole numbers to any place.
1.1.b: Students can: Use decimal notation to express fractions, and compare decimal fractions.
1.1.b.ii: Use decimal notation for fractions with denominators 10 or 100.
1.1.b.iii: Compare two decimals to hundredths by reasoning about their size.
#### 1.2: Understand that equivalence is a foundation of mathematics represented in numbers, shapes, measures, expressions, and equations
1.2: Different models and representations can be used to compare fractional parts
1.2.a: Students can: Use ideas of fraction equivalence and ordering to:
1.2.a.i: Explain equivalence of fractions using drawings and models.
1.2.a.ii: Use the principle of fraction equivalence to recognize and generate equivalent fractions.
1.2.a.iii: Compare two fractions with different numerators and different denominators, and justify the conclusions.
1.2.b: Students can: Build fractions from unit fractions by applying understandings of operations on whole numbers.
1.2.b.i: Apply previous understandings of addition and subtraction to add and subtract fractions.
1.2.b.i.1: Compose and decompose fractions as sums and differences of fractions with the same denominator in more than one way and justify with visual models.
1.2.b.i.2: Add and subtract mixed numbers with like denominators.
1.2.b.i.3: Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators.
1.2.b.ii: Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.
1.2.b.ii.1: Express a fraction a/b as a multiple of 1/b.
1.2.b.ii.2: Use a visual fraction model to express a/b as a multiple of 1/b, and apply to multiplication of whole number by a fraction.
#### 1.3: Are fluent with basic numerical, symbolic facts and algorithms, and are able to select and use appropriate (mental math, paper and pencil, and technology) methods based on an understanding of their efficiency, precision, and transparency
1.3: Formulate, represent, and use algorithms to compute with flexibility, accuracy, and efficiency
1.3.a: Students can: Use place value understanding and properties of operations to perform multi-digit arithmetic.
1.3.a.i: Fluently add and subtract multi-digit whole numbers using the standard algorithm.
1.3.a.ii: Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations.
1.3.a.iii: Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division.
1.3.a.iv: Illustrate and explain multiplication and division calculation by using equations, rectangular arrays, and/or area models.
1.3.b: Students can: Use the four operations with whole numbers to solve problems.
1.3.b.i: Interpret a multiplication equation as a comparison.
1.3.b.ii: Represent verbal statements of multiplicative comparisons as multiplication equations.
1.3.b.iii: Multiply or divide to solve word problems involving multiplicative comparison.
1.3.b.iv: Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted.
1.3.b.v: Represent multistep word problems with equations using a variable to represent the unknown quantity.
1.3.b.vi: Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
1.3.b.vii: Using the four operations analyze the relationship between choice and opportunity cost.
### 2: Patterns, Functions, and Algebraic Structures
#### 2.2: Make claims about relationships among numbers, shapes, symbols, and data and defend those claims by relying on the properties that are the structure of mathematics
2.1: Number patterns and relationships can be represented by symbols
2.1.a: Students can: Generate and analyze patterns and identify apparent features of the pattern that were not explicit in the rule itself.
2.1.a.iii: Complete input/output tables.
2.1.b: Students can: Apply concepts of squares, primes, composites, factors, and multiples to solve problems.
2.1.b.i: Find all factor pairs for a whole number in the range 1–100.
2.1.b.ii: Recognize that a whole number is a multiple of each of its factors.
2.1.b.iii: Determine whether a given whole number in the range 1–100 is a multiple of a given one-digit number.
2.1.b.iv: Determine whether a given whole number in the range 1–100 is prime or composite.
### 4: Shape, Dimension, and Geometric Relationships
#### 4.1: Understand quantity through estimation, precision, order of magnitude, and comparison. The reasonableness of answers relies on the ability to judge appropriateness, compare, estimate, and analyze error
4.1: Appropriate measurement tools, units, and systems are used to measure different attributes of objects and time
4.1.a: Students can: Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit.
4.1.a.i: Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec.
4.1.a.ii: Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two-column table.
4.1.a.iii: Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit.
4.1.a.iv: Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale.
4.1.a.v: Apply the area and perimeter formulas for rectangles in real world and mathematical problems.
4.1.b: Students can: Use concepts of angle and measure angles.
4.1.b.iii: Demonstrate that angle measure as additive.
4.1.b.iv: Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems.
#### 4.2: Make claims about relationships among numbers, shapes, symbols, and data and defend those claims by relying on the properties that are the structure of mathematics
4.2: Geometric figures in the plane and in space are described and analyzed by their attributes
4.2.a: Students can: Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines.
4.2.b: Students can: Identify points, line segments, angles, and perpendicular and parallel lines in two-dimensional figures.
4.2.c: Students can: Classify and identify two-dimensional figures according to attributes of line relationships or angle size.
4.2.d: Students can: Identify a line of symmetry for a two-dimensional figure.
Correlation last revised: 9/22/2020
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
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# Show if $\sum\limits_{k=1}^\infty {a_k}^2$,$\sum\limits_{k=1}^\infty {b_k}^2$ converge, their product converges too
Show if $\displaystyle\sum\limits_{k=1}^\infty {a_k}^2$ and $\displaystyle\sum\limits_{k=1}^\infty {b_k}^2$ both converge, $\displaystyle\sum\limits_{k=1}^\infty {a_kb_k}$ also converge
then
Show if $\displaystyle\sum\limits_{k=1}^\infty {a_n}$ and $\displaystyle\sum\limits_{k=1}^\infty {b_n}$ both converge, $\displaystyle\sum\limits_{k=1}^\infty {a_nb_n}$ also converge
then
Find two convergent series $\displaystyle\sum\limits_{k=1}^\infty {a_k}$ and $\displaystyle\sum\limits_{k=1}^\infty {b_k}$ such that $\displaystyle\sum\limits_{k=1}^\infty {a_kb_k}$ diverges.
My thought and attempt:
For the first part, By the theorem (If the series $\displaystyle\sum\limits_{k=1}^\infty {a_n}$ is convergent, then $\displaystyle\lim_{n\to\infty} a_n=0$),
if $\displaystyle\sum\limits_{k=1}^\infty {a_k}^2$ and $\displaystyle\sum\limits_{k=1}^\infty {b_k}^2$ both converge, then $\displaystyle\lim_{n\to\infty} {a_k}^2=0$ and $\displaystyle\lim_{n\to\infty} {b_k}^2=0$ both equal to zero.
Then we can obtain both $(a_k)_{k=1}^\infty$ and $(b_k)_{k=1}^\infty$ converge
thus $\displaystyle\lim_{n\to\infty} {(a_k)^2\over|a_k|} = \displaystyle\lim_{n\to\infty} |a_k| = 0$, do the same thing for $b_k$
Since limit preserves arithmetic operation, $\displaystyle\sum\limits_{k=1}^\infty {a_kb_k}$ converges.
Second part, I dont have any idea...
Third part is $a_n = b_n = {(-1)^n\over\sqrt{n+1}}$, am I right??
• Are $a_n, b_n$ real? If so, then hint: Cauchy-Schwarz. Feb 12, 2013 at 2:37
• I think you should double-check your second task. If $\sum a$ and $\sum b$ both converge, there is no proof that $\sum ab$ converges because it's not always true. If it were always true, then you would not be able to find the counter-example that you are asked to find for your third question. Are you sure the second question is not meant to be about two absolutely convergent series (as opposed to conditionally convergent)? Then the third part could be fulfilled by two conditionally convergent series. Feb 12, 2013 at 2:39
• I suspect the second part is: $\sum_{k=1}^\infty |a_n|, \sum_{k=1}^\infty |b_n|$ both converges, then $\sum_{k=1}^\infty a_n b_n$ converges. Feb 12, 2013 at 2:41
• To add to @ToddWilcox's comments, remember that it's possible for a sequence to go to zero, but for its sum to diverge, such as $a_k = 1/k$. Feb 12, 2013 at 2:42
Since $(a-b)^2 \geq 0$, you have $|ab| \leq \frac{1}{2}(a^2+b^2)$. Hence you have $\sum_{k=1}^n |a_k b_k| \leq \frac{1}{2} \sum_{k=1}^n (a_k^2+b_k^2) \leq \frac{1}{2} \sum_{k=1}^\infty (a_k^2+b_k^2)$, from which it follows that $\sum_{k=1}^\infty |a_k b_k| < \infty$.
If $\sum_{k=1}^\infty |a_k| < \infty$ then $a_k \to 0$. In particular, the terms are bounded, ie, $|a_k| \leq M$ for some $M$. Then since $a_k^2 \leq M |a_k|$, it follows that $\sum_{k=1}^\infty a_k^2 < \infty$. Similarly $\sum_{k=1}^\infty b_k^2 < \infty$, and using the first result yields the desired result.
However, if the sequences are not absolutely convergent, the result is not true. Take $a_k = b_k = (-1)^k \frac{1}{\sqrt{k}}$. These series converge because they are alternating and the terms converge to zero, but $a_k b_k = \frac{1}{n}$, and it is well known that $\sum_n \frac{1}{n}$ is divergent.
By the Cauchy-Schwarz Inequality, $$\forall n \in \mathbb{N}: \quad \sum_{k=1}^{n} |a_{k} b_{k}| \leq \sqrt{\sum_{k=1}^{n} a_{k}^{2}} \cdot \sqrt{\sum_{k=1}^{n} b_{k}^{2}}.$$ Therefore, $$\sum_{k=1}^{\infty} |a_{k} b_{k}| \leq \sqrt{\sum_{k=1}^{\infty} a_{k}^{2}} \cdot \sqrt{\sum_{k=1}^{\infty} b_{k}^{2}} < \infty.$$
In the language of real analysis, what we have shown is the following.
Theorem If $(a_{k})_{k \in \mathbb{N}},(b_{k})_{k \in \mathbb{N}} \in {\ell^{2}}(\mathbb{N})$, then $(a_{k} b_{k})_{k \in \mathbb{N}} \in {\ell^{1}}(\mathbb{N})$ and $$\left\| (a_{k} b_{k})_{k \in \mathbb{N}} \right\|_{1} \leq \left\| (a_{k})_{k \in \mathbb{N}} \right\|_{2} \cdot \left\| (b_{k})_{k \in \mathbb{N}} \right\|_{2} < \infty.$$
Just to add to copper.hat’s last example.
Both $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}$ and $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k}}{\ln(k)}$ converge by the Alternating Series Test, but $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k \ln(k)}$ diverges by the Integral Test.
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# RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2
## RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2
Other Exercises
Factorize each of the following expressions:
Question 1.
p3 + 27
Solution:
We know that a3 + b3 = (a + b) (a2 – ab + b2)
a3 – b3 = (a – b) (a2 + aft + b2)
p3 + 21 = (p)3 + (3)3
= (p + 3) (p2– p x 3 + 32)
= (p + 3) (p2 – 3p + 9)
Question 2.
y3 + 125
Solution:
y3 + 125 = (p)3 + (5)3
= (p + 5) (p2 – 5y + 52)
= (P + 5) (p2 – 5y + 25)
Question 3.
1 – 21a3
Solution:
1 – 21a3 = (1)3 – (3a)3
= (1 – 3a) [12 + 1 x 3a + (3a)2]
= (1 – 3a) (1 + 3a + 9a2)
Question 4.
8x3y3 + 27a3
Solution:
8x3y3 + 27a3
= (2xy + 3a) [(2xy)2 – 2xy x 3a + (3a)2]
= (2xy + 3a) (4x2y – 6xya + 9a2)
Question 5.
64a3 – b3
Solution:
64a3 – b3 = (4a)3 – (b)3
= (4a – b) [(4a)2 + 4a x b + (b)2]
= (4a – b) (16a2 + 4ab + b2)
Question 6.
Solution:
Question 7.
10x4– 10xy4
Solution:
I0x4y- 10xy4 = 10xy(x3 -y3)
= 10xy(x – y) (x2 + xy + y2)
Question 8.
54x6y + 2x3y4
Solution:
54 x6y + 2x3y4 = 2x3y(27x3 + y3)
= 2x3y[(3x)3 + (y)3]
= 2x3y(3x + y) [(3x)2 -3x x y + y2]
= 2x3y(3x + y) (9x2 -3xy + y2)
Question 9.
32a3 + 108b3
Solution:
32a3 + 108b3
= 4(8a3 + 27b3) = 4 [(2a)3 + (3 b)3]
= 4(2a + 3b) [(2a)2 – 2a x 3b + (3b)2]
= 4(2a + 3b) (4a2 – 6ab + 9b2)
Question 10.
(a – 2b)3 – 512b3
Solution:
(a – 2b)3 – 512b3
= (a – 2b)3 – (8b)3
= (a – 2b- 8b) [(a – 2b)2 + (a – 2b) x 8b + (8b)2]
= (a – 10b) [a2 + 4b2 – 4ab + 8ab – 16b2 + 64b2]
= (a – 10b) (a2 + 4ab + 52b2)
Question 11.
8x2y3 – x5
Solution:
8x2y3 – x5 = x2(8y3 – X3)
= x2(2y)3 – (x)3]
= x2[(2y – x) (2y)2 + 2y x x + (x)2]
= x2(2y – x) (4y2 + 2xy + x2)
Question 12.
1029 -3x3
Solution:
1029 – 3X3 = 3(343 – x3) ‘
= 3 [(7)3 – (x)3]
= 3(7 – x) (49x + 7x + x2)
Question 13.
x3y3+ 1
Solution:
x3y3 + 1 = (xy)3 + (1)3
= (xy + 1) [(xy)2 – xy x 1 + (1)2]
= (xy + 1) (x2y2 – xy + 1)
= (xy + 1) (x2y – xy + 1)
Question 14.
x4y4 – xy
Solution:
x4y4 – xy = xy(x3y3 – 1)
= xy[(xy3-(1)3]
= xy (xy – 1) [x2y2 + 2xy + 1]
Question 15.
a3 + b3 + a + b
Solution:
a3 + b3 + a + b
= (a + b) (a2 – ab + b2) + 1 (a + b)
= (a + b) (a2 – ab + b2 + 1)
Question 16.
Simplify:
Solution:
Question 17.
(a + b)3 – 8(a – b)3
Solution:
(a + b)3 – 8(a – b)3
= (a + b)2 – (2a – 2b)3
= (a+ b – 2a + 2b) [(a + b)2 + (a + b) (2a-2b) + (2a – 2b)2)]
= (3b – a) [a2 + b2 + 2ab + 2a22ab + 2ab – 2b2 + 4a2 – 8ab + 4b2]
= (3b – a) [7a2 – 6ab + 3b2]
Question 18.
(x + 2)3 + (x- 2)3
Solution:
(x + 2)3 + (x – 2)3
= (x + 2 + x – 2) [(x + 2)2 – (x + 2) (x – 2) + (x – 2)2]
= 2x [x2 + 4x + 4 – (x2 + 2x – 2x – 4) + x4x + 4]
= 2x[x2 + 4x + 4- x2-2x + 2x + 4+ x2– 4x + 4]
= 2x[x2 + 12]
Question 19.
x6 +y6
Solution:
x6 + y= (x2)3 + (y2)3
= (x2 + y2) [x4 – x2y2 + y4]
Question 20.
a12 + b12
Solution:
a12 + b12 = (a4)3 + (b4)3
= (a4 + b4) [(a4)2 – a4b4 + (b4)2]
= (a4 + b4) (a8 – a4b4 + b8)
Question 21.
x3 + 6x2 + 12x + 16
Solution:
x3 + 6x2 + 12x + 16
= (x)3 + 3.x2.2 + 3.x.4 + (2)3 + 8 {∵ a3 + 3a2b + 3ab2 +b3 = (a + b)3}
= (x + 2)3 + 8 = (x + 2)3 + (2)3
= (x + 2 + 2) [(x + 2)2 – (x + 2) x 2 + (2)2] {∵ a3 + b2 = (a + b) (a2 – ab + b2}
= (x + 4) (x2 + 4x + 4 – 2x – 4 + 4)
= (x + 4) (x2 + 2x + 4)
Question 22.
Solution:
Question 23.
a3 + 3a2b + 3ab2 + b3 – 8
Solution:
a3 + 3a2b + 3ab2 + b3 – 8
= (a + b)3 – (2)3
= (a + b -2)[(a + b)2 + (a +b)x2 + (2)2]
= (a + b-2) (a2 + b2 + 2ab + 2a + 2b + 4)
= (a + b – 2) (a2 + b2 + 2ab + 2(a + b) + 4]
= (a + b – 2) [(a + b)2 + 2(a + b) + 4}
Question 24.
8a3 – b3 – 4ax + 2bx
Solution:
8a3 – b3 – 4ax + 2bx
(2a)3 – (b)3 – 2x(2a – b)
= (2a-b)[(2a)2 + 2a x b + (b)2]- 2x(2a-b)
= (2a – b) [4a2 + 2ab + b2] – 2x(2a – b)
= (2a – b) [4a2 + 2ab + b2 – 2x]
Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# General questions about integrals (definitions)
• I
• JulienB
In summary: I'm sorry, I don't remember well. It seems that the condition is that the integral exists and is positive, but I'm not sure about the details.
JulienB
Hi everybody! I'm currently studying integrals, and I would like to clarify a few definitions, especially about the criterions of convergence/divergence of an integral. Basically if that's okay for you guys I'm going to list and number a few statements and I'd like to know if they are true or not.
1. If a definite integralab f(x) dx has both its domain of integration and integrand bounded, but its set of locations where it is not continuous is an uncountable set, then f(x) is not integrable.
(or do we have to run other tests to find out if the function is integrable or not?)
2. If a definite integralab f(x) dx has both its domain of integration and integrand unbounded, then it is called an improper integral. While the term improper integral normally designates the limit of an unbounded integral, this "ambiguity is resolved as both the proper and improper integral will coincide in value" (Wikipedia, improper integral).
3. The improper integral of a function f(x) exists, if there exists a function g(x) so that ∫ab g(x) dx converges and |f(x)| ≤ g(x) ∀ x ∈ [a,b). If one of those criterions is not met, then we cannot conclude anything about the existence of the improper integral of f(x) and we must run other tests. (Majorant criterion)
4. The improper integral of a function f(x) doesn't exist, if there exists a function g(x) so that ∫ab g(x) dx diverges and f(x) ≥ g(x) ∀ x ∈ [a,b). If one of those criterions is not met, then we cannot conclude anything about the non-existence of the improper integral of f(x) and we must run other tests. (Minorant criterion)I stop here, it would already mean a lot to me if those simple assumptions would become facts. :)
For 1), consider the function ##f: \mathbb R \to \mathbb R## defined by:
##\left\{\begin{array}{l}
0 \ on \ \mathbb Q\\
1 \ on \ \mathbb R \setminus \mathbb Q
\end{array}
\right.##
For 4), consider ##g(x)=-\frac{1}{x}## and ##f(x)=0## on ##]0,1[##
JulienB
Samy_A said:
For 1), consider the function ##f: \mathbb R \to \mathbb R## defined by:
##\left\{\begin{array}{l}
0 \ on \ \mathbb Q\\
1 \ on \ \mathbb R \setminus \mathbb Q
\end{array}
\right.##
It seems to me that this function is integrable though its set of non continuous locations is uncountable. Is that right? Then I would conclude that my statement was false and that this criterion only let's us know if a function is integrable but not if it is not integrable.
Samy_A said:
For 4), consider ##g(x)=-\frac{1}{x}## and ##f(x)=0## on ##]0,1[##
I see, f(x) converges though it is bigger as g(x).. Would that problem be solved if I had stated |f(x)| ≥ g(x)?
Oh, and also are 2. and 3. correct statements then?
Thank you so much Samy.
Julien.
Another question related to 1. would then be: if f(x) is bounded on both its domain of integrability and integrand but its set of non continuous locations is uncountable, what criterion(s) am I left with to check if f(x) is integrable or not? Julien.
JulienB said:
It seems to me that this function is integrable though its set of non continuous locations is uncountable. Is that right? Then I would conclude that my statement was false and that this criterion only let's us know if a function is integrable but not if it is not integrable.
Statement 1) is indeed false. I don't exactly understand what criterion for being integrable or not it could yield.
JulienB said:
I see, f(x) converges though it is bigger as g(x).. Would that problem be solved if I had stated |f(x)| ≥ g(x)?
No, see the same counterexample.
JulienB said:
Oh, and also are 2. and 3. correct statements then?
3) is correct (provided f is a measurable function). I'm not sure about what 2) is supposed to mean.
JulienB said:
Another question related to 1. would then be: if f(x) is bounded on both its domain of integrability and integrand but its set of non continuous locations is uncountable, what criterion(s) am I left with to check if f(x) is integrable or not?
That's a difficult question to answer. It depends on how you studied integration: Riemann integral, Lebesgue integral?
JulienB
Samy_A said:
Statement 1) is indeed false. I don't exactly understand what criterion for being integrable or not it could yield.
Mmm that comes from the script of my maths teacher. It's in German but I try to translate it:
Bounded functions are integrable, if they are "almost everywhere" continuous: a bounded function f: [a,b] → ℂ is integrable, if the set of its discontinuities is countable.
Samy_A said:
No, see the same counterexample.
Oh yes indeed. Maybe f(x) ≥ |g(x)| then? Or is it that the minorant criterion does not exist at all for integration?
Samy_A said:
3) is correct (provided f is a measurable function). I'm not sure about what 2) is supposed to mean.
About 2: I was under the impression that what we called an improper integral is actually the limit of a definite integral as one of the boundaries tends towards ±∞ or as the integrand tends towards ±∞. Actually I just realized I wrote "both" instead of "either" in my original statement and that's not how I meant it, sorry. And what I wished to check was if that is the correct definition of an improper integral.
Samy_A said:
That's a difficult question to answer. It depends on how you studied integration: Riemann integral, Lebesgue integral?
I study Riemann integrals, as far as I know. Wikipedia seems to say that statement number 1 (or a corrected version of it) is called "Lebesgue's integrability condition", but I still believe it applies to Riemann integrals right?Julien.
JulienB said:
Mmm that comes from the script of my maths teacher. It's in German but I try to translate it:
Bounded functions are integrable, if they are "almost everywhere" continuous: a bounded function f: [a,b] → ℂ is integrable, if the set of its discontinuities is countable.
Ah yes, that is correct.
JulienB said:
Oh yes indeed. Maybe f(x) ≥ |g(x)| then? Or is it that the minorant criterion does not exist at all for integration?
Yes, I think that one is correct.
JulienB said:
About 2: I was under the impression that what we called an improper integral is actually the limit of a definite integral as one of the boundaries tends towards ±∞ or as the integrand tends towards ±∞. Actually I just realized I wrote "both" instead of "either" in my original statement and that's not how I meant it, sorry. And what I wished to check was if that is the correct definition of an improper integral.
That is indeed the definition of an improper integral.
JulienB said:
I study Riemann integrals, as far as I know. Wikipedia seems to say that statement number 1 (or a corrected version of it) is called "Lebesgue's integrability condition", but I still believe it applies to Riemann integrals right?
Yes.
JulienB
@Samy_A Thank you for your answers, that was very helpful. I like to be sure I understand definitions well before I go further :)
Regarding 1 ("bounded functions are integrable, if they are "almost everywhere" continuous"), do you have a suggestion of another criterion to check integrability if the set of discontinuities of the function is uncountable?
I have another one that says:
Functions that are "almost everywhere" equal have the same integrals: If f: [a,b] → ℂ is integrable and g: [a,b] → ℂ bounded, and if the set of all x ∈ [a,b] with f(x) ≠ g(x) is countable, then g is also integrable and ∫ab f(x)dx = ∫ab g(x)dx.
Not sure if that helps in many cases though. What's your opinion about that?
Julien.
JulienB said:
I have another one that says:
Functions that are "almost everywhere" equal have the same integrals: If f: [a,b] → ℂ is integrable and g: [a,b] → ℂ bounded, and if the set of all x ∈ [a,b] with f(x) ≠ g(x) is countable, then g is also integrable and ∫ab f(x)dx = ∫ab g(x)dx.
Not sure if that helps in many cases though. What's your opinion about that?
Is that true for Riemann integrals? It is for Lebesgue integral, but I doubt it is for the Riemann integral.
The Lebesgue integrability condition you mentioned states that for a bounded f: [a,b] → ℂ, f is Riemann integrable if and only if the set of discontinuities of f has Lebesgue measure 0.
But one can easily take a nice continuous function f, change it's values on the rational numbers in such a way that the resulting function g is nowhere continuous.
g will be Lebesgue integrable, but not Riemann integrable.
Samy_A said:
Is that true for Riemann integrals? It is for Lebesgue integral, but I doubt it is for the Riemann integral.
Well what I quoted in my previous post is absolutely all there is in my script about that criterion. But to be honest, the script my teacher made is very unclear and I doubt it is complete. He has proved in the past to have a certain talent for misleading us :(
I hear what you're saying with your example, but not even is "Lebesgue" mentioned once in my script, while Riemann is at many occasions. I will ask other students and search a bit on the internet too.
Thanks a lot again!Julien.
Samy_A said:
Is that true for Riemann integrals?
It is if you assume both are integrable. So if ##f## and ##g## are Riemann-integrable and ##f=g## a.e. then the integrals coincide.
micromass said:
It is if you assume both are integrable. So if ##f## and ##g## are Riemann-integrable and ##f=g## a.e. then the integrals coincide.
Unfortunately that's not what is stated in my script...It says that if:
1. f: [a,b] → ℂ is integrable,
2. g: [a,b] → ℂ is bounded,
3. the set of all x ∈ [a,b] for which f(x) g(x) is countable,
then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.
It does look wrong to me, especially with the example Samy submitted.Julien.
JulienB said:
Unfortunately that's not what is stated in my script...It says that if:
1. f: [a,b] → ℂ is integrable,
2. g: [a,b] → ℂ is bounded,
3. the set of all x ∈ [a,b] for which f(x) g(x) is countable,
then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.
It does look wrong to me, especially with the example Samy submitted.Julien.
Yes, that is incorrect. It becomes correct for the Lebesgue integral though. It is also correct if we demand finite in (3).
micromass said:
Yes, that is incorrect. It becomes correct for the Lebesgue integral though. It is also correct if we demand finite in (3).
Alright, thank you for your answer. I reformulate what you said with my own words to make sure I get it right:
1. f: [a,b] → ℂ is integrable,
2. g: [a,b] → ℂ is bounded,
3. the set of all x ∈ [a,b] for which f(x) g(x) is finite,
then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.
About Lebesgue integrals, we didn't do that yet so I guess that criterion for integrability was meant for Riemann integrals.
Thx a lot,Julien.
## 1. What is an integral?
An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value or accumulated quantity of a function over a given interval.
## 2. What is the difference between a definite and indefinite integral?
A definite integral has specific limits of integration, meaning it calculates the area under a curve between two specific points on the x-axis. An indefinite integral does not have specific limits, and instead represents a general antiderivative of a function.
## 3. How do you calculate an integral?
To calculate an integral, you can use various integration techniques such as substitution, integration by parts, or partial fractions. You can also use software or calculators to numerically approximate the integral.
## 4. What are the applications of integrals?
Integrals have many real-world applications, such as calculating the area under a curve in physics, finding the volume of a solid in engineering, and determining the average value of a function in economics. They are also used in optimization problems and probability calculations.
## 5. Can any function be integrated?
Not all functions can be integrated analytically, meaning with a closed-form solution. However, most functions can be numerically integrated using numerical methods such as the trapezoidal rule or Simpson's rule. Some functions, such as those with infinite discontinuities, are not integrable at all.
• Calculus
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# capacitor value to measure rise time and fall time
1. ## capacitor rise time
how to get the value of the capacitor across the output of a driver? Please see the ckt attach.
Thank you very much!
2. ## rise time of capacitor
Why Do you want to use a capacitor?
•
3. ## rise time capacitor
myql,
Your question is not clear to me.
.
Are you trying to measure the value of the capacitor by measuring the rise time?
.
Are you trying to determine the value of capacitor required to get a specified rise time?
Regards,
Kral
4. ## rise time capacitance
kral,
Yes, I am trying to determine the value of the capacitor required to get a specified rise time.
could you help me on this matter?
Thank you very much!
Regards,
myql
5. ## capacitor rise time calculation
try this method.
t= k. R.c
t = time
k = constant
R = Resistence
c = capasitance
Im not sure with the value of k (may be 0,6)
93_1201065577.jpg][/url]
1 members found this post helpful.
•
6. ## capacitor fall time
Hi,
You can do one thing. Suppose you put a known capacitor C_kn_1 at the output. The driver may have some capacitance between its o/p terminal and ground.Lets say it is C_out. Then the total capacitance at the o/p is C_out + C_kn_1. Measure the rise time t_r_1. The rise time will be
t_r_1 = R_out * (C_kn_1 + C_out)
Similarly put another known resistor C_kn_2 and the rise time can be measured. This will be
t_r_2 = R_out * (C_kn_2 + C_out)
Subtracting these two you get
t_r_1 ~ t_r_2 = R_out (C_kn_1 ~ C_kn_2)
where only unknown term is R_out. You can find it out and from the expression of either t_r_1 or t_r_2 you can find out C_out.
Therefore for any predetermined rise time t_r the capacitance required is found from the equation
t_r = R_out* (C_out + C_unkn)
or C_unkn = t_r / R_out.
I don't know how accurately you can find it out with this. the accuracy of course depends on a lot other things, like, when you load something to the driver the load will also have some input cvapacitance which adds to the total capacitance to reduce the rise time further.
where R_out is the output resistance of the driver. Then, knowing the output resistance
1 members found this post helpful.
•
7. ## how to measure rise time and fall time
myql
There are 2 commonly used definitions of rise time:
.
1) Time required for output to undergo 63% (X.63) of its total swing (the exact number is [1-(1/e)]
For this definition, C = t/R
.
2) Time required for the output to go from 10% of its total swing to 90% of its total swing. This is the more commonly used definition for digital systems. For this definition, C =.455t/R.
.
I’m assuming that you know the source resistance R. Let me know if you want the derivations.
.
Regards,
Kral
8. ## capacitor rise time formula
rise time =.35/F_H(f=freuq.)
1 members found this post helpful.
9. ## measure output rise time
myql,
a-tek7 is correct. This results in C =2.85t/R. Ignore my post regarding the 10-90 equation.
Regards,
Kral
10. ## 33 capacitor value
low rise time means high bandwidth. A parallel cap means
filtering high frequencies, sort of low past filter, limiting
bandwidth. So a cap in parallel will lower the rise time.
Unless what you are looking for is to limit the rise time to a
certain value the question does not make sense.
11. ## trying to measure capacitor values
Kral,
can you send me the derivation?
Thank you very much!
Regards,
myql
12. ## rising time of the capacitor
The requirement of the controlled rise time or fall time is crutial. There are devices, in fact a large number of analog devices, which shows capacitive load to the drivers. Then any sharp change in the voltage levels in the input to it can give rise to a large spike of curent depending on the output resistance of the driver. A smaller output resistance makes the driver wider in bandwidth of course but there are chances that the large spike f current can even exced the allowable input current limit of the capacitive load. In that case the device can get damaged. Especially sensitive devices like CCD, APS all have these sort of problems. The rise time spec is very important for the driver.
1 members found this post helpful.
•
13. ## rise time and fall time
F_H=1/(2ΠRC)& T_R=2.2RC.
RC=RC so T_R/2.2=1/(2ΠF_H) so T_R=(2.2/2Π)/F_H=.35/F_H
1 members found this post helpful.
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