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## What is the formula for lift?
The lift equation states that lift L is equal to the lift coefficient Cl times the density r times half of the velocity V squared times the wing area A. For given air conditions, shape, and inclination of the object, we have to determine a value for Cl to determine the lift.
## What is the unit of lift?
L signifies lift force, measured in Newtons; ρ signifies air density, measured in kilograms per cubic meter; v2 signifies true airspeed squared, which is the square of the speed of the helicopter relative to the oncoming air, expressed in meters per second.
Daniel Bernoulli
## How do you calculate lift speed?
Lift speed is usually expressed in m/sec (metres per second). Platform Lifts under the Machinery Directive are limited to a maximum speed of 0.15m/sec (150mm per second). In order to calculate the time that the lift will take to complete a single journey you will need to divide the total lift travel by the speed.
## How do you make a plane lift?
Airplane wings are shaped to make air move faster over the top of the wing. When air moves faster, the pressure of the air decreases. So the pressure on the top of the wing is less than the pressure on the bottom of the wing. The difference in pressure creates a force on the wing that lifts the wing up into the air.
## What gives a wing lift?
An airfoil generates lift by exerting a downward force on the air as it flows past. According to Newton’s third law, the air must exert an equal and opposite (upward) force on the airfoil, which is lift. The airflow changes direction as it passes the airfoil and follows a path that is curved downward.
## What does downwash mean?
Wiktionary. downwash(Noun) downward air turbulence caused by a propeller or jet, but especially by helicopter blades. downwash(Noun) The downward motion of air as a result of eddies behind a wing or chimney.
## How do you measure wing lift?
The modern lift equation states that lift is equal to the lift coefficient (Cl) times the density of the air (r) times half of the square of the velocity (V) times the wing area (A).
## What angle of attack produces the greatest lift?
critical angle
You might be interested: Quadratic equation example
## What are the 4 principles of flight?
The four forces are lift, thrust, drag, and weight. As a Frisbee flies through the air, lift holds it up.
## What does aerofoil mean?
An airfoil (American English) or aerofoil (British English) is the cross-sectional shape of a wing, blade (of a propeller, rotor, or turbine), or sail (as seen in cross-section). An airfoil-shaped body moving through a fluid produces an aerodynamic force.
## How high can a helicopter fly?
Turbine-engined helicopters can reach around 25,000 feet. But the maximum height at which a helicopter can hover is much lower – a high performance helicopter like the Agusta A109E can hover at 10,400 feet.
## How much lift does a helicopter produce?
In the hover, it must produce a lift force equal to the weight of the helicopter, in order for the helicopter to remain at constant altitude. A helicopter with a mass of 700 kg must produce a lift of 700*g = 6,867 N to hover at constant altitude.
### Releated
#### Depreciation equation
What are the 3 depreciation methods? There are three methods for depreciation: straight line, declining balance, sum-of-the-years’ digits, and units of production. What do you mean by depreciation? Definition: The monetary value of an asset decreases over time due to use, wear and tear or obsolescence. This decrease is measured as depreciation. How do you […]
#### Polar to cartesian equation calculator wolfram
How do you convert polar to Cartesian? Summary: to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you find the polar Cartesian equation? Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding […] | HuggingFaceTB/finemath | |
# Memory Overview
```• Memory stores the program running and the data on which the
program operates
•Data store in binary code.
• Terminology:
Memory Cell – A device or an electrical circuit used to store
a single bit (0 or 1). Ex: flip-flop
Memory Word – A group of bits (cells) I memory that
represents instructions or data of some
type. Ex: Index Register consisting of 16 bit
can be considered to be a memory word
Byte – A special term used for a group of 8 bits
Nibble – Half of Byte ( 4bit).
Capacity –A way of specifying how many bits can be stored
in a particular memory device or complete memory
system.
Capacity metrics unit: 1K = 210 = 1024.
1M = 220 = 1,048,576.
1G = 230 = 1,073,741,824
A certain semiconductor memory chip is specified as
2K x 8.
a. How many words can be stored on this chip?
b. What is the word size?
c. How many capacity can this chip store?
Solution:
a. 2K = 2 x 1024 = 2048 location.
b. Word size is 8-bits (one byte).
c. The total capacity is 2048 x 8bit = 16,384 bits
Which memory stores the most bits:
a 5M x 8bits memory OR
a 1M x 16bits memory?
Solution: Capacity
5M x 8 = 5 x 1,048,576 x 8 = 41,943,040 bits
1M x 16 = 1,048,576 x 16 =16,777,216 bits
The 5M x 8 memory stores more bits.
Density- Another term for capacity
Address – A number that identifies the location of a word
in memory.
Read operation – the operation whereby the binary word
stored in a specific memory location (address) is sense and
then transferred to another device.
Write operation – The operation whereby a new word is
placed into a particular memory location.
Access Time – A measure of memory device’s operating
speed. It is the amount of time required to perform a read
operation.
Density- Another term for capacity
Address – A number that identifies the location of a word
in memory.
Read operation – the operation whereby the binary word
stored in a specific memory location (address) is sense and
then transferred to another device.
Write operation – The operation whereby a new word is
placed into a particular memory location.
Access Time – A measure of memory device’s operating
speed. It is the amount of time required to perform a read
operation.
Main Memory – Also referred to as the computer’s working
memory. It stores instructions and data the CPU is
currently working on. It is the highest-speed memory in
the computer and is always a semiconductor memory.
Auxiliary Memory – Also referred to as mass storage
because it stores massive amounts of information external
to the main memory. It is slower in speed than main
memory and is always nonvolatile. CDs are common
auxiliary devices.
Memory Type
ROM
1. MROM : Mask-programmed ROM.
2. PROM : Programmable ROM
3. EPROM : Erasable PROM
4. EEPROM : Electrically-erasable PROM or
EAROM: Electrical Alterable ROM
6. FLASH MEMORY
RAM
(Random Access Memory)
1. SRAM : Static RAM
2. DRAM : Dynamic RAM
The read-only memory is type of semiconductor memory
designed to hold data that either are permanent or will not
change frequently. (Non-volatile)
During normal operation data can be read from ROM.
Data can be entered electrically –programming or burningin the ROM.
Some ROMs cannot have their data changed once they
have been programmed; others can be erased and
reprogrammed as often as desired.
A major use for ROMs is in the storage of programs in
microcomputers. When the microcomputer is turned on, it
can immediately begin executing the program stored in
ROM
Has 3 sets of signals: address inputs, control inputs, and data
outputs.
Store 16 words because it has 2^4=16 possible addresses, and
each word contains 8-bit because there are 8 data outputs.
This is a 16 x 8 ROM.
The most common numbers of data outputs for ROMs are 4,
8,16 bits with 8-bit word being the most common.
Control input CS-Chip Select – an enable input that enables or
disabled the ROM outputs
Many ROMs have two or more control inputs that must be
active in order to enable the data outputs so that data can be
CS input shown in figure is active-LOW; therefore, it must
be in the LOW state to enable the ROM data to appear at
the data outputs
Notice that there are no R/W input because the ROM
cannot be written into during normal operation.
16 different data words are stored at the 16 different
In order to read a data word from ROM, we need to do 2
things :
Apply the appropriate address inputs
Activate the control inputs.
Ex: if we want to read the data stored at location 0111 of the
ROM, we must apply A3A2A1A0=0111 to the address inputs and
then apply a LOW to CS. The address inputs will be decoded
inside the ROM to select the correct data word, 11101101, that
will appear at outputs D7 to D0. If CS is kept HIGH the ROM
outputs will be disabled and will be in the Hi-Z state.
Has its storage location written into by the manufacturer
according to the customer’s specifications.
A mask is used to control the electrical interconnections on the
chip.
A special mask is required for each different set of information to
be stored in the ROM.
Disadvantage – of this type of ROM is that cannot be
reprogrammed in the event of a design change requiring a
modification of the stored data
Is the most economical approach when a large quantity of
identically programmed ROMs are needed.
For lower-volume applications, manfacturers have
developed fusible-link PROMs that are userprogrammable; that is, they are not programmed
during the manufacturing process but are customprogrammed by the user.
Once programmed, cannot be erased and
reprogrammed
If the programmed in the PROM must be changed,
the PROM must be thrown away.
EPROM programing can be done by
charging floating gate in side it.
The programming of an EPROM can be
done in special programmer unit circuit
and erasing using UV light source .
Can be programmed by the user and can be
erased and reprogrammed as often as desired.
Nonvolatile memory that will hold its stored
data indefinitely
The programming process is usually performed
by a special programming circuit that is separate
from he circuit in which the EPROM will
eventually be working.
EPROMs are available in a wide range of
capacities and access times; devices with a
capacity of 512K x 8 and can access time of 20 ns
are commonplace
1.
2.
3.
They must be removed from their circuit to be erased
and reprogrammed
The erase operation erases the entire chip-there is no
way to select only certain addresses to be erased
The erase and reprogramming process can typically take
20 minutes or more.
Fused
quartz
The disadvantages of the EPROM were overcome by
the development of the electrically erasable PROM
(EEPROM) as an improvement over the EPROM.
The erasing and programming of an EPROM can be
done in circuit ( without UV light source or a special
programmer unit)
Advantages: ability to erase and rewrite individual
bytes (8-bit words) in the memory array electrically.
During a write operation, internal circuitry
automatically erases all of the cells at an address
location prior to writing in the new data. This byte eras
ability makes it much easier to make changes in the
data stored in an EEPROM
From EEPROM to Flash memory cell, is like the simple singe
transistor EPROM cell, being only slightly larger.
Allows electrical erasability but can be built with much higher
densities than EEPROMs.
The cost of flash memory is considerably less than for EEPROM
Rapid erase and write times.
Use bulk erase operation in which all cells on the chip are erase
simultaneously
This bulk erase process typically requires hundreds of
milliseconds compares to 20 minutes for UV EPROMs
Any memory address location is as easily accessible as
any other.
Is used in computers for the temporary storage of
programs and data.
The contents of many RAM address locations will be
read from and written to as the computer executes a
program. This requires fast read and write cycle times
for the RAM so as not to slow down the computer
operation
Disadvantage – it is volatile and will lose all stored
information if power is interrupted or turned
off.(volatile)
Advantage- can be written into and read from rapidly
with equal ease
SRAM use bistable latching
circuitry for single bit storage.
Using BJT and MOS technology.
Advantage of BJT is high speed device.
Advantage of CMOS is high capacity
and low power consumption.
SRAM CELL
Can store data as long as power is applied to the
chip.
SRAM memory cells are essentially flip-flops that
will stay in a given state (store a bit) indefinitely
provide that power to the circuit is not interrupted.
Main applications of SRAM are used in various
electronic applications including toys,
automobiles, digital devices and computers.
Cip piawai industri
6116
2k x 8 bit (16 kilobit)
6164/6264
8k x 8 bit (64 kilobit)
43256/66256 32k x 8 bit (256
kilobit)
Pin description
A0 – An
connect to address bus
D0 – Dn
- bus line
connect to bus data
CS*
- Chip select atau CE* - Chip enable
to active device
OE*
- Output enable
RAM give data to data bus
WE*
- Write enable
to active write data bus
High capacity, low power requirement, moderate
operating speed.
DRAM stores 1s and 0s as charges on a small MOS
capacitor. Because of the tendency for these charges
to leak off after a period of time, DRAM require
periodic recharging or the memory cells; this called
refreshing the DRAM.
Have 4 times the density of SRAM
The main internal memory of the most personal
microcomputers uses a DRAM because of its high
capacity and low power consumption
DRAM - Operation principle
DRAM is usually arranged in a square array of one capacitor and transistor
per cell.
The illustrations to the right show a simple example with only 4 by 4 cells
(modern DRAM can be thousands of cells in length/width).
The long lines connecting each row are known as word lines. Each
column is actually composed of two bit lines, each one connected to every
other storage cell in the column.
Principle of operation of DRAM read, for simple 4 by 4 array.
1. FPM DRAM – Fast page mode DRAM
membolehkan data dicapai dengan cepat pada
‘page’ yang sama (beberapa alamat dalam julat
tertentu)
2. EDO DRAM – Extended Data Output DRAM
membaiki ciri FPM dari segi cara membaca
3. SDRAM – Synchronous DRAM mempunyai ciri
membaca data dengan lebih laju.
ROM : Read-Only Memory
Non-volatile (data retained even without power)
Exists on all computers
Functions on general-purpose computer: power-on self test, basic
input/output system (BIOS), monitor program, etc.
Functions on embedded systems: power-on self test, monitor
program, application program.
RAM : Random Access Memory
Volatile (data disappears without power)
Functions on general purpose computer: main memory for running
operating system and application program
Functions on embedded systems: scratch-pad memory
May not be required on very simple embedded systems
Introduction to address decoding
Full address decoding
Partial address decoding
Implementing address decoders
Examples
Memory Map and Address
Decoding
Different portions of memory are used for different purposes:
RAM, ROM, I/O devices
Even if all the memory was of one type, we still have to
implement it using different and unique addressing.
This means that for a given valid address, one and only one
memory-mapped component must be accessed.
Address decoding is the process of generating chip select
(CS*) signals from the address bus for each device in the
system
Contoh :
Let’s assume a very simple system like that:
> CPU 8 bit data bus line
> 16 bit address bus line
> 12 Kbyte ROM
> 4 Kbyte for I/O ports
> 16 Kbyte RAM
Make a sample memory map for that system.
SOLUTION
i.
What is the entire range for system addresses?
ii.
What is the entire range for every component
ROM, I/O and RAM
iii. Assume that
memory map figure like that:
ROM
I/O
RAM
unused
i.
System Size = 2n (n= address pin)
= 216
= 65536 Byte
Start Address = 0
End Address = 65536 – 1 (size -1)
= 65535
Range Address System :
0 -- 65535 OR 0000 - FFFF (hexadecimal)
Start
0000 H
ROM
I/O
RAM
unused
End
FFFF H
ii.
Range Address for ROM
Given Size of ROM
=
=
=
=
> Start Address for ROM
> End Address for ROM
12 Kbyte
12 x 1024 byte
12288 byte
3000 (hex)
= 0000
= 3000 – 1
= 2FFF
0000 H
2FFFF H
ROM
I/O
RAM
unused
FFFF H
iii.
Range Address for I/O
Given Size of I/O =
=
=
=
4 Kbyte
4 x 1024 byte
4096 byte
\$1000 (hex)
Start Address I/O = End Address for ROM + 1
= \$2FFF + 1
= \$3000
0000 H
End Address I/O = \$3000 + \$1000 – 1
= \$3FFF
2FFF H
ROM
3000 H
I/O
3FFF H
RAM
unused
FFFF H
iv.
Range Address for RAM
Given Size of RAM
Start Address for RAM
End Address for RAM
= 16 Kbyte
= 16 x 1024 byte
= 16384 byte
= \$4000
= End Address for I/O + 1
= \$3FFF + 1
= \$4000
= \$4000 + \$4000 – 1
= \$7FFF
0000 H
2FFF H
ROM
3000 H
I/O
3FFF H
4000 H
RAM
7FFF H
FFFF H
unused
Memory Map for the System is Figure below:
0000
ROM
2FFF
3000
I/O
3FFF
4000
RAM
7FFF
8000
UNUSED
FFFF
Latihan 1 :
1.
Lukiskan pemetaan alamat suatu sistem mikropemproses
Spesifikasi luaran adalah seperti berikut :
- EPROM bersaiz 2 MB bermula dari alamat \$000000
- RAM bersaiz 4MB berakhir di alamat \$7FFFFF
- I/O bersaiz 256 KB bermula dari alamat \$800000
68000.
Latihan 1 :
2.
Jika pemetaan alamat suatu mikropemproses 8 bit diberi seperti
berikut, tentukan saiz ROM, RAM dan I/O
0000
7FFF
RAM
8000
I/O
8FFF
9000
UNUSED
BFFF
C000
FFFF
ROM
Dalam suatu sistem komputer, terdapat beberapa
peranti yang berada di bawah kawalan pemproses.
Pada satu masa, pemproses hanya boleh bertukar
data atau berinteraksi hanya dengan satu peranti
sahaja.
Pemilihan peranti ditentukan oleh kedudukannya
dalam peta ingatan. Jadi penyahkod alamat
(address decoder) diperlukan bagi memilih
peranti yang hendak diaktifkan.
Untuk merekabentuk penyahkod alamat,
terdapat beberapa langkah iaitu :
i. Tentukan julat alamat untuk peranti
(rujuk peta ingatan)
ii. Bilangan cip yang diperlukan
iii. Bilangan talian alamat pada cip (talian
alamat rendah dari pemproses ke cip)
iv. Baki talian alamat masuk ke penyahkod
alamat
v. Lukis litar penyambungan antara komponenkomponen berkaitan.
Contoh :
Lukiskan sambungan penyahkod alamat bagi suatu sistem komputer yang
mempunyai kapasiti ingatan 256 x 4 bit. Diberi satu cip ingatan RAM 64 x
4 bit dan peta ingatan seperti berikut :
Peta Ingatan
00
RAM
FF
terdapat 256 ruang
alamat
1) Bilangan cip
= saiz sistem
saiz cip
= 256 x 4 =
4 cip RAM
64 x 4
2) Talian alamat sistem :
2n = 256
n = log 256 = 8 talian alamat sistem (A0 – A7)
log 2
3) Talian alamat cip :
2n = 64
n = log 64
log 2
= 6 talian iaitu A0 – A5
( talian alamat rendah dari pemproses terus ke cip ingatan )
4) Baki talian = 8 – 6 = 2 iaitu A6 dan A7
(masukan ke penyahkod alamat)
5. Penyambungan litar penyahkod alamat adalah seperti berikut :
D0-D3
Talian teratas A6 dan A7 disambungkan ke
penyahkod alamat. Litar penyahkod berfungsi
memilih satu daripada 4 cip ingatan tersebut.
Penyahkod yang digunakan adalah ‘ 2- line to 4
- line’
Setiap cip mempunyai CS input masing-masing.
Jika A6 dan A7 berlogik ‘0’,
maka pin CS0 akan aktif iaitu logic ‘0’ .
Ini bermakna RAM 1 akan dipilih.
Lokasi ingatan yang digunakan antara setiap cip
ingatan ditentukan oleh talian A0 hingga A5.
Disamping itu terdapat satu lagi kaedah untuk
menentukan talian yang masuk ke cip dan baki talian
yang masuk ke penyahkod alamat iaitu dengan merujuk
Alamat mula ROM 1 = \$ 00 0 0 0 0 0 0 0 0
Alamat akhir ROM 1 = \$ 3F 0 0 1 1 1 1 1 1
Mula dari kanan :
Bit yang bernilai 0 di alamat mula dan 1 di alamat akhir
pergi terus ke ingatan.
Di sini A0 hingga A5 pergi terus ke ingatan.
Baki talian iaitu A6 – A7 pergi ke penyahkod.
Perhatikan nilai-nilai A6 - A7 dimana setiap cip
mempunyai nilai yang berbeza dan nilai ini yang
menentukan cip yang akan diaktifkan.
1. Kirakan jumlah cip EPROM 27128 bersaiz 16K x
8 bit yang diperlukan bagi suatu sistem komputer
64K x 8 bit. Lukis sambungan penyahkod alamat
bagi sistem ingatan tersebut.
2. Kirakan jumlah cip RAM 2114 bersaiz 1024 x 4
bit yang diperlukan bagi suatu sistem komputer
3K x 8 bit. Lukis sambungan penyahkod alamat
bagi sistem ingatan tersebut.
```
17 cards
14 cards
11 cards
19 cards | HuggingFaceTB/finemath | |
# Circular motion question
1. Nov 10, 2005
### donjt81
This is the question...
A small wheel of radius 1.4cm drives a large wheel of radius 15cm by having their circumferences pressed together. If the small wheel turns at 407 rad/s, how fast does the larger one turn? Answer in rad/s
This is what I was thinking...
radius of smaller wheel = .014m
radius of larger wheel = .15m
circumference of smaller wheel = 2*pi*r = 2*3.14*.014 = .08792
angular velocity of smaller wheel (given) = 407 rad/s
angular velocity = circumference/time
time = circumference/angular velocity
=.08792/407 = .000216s
circumference of larger wheel = 2*pi*r = 2*3.14*.15 = .942
angular velocity = circumference/time
Does this approach look right?
2. Nov 10, 2005
### BerryBoy
I disagree... By intuition, you can predict that the larger wheel is going to turn more slowly.. Try another approach..
Hint: Consider the fact that the speeds of circumferences are equal.
Eq: speed = w x r
w = angular velocity
Does this help?
Sam
Last edited: Nov 10, 2005
3. Nov 10, 2005
### donjt81
You are right... the larger wheel should go slower.
but since the speed of smaller wheel is 407 rad/s wont the larger wheel speed be the same?
so is the answer to the problem 407 rad/s for the larger wheel? but that doesnt make sense because the larger wheel is supposed to go slower...
I am confused...
4. Nov 10, 2005
### BerryBoy
OK, so if the speed at the circumfrence is:
v = w x r (as I stated above).
If the wheels are in contact this speed is equal on both wheels (not the angular velocity). Therefore:
wsmall x rsmall = wlarge x rlarge
I can't give you anymore hints without doing it now.
Good Luck... Let me know what you get for an answer.
Sam
Last edited: Nov 10, 2005
5. Nov 10, 2005
### donjt81
ohh i got it. i was confused between angular velocity (w) and tangential velocity(v).
Hey another question. do you know if i did the right thing by converting the radius to (m) or should i have left it as (cm)?
6. Nov 10, 2005
### BerryBoy
You could have left it as cm because you're dividing by the other length (which should have the same units).
I usually convert thing to metres at the start though, its good practice in my experience.
Sam
7. Nov 10, 2005
### donjt81
ok thanks
8. Nov 10, 2005
### BerryBoy
Exactly what I got
9. Nov 10, 2005 | HuggingFaceTB/finemath | |
# How to factor equations with an exponent of 3
I know there's a formula somewhere, but how do you factor an equation with an exponent of three.
In my solution's manual it says: x^3 - x^2 + 11x - 6 = (x-1)(x-2)(x-3)
And i'm just trying to figure out how they got that.
Thank you.
dextercioby
Homework Helper
That's impossible
$$x^{3}-x^{2}+11x-6\neq (x-1)(x-2)(x-3)$$
Daniel.
Hurkyl
Staff Emeritus
Gold Member
Practice.
I find that looking for the roots of the equation is often the easiest... you should know a method that lets you identify all of the rational numbers that could possibly be the solution of a polynomial.
(At least polynomials whose coefficients are integers)
dextercioby
Homework Helper
You misstyped it
$$(x-1)(x-2)(x-3)=x^{3}-6x^{2}+11x-6$$
Look at the roots of the polyomial.U can see that 1 is a root.
Now u'll have to divide the ployonmial $x^{3}-6x^{2}+11x-6$ through $x-1$.
Daniel.
There is a way to factorize third degree polynomials by finding the roots :
Without restricting generality, a 3rd degree polynomial can always be put in the form $$p(x)=x^3+px+q$$
$$\textrm{let } x=r+s$$
$$\Rightarrow p(x)=r^3+s^3+(3rs+p)(r+s)+q=0$$
this can be solved by the trivial system of equations :
$$r^3+s^3=-q$$
$$3rs=-p$$
which can be transformed in a quadatic equation
$$y= \frac{2ax^3+bx^2-d}{3ax^2+2bx+c}$$
this recrusive equation will converge to the real roots of any 3rd degree polynum ($$ax^3+bx^2+cx+d=0$$)
at first pick a random number x and solve the equation then use the result as x to solve it again and so on untill you get a good enough approximation of the root.
Last edited:
This is how I learned it. Say you have some cubic (let's pick an easy one):
x^3+x^2-x-1 = (x^3-x) + (x^2-1) = x(x^2-1) + (x^2-1), and you can factor out an x^2-1 here: (x^2-1)(x+1).
Obviously this does not work for all expressions that are possibly factorable, but it is a good way to make a quick check. If you find that you can immediately do it this way, then that's great. Otherwise, try one of the other ways listed above.
I was taught to find one root of a cubic (by trial and error) and then divide that root (i.e. x-1) into the cubic to find what remains. The quadratic that is left can be factorised or be applied to the quadratic equation to find the other two roots.
Basically it is the factor theorm.
HallsofIvy
Homework Helper
If by "factor" you mean "factor into terms with integer coefficients", the "rational root theorem" is useful: if x= m/n is a rational root of the polynomial axn+ bxn-1+ ...+ cx+ d= 0 (where all coefficients are integers) then the numerator m is a factor of the constant term d and the denominator n is a factor of the leaing coefficient a". Of course, if x= m/n is a root, then (x-m/n) is a factor and so is (nx- m).
In the example given x3 - 6x2+ 11x - 6 , we know that any rational root must have denominator that divides 1 (and so is an integer) and m (the integer root) must divide 6: it must be 1, 2, or 3. Of course, we still need to check to see IF each IS a root. 1- 6+ 11- 6= 12-12= 0, 8- 24+ 22- 6= 30-30= 0, and 27- 54+ 33- 6= 60-60= 0. Yes, each is a root and so x3 - 6x2+ 11x - 6 = (x-1)(x-2)(x-3).
In the example x3 - x2 + 11x - 6 , any rational (really integer) roots MUST divide 6 and so must be 1, 2, or 3. But 1- 1+ 11- 6 is NOT 0,
8- 4+ 22- 6 is NOT 0, 27- 9+ 33- 6 is NOT 0 so 1, 2, 3, are NOT roots. Since those are the only possible integer (or rational) roots, x3 - x2 + 11x - 6 cannot be factored using only integer (or rational) coefficients.
Of course, if you can find 3 real roots (perhaps by using the cubic formula kleinwolf referred to) you can factor using real (irrational) coefficients. It might happen that the cubic does not have 3 real roots (it must have at least one). In that case you could factor into a product of a linear term and a quadratic term with real coefficients or into linear terms using complex coefficients.
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I'm trying to find the roots of an equation in the form of Ax^3 + Bx^2 + Cx + D
I've talked to other people at my school and they say to synthetically divide by a 1, 2, or 3 and see if it goes in evenly then factor the remaining equation.
Thanks for everyone's response.
^^ thats the way to do it (generally if you get this question in an exam or classwork then at least one of the roots will be a small integer - allowing you to use this method)
HallsofIvy
Homework Helper
Cafka said:
I'm trying to find the roots of an equation in the form of Ax^3 + Bx^2 + Cx + D
I've talked to other people at my school and they say to synthetically divide by a 1, 2, or 3 and see if it goes in evenly then factor the remaining equation.
Thanks for everyone's response.
For this particular problem, divide by 1, 2, 3 and "see if it goes in evenly". The reason for choosing 1, 2, 3 is, as I said before because they divide the constant term D= 6.
don't you have to look at -1, -2, and -3 also?
HallsofIvy
Homework Helper
Good point- thanks. -1, -2, and -3 are also factors of 6.
all those that said use a trial and error method or substitute 1,2,3 and ten factorise are mistaken ,wat if i give a polynomial with highpower and large quoeficients that has neither 1,2,3,4,5or any samll number as its factor watdo u do,and for u cafka dont bother trying 2 derive it urself cos u might not get it 4 a long time ,it took many mathematicians 2 continue from where one stopped so u see .gregmead u are right i bet they dont wanna see the formula,also a poly greater than 4 cant be resolved, that was proven by abel in his impossibility theorem.see ya
HallsofIvy | HuggingFaceTB/finemath | |
# Sequences and series
Consider the following sum: The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite sum? While at first it may seem difficult or impossible, we have certainly done something similar when we talked about one quantity getting “closer and closer” to a fixed quantity. Here we could ask whether, as we add more and more term, the sum gets closer and closer to some fixed value. That is, look at And so on, ask whether these values have a limit. It seems pretty clear that they do, namely 1. In fact, as we will see, it’s not hard to show that And then There is one place that you have accepted this notion of infinite sum without really thinking of it as a sum: For example, or Our first task, then, to investigate infinite sums, called series, is to investigate limits of sequences of numbers. That is, we officially call A series, while Is a sequence, and That is, the value of a series is the limit of a particular sequence.
11.1 Sequences
While the idea of sequence of numbers, a1,a2,a3,… is straightforward, it is useful to think of a sequence as a function. We have up until now dealt with functions whose domains are the real numbers, or a subset of the real numbers, like f(x) = sin x. A sequence is a function with domain the natural numbers N ={1,2,3,…} or the non-negative integers, ℤ 0 {0,1,2,3,…}. The range of the function is still allowed to be the real numbers; in symbols, we say that a sequence is a function f: ℕ→ ℝ . Sequences are written in a few different ways, all equivalent; these all mean the same thing:
a1, a2, a3 , As with functions on the real numbers, we will most often encounter sequences that can be expressed by a formula. We have already seen the sequence ai = f (i) = ,
And others are easy to come by: Frequently these formulas will make sense if thought of either as functions with ℝ or ℕ, though occasionally one will make sense only for integer values.Faced with a sequence we are interested in the limit When x is a real valued variable ; now we simply want to restrict the “input” values to be integers. No real difference is required in the definition of limit, except that we specify, perhaps implicitly, that the variable is an integer. Compare this definition to definition 4.10.2.
DEFINITION 11.1.1 Suppose that is a sequence. We say that = L if for every €>0 there is an N > 0 so that whenever n >N, If we say that the sequence converges, otherwise it diverges.
If f(i) defines a sequence, and f(x) makes sense, and , then it is clear that as well, but it is important to note that the converse of this statement is not true. For example, since it is clear that also that is, the numbers Get closer and closer to 0. Consider this, however: Let ƒ(n) = sin(nπ ). This is the sequence
Sin(0π ), sin(1π ),sin (2π ),sin (3 π), . .…. = 0,0,0,0,…..
Since sin(nπ )=0 when n is an integer. Thus But , when x is real, does not exist: as x gets bigger and bigger, the values sin(xπ ) do not get closer and closer to a single value, but take on all values between -1 and 1 over and over. In general whenever you want to know you should first attempt to compute , since if the latter exists it is also equal to the first limit. But if for some reason does not exist, it may still be true that exists, but you’ll have to figure out another way to compute it.
It is occasionally useful to think of the graph of a sequence. Since the function is defined only for integer values, the graph is just a sequence of dots. In figure 11.1.1 we see the graphs of two sequences and the graphs of the corresponding real functions.
Figure 11.1.1 Graphs of sequences and their corresponding real functions
Not surprisingly, the properties of limits of real functions translate into properties of sequences quite easily. Theorem 2.3.6 about limits becomes.
THEOREM 11.1.2
Suppose that and k is some constant.
Then Likewise the Squeeze Theorem (4.3.1) becomes.
THEOREM 11.1.3
Suppose that , for some N. If And a final useful fact:
THEOREM 11.1.4 This says simply that the size of ” an” gets close to zero if and only if “an” gets close to zero.
EXAMPLE 11.1.5
Determine whether converges or diverges. If it converges, compute the limit. Since this makes sense for real numbers we consider Thus the sequence converges to 1.
EXAMPLE 11.1.6
Determine whether converges or diverges. If it converges, compute the limit. We compute Using L’Hopital’s Rule. Thus the sequence converges to 0.
EXAMPLE 11.1.7
Determine whether converges or diverges. If it converges, compute the limit. This does not make sense for all real exponents, but the sequence is easy to understand: it is
1,-1,1,-1,1… And clearly diverges.
EXAMPLE 11.1.8
Determine whether converges or diverges. If it converges, compute the limit. We consider the sequence so by theorem 11.1.4 the sequence converges to 0
EXAMPLE 11.1.9
determine wether converges or diverges .If it converges compute the limite. since and we can use theorem 11.1 .3 EXAMPLE 11.1.10
A particularly common and useful sequence is , for various values of r. Some are quite easy to understand: If r = 1 the sequence converges to 1 since every term is 1, and likewise if r =0 the sequence converges to 0. If r =-1 this is the sequence of example 11.1.7 and diverges. If r> 1 or r< -1 the terms r n get large without limit, so the sequence diverges. If 0 < r < 1 then the sequence converges to 0.
If -1< r<0 then |r n| = |r |n and 0 < |r|< 1, so the sequence converges to 0, so also converges to 0. In summary, converges precisely when -1 < r< 1 in which case.
Sometime we will not be able to determine the limit of a sequence, but we still would like to know whether it converges. In some cases we can determine this even without being able to compute the limit.
A sequence is called increasing or something strictly increasing if for all i. It is called non-decreasing or sometimes (unfortunately) increasing if for all i. Similarly a sequence is decreasing if for all i and non-increasing if for all i. If a sequence has any of these properties it is called monotonic.
EXAMPLE 11.1.11 Is increasing and Is decreasing.
A sequence is bounded above if there is some number N such that for every n, and bounded below if there is some number N such that for every n. If a sequence is bounded above and bounded below it is bounded. If a sequence is increasing or non-decreasing it is bounded below (by ), and if it is decreasing or non-increasing it is bounded above (by ). Finally, with all this new terminology we can state an important theorem.
THEOREM 11.1 12
If a sequence is bounded and monotonic then it converges.
We will not prove this; the proof appears in many calculus books. It is not hard to believe: suppose that a sequence is increasing and bounded, so each term is larger than the one before, yet never larger than some fixed value N. The terms must then get closer and closer to some value between and N. It need not be N, since N may be a “too-generous” upper bound; the limit will be the smallest number that is above all of the terms .
EXAMPLE 11.1.13
All of the terms are less than 2, and the sequences is increasing. As we have seen, the limit of the sequence is 1, 1 is the smallest number that is bigger than all the terms in the sequence. Similarly, all of the terms (n+1)/n are bigger than ½, and the limit is 1 1 is the largest number that is smaller than the term of the sequence.
We don’t actually need to know that a sequence is monotonic to apply this theorem; it is enough to know that the sequence is “eventually” monotonic, that is, that at some point is becomes increasing or decreasing. For example, the sequence 10, 9, 8, 15, 3, 21, 4, 3/4, 7/8, 15/16, 31/32,… is not decreasing, because among the first few terms it is not. But starting with the term ¾ it is increasing, so the theorem tells us that the sequence 3/4, 7/8, 15/16, 31/32,… converges. Since convergece depends only on what happens as n gets large, adding a few terms at the beginning can’t turn a convergent sequence into a divergent one.
EXAMPLE 11.1.14
We first show that this sequence is decreasing, that is, that Consider the real function when We can compute the derivative, and note that when this is negative. Since the function has negative slope, when . Since all terms of the sequence are positive, the sequence is decreasing and bounded when and so the sequence converges. (As it happens, we can compute the limit in this case, but we know it converges even without knowing the limit; see exercise 1.)
EXAMPLE 11.1.15
Again we show that the sequences is decreasing, and since each term is positive the sequence converges. We can’t take the derivative this time, as x! Doesn’t make sense for x real. But we note that if then , which is what we want to know. So we look at : (Again it is possible to compute the limit; see exercise 2.)
11.2 SERIES
While much more can be said about sequences, we now turn to our principal interest, series. Recall that a series, roughly speaking, is the sum of a sequence: if is a sequence then the associated series is Associated with a series is a second sequence, called the sequence of partial sums
A series converges if the sequence of partial sums converges, and otherwise the series diverges.
EXAMPLE 11.2.14
Is called a geometric series. A typical partial sum is
.
We note that
So
If
Thus, when the geometric series converges to . When, for example, k=1 and x=1/2:
We began the chapter with the series namely, the geometric series without the first term 1. Each partial sum of this series is 1 less than the corresponding partial sum for the geometric series, so of course the limit is also one less than the value of the geometric series, that is,
It is not hard to see that the following theorem follows from theorem 11.1.2.
THEOREM 11.1.2
Suppose that and are convergent series, and c is a constant. Then
1. is convergent and
2. is convergent and
The two parts of this theorem are subtly different. Suppose that diverges; does also diverge if c is non-zero? Yes: suppose instead that converges; then by the theorem, converges, but this is the same as which by assumption diverges. Hence also diverges. Note that we are applying the theorem with replaced by and c replaced by (1/c).
Now suppose that and diverge; does also diverge? Now the answer is no: let and so certainly and diverge. But Of course, sometimes will also diverge, for example, if , then diverges.
In general, the sequence of partial sums is harder to understand and analyze than the sequence of terms , and it is difficult to determine whether series converge and if so to what. Sometimes things are relatively simple, starting with the following.
THEOREM 11.2.3
If converges then
Proof. Since converges, and , because this really says the same thing but “renumbers” the terms. By theorem 11.1.2,
But , so as desired
This theorem presents an easy divergence test: if given a series the limit does not exist or has a value other than zero, the series diverges. Note well that the converse is not true: if then the series does not necessarily converge.
EXAMPLE 11.2.4
Show that diverges.
We compute the limit:
Looking at the first few terms perhaps makes it clear that the series has no chance of converging:
will just get larger and larger; indeed, after a bit longer the series starts to look very much like…+1+1+1+1…, and of course if we add up enough 1’s we can make the sum as large as we desire.
EXAMPLE 11.2.5
Show that diverges.
Here the theorem does not apply: , so it looks like perhaps the series converges. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms you will find that so it might be reasonable to speculate that the series converges to something in the neighborhood of 10. But in fact the partial sums do go infinity; they just get big very, very slowly. Consider the following:
and so on. By swallowing up more and more terms we can always manage to add at least another ½ to the sum, and by adding enough of these we can make the partial sums as big as we like. In fact, it’s not hard to see from this pattern that so to make sure the sum is over 100, for example, we’d add up terms until we get to around , that is, about 4 terms. This series, , is called the harmonic series.
Exercises 11.2.
1 Explain why 2 Explain why
3 Explain why 4 Compute
5 Compute 6 Compute
7 Compute 8 Compute
9 Compute
11.3 THE INTEGRAL TEST
It is generally quite difficult, often impossible, to determine the value of a series exactly. In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem.
If all of the terms in a series are non-negative, then clearly the sequence of partial sums is non-decreasing. This means that if we can show that the sequence of partial sums is bounded, the series must converge. We know that if the series converges, the terms approach zero, but this does not mean that for every n. many useful and interesting series do have this property, however, and they are among the easiest to understand. Let’s look at an example.
EXAMPLE 11.3.1 Show that converges.
The terms are positive and decreasing, and since the terms 1/n2approach zero. We seek an upper bound for all the partial sums, that is, we want to find a number N so that sn N for every n. the upper bound is provided courtesy of integration, and is inherent in figure 11.3.1.
Figure 11.3.1 Graph of y=1/x2 with rectangles.
The figure shows the graph of y=1/x2 together with some rectangles that lie completely below the curve and that all have base length one. Because the heights of the rectangles are determined by the height of the curve, the areas of the rectangles are 1/12, 1/22, 1/32, and so on –in other words, exactly the terms of the series. The partial sum sn is simply the sum of the areas of the first n rectangles. Because the rectangles all lie between the curve 2any sum of rectangle areas is less than the area under the entire curve, that is, all the way to infinity, there is bit of trouble at the left end, where there is an asymptote, but we can work around that easily. Here it is:
Sn = dx < 1+ dx = 1+ ,
Recalling that we computed this improper integral in section 9.7. Since the sequence of partial sums sn is increasing and bounded above by 2, we know that n = L <2, and to show that L = 2/6 1.6.
We already know that diverges. What goes wrong if we try to apply this technique to it? Here’s the calculation:
Sn=
The problem is that the improper integral doesn’t converge. Note well that this does not prove that diverges, just that this particular calculation fails to prove that it converges. A slight modification, however, allows us to prove in a second way that diverges.
Example 11.3.2 consider a slightly altered version of figure 11.3.1, shown figure 11.3.2.
Figure 11.3.2 Graph of y =1/x with rectangles.
The rectangle this time are above the curve, that is, each rectangle completely contains the corresponding area under the curve. This means that
Sn =
As n gets bigger, In (n+1) goes to infinity, so the sequence of partial sums sn must also go to infinity, so the harmonic series diverges.
The important fact that clinches this example is that
Which we can rewrite as
So these two examples taken together indicate that we can prove that a series converges or prove that it diverges with a single calculation of an improper integral. This is known as the integral test, which we state as a theorem.
THEOREM 11.3.3 suppose that f (x) > 0 and is decreasing on the infinite interval [k, (for some k and that an = f (n). Then the series converges if and only if the improper integral converges.
The two examples we have seen are calls p-series is any series of the from np. if p , ,
So the series diverges. For positive values of p we can determine precisely which converge.
THEOREM 11.3.4 A p-series with p>0 converges if p> 1.
Proof. We use the integral test; we have already done p = 1, so assume that p
If p>1 then 1-p<0 and = 0, so the integral converges. If 0<p<1 then 1-p>0 and = , so the integral diverges.
EXAMPLE 11.3.5 Show that converges.
We could of course use the integral test, but now that we have the theorem we may simply note that this is a p-series with p .
EXAMPLE 11.3.6 Show that converges.
We know that if converges then also converges, by theorem 11.2.2. Since is a convergent p-series, converges also.
EXAMPLE 11.3.7 Show that diverges.
This also follows from theorem 11.2.2: Since is a p-series p= , it diverges, and so does
Since it is typically difficult to compute the value of a series exactly, a good approximation is frequently required. In a real sense, a good approximation is only as good as we know it is, that is, while an approximation may in fact be good, it is only valuable in practice if we can guarantee its accuracy to some degree. This guarantee is usually easy to come by for series with decreasing positive terms.
EXAMPLE 11.3.8 Approximate to two decimal places.
Referring to figure 11.3.1, if we approximate the sum by the error we make is the total area of the remaining rectangles, all of which lie under the curve from x=N out to infinity. So we know the true value of the series is larger than the approximation plus the area under the curve from N to infinity. Roughly, then, we need to find N so that
We can compute the integral: , so N=100 is a good starting point. Adding up the first 100 terms gives approximately 1.634983900, and that plus 1/100 is 1.644983900, so approximating the series by the value halfway between these will be at most 1/200=0.005 in error. The midpoint is 1.639983900, but while this is correct to 0.005, we can’t tell if the correct two-decimal approximation is 1.63 or 1.64. We need to make N big enough to reduce the guaranteed error, perhaps to around 0.004 to be safe, so we would need 1/N 0.008, or N=125. Now the sum of the first 125 terms is approximately 1.636965982, and that plus 0.008 is 1644965982 and the point halfway between them is 1.640965982.the true value is then 1.640965982 0.004, and all numbers is this range round to 1.64, so 1.64 is correct to two decimal places. We have mentioned that the true value of this series can be shown to be / 6 1.644934068 which rounds down to 1.64 (just barely) and is indeed below the upper bound of 1.644965982, again just barely. Frequently approximations will be even better than “guaranteed” accuracy, but not always, as this example demonstrates.
Exercises 11.3.
Determine whether each series converges or diverges.
1. 2.
3. 4.
5. 6.
7. 8.
9. Find an N so that is between +0.005.
10. Find an N so that is between and +10-4
11. Find an N so that is between and +0.005.
12. Find an N so that is between and + 0.005.
11.4 Alternating series
Next we consider series with both positive terms, but negative terms, but in a regular pattern: they alternate, as in the alternating harmonic series for example:
In this series the sizes of the terms decrease, that is, forms a decreasing sequence, but this is not required in an alternating series. As with positive term series, however, when the terms do have decreasing sizes it is easier to analyze the series, much easier, in fact, than positive term series. Consider pictorially what is going on in the alternating harmonic series, shown in figure 11.4.1. Because the sizes of the terms are decreasing, the partial sums , , , and so on, form a decreasing sequence that is bounded below by , so this sequence must converge. Likewise, the partial sums , , and so on, from an increasing sequence that is bounded above by , so this sequence also converges. Since all the even numbered partial sums are less than all the odd numbered ones, and since the “jumps” (that is, the terms) are getting smaller and smaller, the two sequences must converge to the same value, meaning the entire sequence of partial sums , , ,… converges as well.
Figure 11.4.1 The alternating harmonic series.
There’s nothing special about the alternating harmonic series-the same argument works for any alternating sequence with decreasing size terms. The alternating series test is worth calling a theorem.
THEOREM 11.4.1 Suppose that { } =1 is a non-increasing sequence of positive numbers and =0. Then the alternating series converges.
Proof. The odd numbered partial sums, , , , and so on, from a non-increasing sequence, because
= – + , since . This sequence is bounded blew by , so tit must converge, say = L. likewise, the partial sums , , , and so on, form a non-decreasing sequence that is bounded above by , so this sequence also converges, say = M. Since = 0 and
= + ,
L= = + +1) = + = M+0 =M,
So L= M, the two sequences of partial sums converge to the same limit, and this means the entire sequence of partial sums also converges to L.
Another useful fact is implicit in this discussion. Suppose that
And that we approximate L by a finite part of this sum, say
Because the terms are decreasing in size, we know that the true value of L must be between this approximation and the next one, that is, between
and
Depending on whether N is odd or even, the second will be larger or smaller than the first.
EXAMPLE 11.4.2 Approximate the alternating harmonic series to one decimal place. We need to go roughly to the point at which the next term to be added or subtracted is 1/10. Adding up the first nine and the first ten terms we get approximately 0.746 and 0.646. These are 1/10apart, but it is not clear how the correct value would be rounded. It turns out that we are able to settle the question by computing the sums of the first eleven and twelve terms, which give 0.737 and 0.653, so correct to one place the value is 0.7.
We have considered alternating series with first index 1, and in which the first term is positive, but a little thought shows this is not crucial. The same test applies to any similar series, such as , , , etc.
Exercises 11.4.
Determine whether the following series converge or diverge.
1. 2.
3. 4.
5. Approximate to two decimal places.
6. Approximate to two decimal places.
11.5 COMPARISON TESTS
As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones that we already understand.
EXAMPLE 11.5.1 Dose converge?
The obvious first approach, based on what we know, is the integral test. Unfortunately, we can’t compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a p-series, that is,
<
When n 3. Since adding up the terms 1/ doesn’t get “too big”, the new series “should” also converge. Let’s make this more precise.
The series converges if and only if converges-all we’ve done is dropped the initial term. We know that converges. Looking at two typical partial sums:
+ + + … + < + + + … + =
Since the p-series converges, say to L, and since the terms are positive, < L. since the terms of the new series are positive, the from an increasing sequence and < < L for all n. Hence the sequence { } is bounded and so converges.
Sometimes, even when the integral test applies, comparison to a known series is easier, so it’s generally a good idea to think about doing a comparison before doing the integral test.
EXAMPLE 11.5.2 Does converge?
We can’t apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example, however,
Because 1. Once again the partial sums are non-decreasing and bounded above by = L, so the new series converges.
Like the integral test, the comparison test can be used to show both convergence and divergence. In the case of the integral test, a single calculation will confirm whichever is the case. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly.
EXAMPLE 11.5.3 Does converge?
We observe that the -3 should have little effect compared to the inside the square root, and therefore guess that the terms are enough like 1/ = 1/n that the series should diverge. We attempt to show this by comparison to the harmonic series. We note that
> =
So that
= + + … + > + + … + =
Where is 1 less than the corresponding partial sum of the harmonic series (because we start at n=2 instead of n=1). Since = , = as well.
So the general approach is this: if you believe that a new series is convergent, attempt to find a convergent series whose terms are larger than the terms of the new series; if you believe that a new series is divergent, attempt to find a divergent, attempt to find a divergent series whose terms are smaller than the terms of the new series.
EXAMPLE 11.5.4 Does converge?
Just as in the last example, we guess that this is very much like the harmonic series and so diverges. Unfortunately, , so we can’t compare the series directly to the harmonic series. A little thought leads us to , so if diverges then the given series diverges. But since , theorem 11.2.2 implies that it does indeed diverge.
For reference we summarize the comparison test in a theorem.
THEOREM 11.5.5 Suppose that and are not-negative for all n and that when n N, for some N.
If converges, so does .
If diverges, so does .
Exercises 11.5
Determine whether the series converge or diverge.
11.6 ABSOLUTE CONVERGENCE
Roughly speaking there are two ways for a series to converge: As in the case of , the individual terms get small very quickly, so that the sum of all of them stays finite, or, as in the case of , the terms don’t get small fast enough ( diverges), but a mixture of positive and negative terms provides enough cancellation to keep the sum finite. You might guess from what we’ve seen that if the terms get small fast enough to do the job, then whether or not some terms are negative and some positive the series converges.
THEOREM 11.6.1 If converges, then converges.
Proof. Note that 0 so by the comparison test converges.
Now converges by theorem 11.2.2.
So given a series with both positive and negative terms, you should first ask whether converges. This may be an easier question to answer, because we have tests that apply specifically to terms with non-negative terms. If converges then you know that converges as well. If diverges then it still may be true that converges; you will have to do more work to decide the question. Another way to think of this result is: it is (potentially) easier for to converge than for to converge, because the latter series cannot take advantage of cancellation.
If converge we say that converges absolutely; to say that converges absolutely is to say that any cancellation that happens to come along is not really needed as the terms already get small so fast that convergence is guaranteed by that alone. If converges but does not, we say that converges conditionally. For example converges absolutely, while converges conditionally.
EXAMPLE 11.6.2 Does converge?
In example 11.5.2 we saw that converges, so the given series converges absolutely.
EXAMPLE 11.6.3 Does converge?
Taking the absolute value, diverges by comparison to , so if the series converges it does so conditionally. It is true that , so to apply the alternating series test we need to know whether the terms are decreasing. If we let f(x)=(3x+4)/( +3x+5) then f’(x)=-(, and it is not hard to see that this is negative for so the series is decreasing and by the alternating series test it converges.
Exercises 11.6
Determine whether each series converges absolutely, converges conditionally, or diverges.
4.
5.
6.
7.
8.
11.7 THE RATIO AND ROOT TESTS
Does the series converge? It is possible, but a bit unpleasant, to approach this with the integral test or the comparison test, but there is an easier way. Consider what happens as we move from one term to the next in this series:
The denominator goes up by a factor of 5, but the numerator goes up by much less: which is much less than when n is large, because is much less than . So we might guess that in the long run it begins to look as if each term is 1/5 of the previous term. We have seen series that behave like this: , a geometric series. So we might try comparing the given series to some variation of this geometric series. This is possible, but a bit messy. We can in effect do the same thing, but bypass most of the unpleasant work.
The key is to notice that .
This is really just what we noticed above, done a bit more officially: in the long run, each term is one fifth of the previous term. Now pick some number between 1/5 and 1, say 1/2.
Because , then when n is enough, say N for some N, and .
So and so on. The general form is So if we look at the series
Its terms are less than or equal to the terms of the sequence
.
So by the comparison test, converges, and this means that converges, since we’ve just added the fixed number .
Under what circumstances could we do this? What was crucial was that the limit of , say L, was less than 1 so that we could pick a value r so that L The fact that in our example) means that we can compare the series , and the fact that guarantees that converges. That’s really all that is required to make the argument work. We also made use of the fact that the terms of the series were positive; in general we simply consider the absolute values of the terms and we end up testing for absolute convergence.
THEOREM 11.7.1 The Ratio Test Suppose that . If the series converges absoluatnely, if the series diverges, and if L=1 this test gives no information.
Proof. The example above essentially proves the first part of this, if we simply replace 1/5 by L and 1/2 by r. Suppose that , and pick r so that . Then for , for some N,
and .
This implies that , but since this means that , which means also that . By the divergence test, the series diverges.
To see that we get no information when L=1, we need to exhibit two series with L=1, one that converges and one that diverges. It is easy to see that and do the job.
EXAMPLE 11.7.2 The ratio test is particularly useful for series involving the factorial function. Consider
Since , the series converges.
A similar argument, which we will not do, justifies a similar test that is occasion ally easier to apply.
THEOREM 11.7.3 The Root Test Suppose that 1/n = L. If L<1 the series converges absolutely, L>1 the series diverges, and if L =1 this test gives no information.
The proof of the root test is actually easier that of the ratio test, and is a good exercise.
EXEMPLE 11.7.4 Analyze
The ratio turns out to be a bit difficult on this series (try it). Using the root test:
1/n = = =0.
Since 0<1, the series converges.
The root test is frequently useful when n appears as an exponent in the general term of the series.
Exercises 11.7.
1. Compute for the series
2. Compute for the series
3. Compute 1/n for the series
4. Compute 1/n for the series
Determine whether the series converge.
1. n 6.
2. 8.
3. Prove theorem 11.7.3, the root test.
11.8 Power Series
Recall that we were able to analyze all geometric series “simultaneously” to discover that
=
If <1, and that the series diverges when 1. At the time, we thought of x as an unspecified constant, but we could just as well think of it as a variable, in which case the series
Is a function, namely, the function k/ (1-x), as long as <1. While k/ (1-x) is a reasonably easy function to deal with, the more complicated does have its attractions:
It appears to be an infinite version of one of the simplest function types — a polynomial.
This leads naturally to the questions: do other functions have representations as series?
Is there an advantage to viewing them in this way?
The geometric series has a special feature that makes it unlike a typical polynomial the coefficients of the powers of x are the same, namely k. We will need to allow more general coefficients if we are to get anything other than the geometric series.
DEFINITION 11.8.1 A power series has the form
With the understanding that may depend on n but not on x.
EXAMPLE 11.8.2 is a power series. We can investigate convergence using the ratio test:
Thus when < 1 the series converges and when > 1 it diverges, leaving only two values in doubt. When x = 1 the series the harmonic series and diverges: when x = -1 it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of as a function form the interval to the real numbers.
A bit of though reveals that the ratio test applied to a power series will always have the same nice form. In general, we will compute
Assuming that lim / exists. Then the series converges if <1 , that is, if , and diverges if
Only the two values require further investigation. Thus the series will definitely define a function on the interval (, and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if L=0 the limit is 0 no matter what value x takes, so the series converges for all x and the function is defined for all real numbers. If , then no matter what value x takes the limit is infinite and the series converges only when x=0. The value 1/L is called the radius of convergence of the series, and the interval on which the series converges is the interval of converge.
Consider again the geometric series, .
Whatever benefits there might be in using the series from of this function are only available to us when x is between -1 and 1. Frequently we can address this shortcoming by modifying the power series slightly. Consider this series:
Because this is just a geometric series with x replaced by (x+2)/3. Multiplying both sides by 1/3 gives , the same function as before. For what values of x does this series converges? Since it is a geometric series, we know that is converges when
So we have a series representation for 1/(1-x) that works on a larger interval than before, at the expense of a somewhat more complicated series. The endpoints of the interval of convergence now are -5 and 1, but note that they can be more compactly described as We say that 3 is the radius of convergence, and we now say that the series is centered at -2.
DEFINITION 11.8.3 A power series centered at a has the form with the understanding that may depend on n but not on x.
Exercises 11.8
Find the radius and interval of convergence for each series. In exercises 3 and 4, do not attempt to determine whether the endpoints are in the interval of convergence.
1. 4.
2. 5.
3. 6.
11.9 CALCULUS WITH POWER SERIES
Now we know that some function can be expressed as power series, which look like infinite polynomials. Since calculus, that is, computation of derivatives and antiderivatives, is easy for polynomials, the obvious question is whether the same is true for infinite series. The answer is yes.
THEOREM 11.9.1 Suppose the power series has radius of convergence R. Then and these two series have radius of convergence R as well.
EXAMPLE 11.9.2 Starting with the geometric series:
when
The series does not convergence when x=-1 or 1-x=2. The interval of convergence is , or 0 so we can use the series to represent when For example and so
Because this is an alternating series with decreasing terms, we know that the true value is between 909/2240 and 909/2240-1/2048=29053/71680 0,4053 so correct to two decimal places the value is 0,41.
What about Since 9/4 is larger than 2 we cannot use the series directly, but so in fact we get a lot more from this one calculation than first meets the eye. To estimate the true value accurately we actually need to be a bit more careful. When we multiply by two we know that the true value is between 0,8106 and 0,812, so rounded to two decimal places the true value is 0,81.
Exercises 11.9
1. Find a series representation for ln 2.
2. Find a power series representation for 1/
3. Find a power series representation for 2/
4. Find a power series representation for 1/ What is the radius of convergence?
5. Find a power series representation for
11.10 TAYLOR SERIES
We have seen that some function can be represented as series, which may give valuable information about the function. So far, we have seen only those examples that result from manipulation of our one fundamental example, the geometric series. We would like to start with a given function and produce a series to represent it, if possible.
Suppose that on some interval of convergence. Then we know that we can compute derivatives of f by taking derivatives of the terms of the series. Let’s look at the first few in general:
By examining these it’s not hard to discern the general pattern. The kth derivative must be | HuggingFaceTB/finemath | |
# ALEX Classroom Resource
## Circuit Analysis: Crash Course Physics #30
Classroom Resource Information
Title:
Circuit Analysis: Crash Course Physics #30
URL:
https://www.pbs.org/video/crash-course-physics-30/
Content Source:
PBS
Type: Audio/Video
Overview:
In this episode of Crash Course Physics, Dr. Shini walks us through the differences between series and parallel circuits and how that makes Christmas lights work the way they work.
Content Standard(s):
Science SC2015 (2015) Grade: 9-12 Physics 12 ) Use the principles of Ohm's and Kirchhoff's laws to design, construct, and analyze combination circuits using typical components (e.g., resistors, capacitors, diodes, sources of power). Unpacked Content Scientific And Engineering Practices:Analyzing and Interpreting DataCrosscutting Concepts: Cause and EffectDisciplinary Core Idea: Waves and Their Applications in Technologies for Information TransferEvidence Of Student Attainment:Students: Design combination circuits using typical components. Construct combination circuits using typical components. Analyze combination circuits using Ohm's and Kirchhoff's laws.Teacher Vocabulary:ammeter voltmeter series parallel model Kirchhoff's laws Ohm's law resistance current electric potential multimeter positive negative electrical components circuit voltage source conductors resistor color code circuit diagram heat charge static electricityKnowledge:Students know: The color code for the resistance of resistors. The basic principles of static electricity. How to construct electrical circuits. Several different components can be used to build an electrical circuit.Skills:Students are able to: Design and use models. Develop an appropriate experimental procedure. Create a data sheet. Collect and organize experimental data. Follow written and verbal instructions. Make measurements using standard units. Effectively manipulate laboratory equipment. Work safely in collaborative lab groups. Manipulate equations. Interpret graphical data. Solve mathematical equations. Use a multimeter.Understanding:Students understand that: Circuits are complete pathways through which current will flow predictably and will provide energy to the connected component(s). Circuits may be simple or complex.AMSTI Resources:ASIM Module: This standard could be met by minimal use of mathematics, however, the incorporation of more rigorous mathematics would deepen the students' knowledge. Circuit Basics; Ohm's Law; Resistors in Series; Resistors in Parallel; Kirchhoff's Rules.
Tags: circuit, circuit analysis, parallel circuit, physics, series circuit, voltage | HuggingFaceTB/finemath | |
# 7th Quiz Science Process Skills
15 Questions | Total Attempts: 295
Settings
• 1.
Of the following which is a qualitative observation
• A.
It is blue
• B.
It is 5 cm long
• C.
I like the bue color
• D.
It has a mass of 10.1g
• 2.
Of the following which is a quantitative observation
• A.
It is 2mm in length
• B.
It has a mass of 505 kg
• C.
It has a density 5.3 g/cm cubed
• D.
It is very soft
• 3.
To find the density of an object you will need to know the mass and the
• A.
Volume
• B.
Weight
• C.
Length
• D.
Grams
• 4.
The scientific method always begins with some sort of
• A.
• B.
An experiment
• C.
A hypothesis
• D.
Conclusions
• 5.
The traditional scientific method is basically linear starting with the Question and ending with the...
• A.
Conclusion
• B.
Experiment
• C.
Observations
• D.
Try agains
• 6.
In an experiment there are many factors that change, we refer to these as different types of
• A.
Variables
• B.
Questions
• C.
Experiments
• D.
• 7.
Making conversions within the metric system is easy. All you have to do is ...
• A.
Move the decimal the proper number of spaces
• B.
Multiply by the conversion factor
• C.
Subtract by the number of steps you take
• D.
Add the number of steps you take
• 8.
Using body parts for a measurement system is...
• A.
A bad idea because everyone is different sizes
• B.
A great idea because everyone carries measuring tools around with them
• C.
A great idea because it makes measuring personal
• D.
Still used by most countries in the world
• 9.
When measuring the amount of liquid in a graduated cylinder you should always
• A.
• B.
Hold it up to the light
• C.
Read it from the bottom up
• D.
• 10.
Which of the following is the base unit for length in the metric system
• A.
Meter
• B.
Liter
• C.
Gram
• D.
Inch
• 11.
0.81 g =_____ cg
• A.
81
• B.
8.1
• C.
0.0081
• D.
2.81
• 12.
4.6 Km = _____m
• A.
4,600
• B.
0.046
• C.
0.46
• D.
460
• 13.
218 cm = _____ Km
• A.
0.00218
• B.
21,800
• C.
0.218
• D.
2.18
• 14.
3.5 l = ______ ml
• A.
3,500
• B.
35
• C.
350
• D.
0.035
• 15.
The independent variable in an experiment is the variable that
• A.
You intentionally change
• B.
You can have 3 or 4 in an experiment
• C.
Responds to what you change
• D.
Stays the same no matter what
Related Topics | HuggingFaceTB/finemath | |
# Triangle Angle Word Problem 4
Taught by YourMathGal
• Currently 4.0/5 Stars.
5761 views | 1 rating
Part of video series
Meets NCTM Standards:
Lesson Summary:
In this lesson, Julie Harland teaches about the sum of angles in a triangle and provides word problems involving isosceles and right triangles. She explains that the angles in an isosceles triangle have the same measure, while in a right triangle, one angle is 90 degrees, making the other two angles complementary. Harland provides step-by-step instruction on how to solve word problems involving these types of triangles, including defining variables and writing equations to find the measure of each angle.
Lesson Description:
Sum of angles in a triangle, isosceles triangle, right triangle word problems involving angles in a triangle.
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
Questions answered by this video:
• What is a right triangle?
• What is special about a right triangle?
• If one angle in an isosceles triangle is 16 degrees more than twice the measure of one of the equal angles, what is the measure of each angle?
• If one angle measure in a right triangle is 38 degrees, what is the measure of the other angles?
• If the angles of a triangle are x, x, and 2x + 16, what does x equal, and what is the measure of each angle?
• What is an easy way of solving 4x + 16 = 180?
• If the angles of a triangle are 38, 90, and x, what does x equal?
• #### Staff Review
• Currently 4.0/5 Stars.
This lesson introduces right triangles and goes through two more problems where you need to write and solve an equation to find the measures of the angles in the triangle. All steps involved in finding the angles measures in the triangles are explained. | HuggingFaceTB/finemath | |
# Linear Regression Simulation Study
In general, linear regression is a linear approach of modelling the relationship of a numerical response (dependent) variable and one or more explanatory (independent) variables.
# Theories
## Model Assumptions
Let:
• $y_i \in \mathbb{R}^n$ as the measured response for the $i$th subject.
• $x_i = (1, x_{i1}, ..., x_{ip}) \in \mathbb{R}^{p+1}$, $i = 1, 2, ..., n$, be the values of the explanatory values for the $i$th subject. The first entry is used for the intercept term.
Using as transpose operator, the linear regression model assumes that:
\begin{align*} Y_{i} &= x_i'\beta + \epsilon_i,\ i = 1, ..., n\\ Y &= X \beta + \epsilon \end{align*}
where:
• $\beta = (\beta_0, \beta_1, ..., \beta_p)' \in \mathbb{R}^{p+1}$ is the unknown regression coefficient vector.
• Design matrix $X \in \mathbb{R}^{n\times {p+1}}$ have $x_i'$ at the $i$th row.
• $Y = (Y_1, ..., Y_n)'$ is a random vector.
• $\epsilon = (\epsilon_1, \epsilon_2, ..., \epsilon_n)' \stackrel{i.i.d.}{\sim} (0, \sigma^2)$, are i.i.d error terms from unknown random distribution with mean 0 and unknown fixed variance $\sigma^2$.
• $cov(\epsilon_i, \epsilon_i) = var (\epsilon_i) = \sigma^2$
• Due to i.i.d. assumption, $cov(\epsilon_i,\ \epsilon_j) = 0,\ i \neq j$
• Thus, the diagonal elements of $var(\epsilon)$ are equal to $\sigma^2$ and the off-diagonal elements are zero. $var(\epsilon) = \begin{bmatrix} \sigma^2 & 0 & \cdots & 0 \\ 0 & \sigma^2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sigma^2 \end{bmatrix}$
• We can write $var(\epsilon) = \sigma^2 I_n$, where $I_n$ is a $n \times n$ identity matrix, with ones on its diagonal and zero everywhere else.
Note that no assumptions are imposed onto the design matrix X: the values can be chosen by experimenter or they could be randomly realised from any distributions.
## Ordinary Least Square Estimator (OLS) of $\beta$
Assuming that $(X'X)^{-1}$ exists, that is $(X'X)$ is invertible, the ordinary least square estimator of $\beta$ is define by minimizing the error terms, the $\epsilon = Y - XB$. Specifically, we are to minimizing the squared error terms.
$\hat\beta = \underset{b\ \in\ \mathbb{R}^p}{\mathrm{argmin}}\ min(Y-Xb)'\ (Y- Xb)$
It can be shown that the solution is $\hat \beta = (X'X)^{-1} X'Y$ given that the $(X'X)^{-1}$ exists.
## Estimator of $\sigma^2$
An unbiased estimator of $\sigma^2$ is \begin{align*} \hat\sigma^2_{\epsilon} &= \frac{1}{n-p} \sum\limits_{i=1}^n \epsilon^2\\ &=\frac{1}{n-p} (Y-X\hat\beta)'\ (Y- X\hat\beta) \end{align*}
# Simulation
Let’s simulate the following linear model: $y_i = \beta_0 + \beta_1x_{i1} + \beta_2x_{i2} + \epsilon_i$ where $i = 1,2,...,100$, $\epsilon \sim N(0, \sigma^2)$, and $\sigma^2 = 4$
First, create the vector of Y. The parameters are set as $\beta =(10, 2, 0.5)'$; $n = 1000$; $\sigma = 2$
set.seed(001)
n = 1000; p = 2; sigma = 4;
# Betas
b0 = 10
b1 = 2
b2 = 0.5
beta = c(b0, b1,b2)
# Xs
x1 = runif(n, 0, 10)
x2 = rnorm(n, mean = 4, sd = 10)
xmat = cbind(rep(1,n), x1, x2)
# errors
eps = rnorm(n,0,sigma)
# calculate Y using linear model
y = xmat %*% beta + eps
#bind data
df_lm = as.data.frame(cbind(y,xmat))
colnames(df_lm) = c("y", "intercept", "x1", "x2")
head(df_lm)
## y intercept x1 x2
## 1 21.09686 1 2.655087 4.773031
## 2 14.25688 1 3.721239 1.031314
## 3 21.11518 1 5.728534 -7.832422
## 4 26.45658 1 9.082078 4.112927
## 5 23.14745 1 2.016819 13.916010
## 6 37.20973 1 8.983897 19.939675
## Visualize data points
Let’s visualize the data.
library(plotly)
pcolor = rep("purple", length(x1))
plot_ly(df_lm,
x = ~x1,
y = ~x2,
z = ~y,
type = "scatter3d",
mode = "markers",
marker = list(color = pcolor,
opacity = 0.4))
Figure 1: Data points generated from an underlying linear model.
## Estimating $\beta$ and $\epsilon$
The $\beta$s can be calculated through two ways: using formulas and using R function.
# Using formula
beta_form = qr.solve(crossprod(xmat)) %*% crossprod(xmat,y)
# USing r function
beta_r = lm.fit(x = xmat, y = y)$coefficients compare = cbind(beta_form, beta_r, beta) colnames(compare)= c("formula", "r_function", "actual") compare ## formula r_function actual ## 9.8791689 9.8791689 10.0 ## x1 2.0064254 2.0064254 2.0 ## x2 0.5127404 0.5127404 0.5 The calculated (estimated) value for the $\beta$s are similar to the one that was used to generate the model, that is 10, 2, and 0.5 for $\beta_0$, $\beta_1$ and $\beta_2$ respectively. The realization of $\hat\sigma^2_{\epsilon}$ is: eps_hat = 1/(n - p)*sum((y - xmat %*% beta_form)^2) eps_hat ## 17.76159 The original $\sigma^2$ is 16 and the realized value is 17.7615874. ## Visualize fitted plane So what’s is the fitted plane on the data points? Let’s visualize it. library(reshape2) # predict over sensible grid of values x1s <- seq(min(x1), max(x1), length.out = 100) x2s <- seq(min(x2), max(x2), length.out = 100) surface <- expand.grid(x1 = x1s,x2 = x2s,KEEP.OUT.ATTRS = F) # Making prediction surface$y <- apply(surface[,c("x1", "x2")], 1, function(x) cbind(1,x, x) %*% beta_form)
surface <- acast(surface, x2 ~ x1, value.var = "y")
pcolor = rep("purple", length(x1))
plot_ly(df_lm,
x = ~x1,
y = ~x2,
z = ~y,
type = "scatter3d",
mode = "markers",
marker = list(color = pcolor,
opacity = 0.7)) %>%
opacity = 0.6)
We can see that the plane fitted from $\hat\beta$ follows the data points pretty closely. However, note that Linear models are very rigid (i.e., a plane). | HuggingFaceTB/finemath | |
1. Express 30 minutes per day in simplest form. A. B. C. 1 to 24 D. 30:1 Hint 2. Eric uses 7 gallons of gas to drive 196 miles. How many miles can he drive using 1 gallon of gas? A. 30 B. 14 C. 28 D. 22 Hint 3. Solve . A. 10 B. 2.5 C. 40 D. 320 Hint 4. Katie is on a hiking trip. She wants to go from Panther's Peak to Coyote Caverns. On her map, the two sites are 4 inches apart and the scale is 1 inch = 0.5 mile. How far will she have to walk? A. 8 mi B. 6 mi C. 4 mi D. 2 mi Hint 5. Which of the following is not reasonable? A. The weight of a mouse is 0.60% the weight of a lion. B. The cost of gasoline is 120% of last year's cost. C. The number of household microwaves today are 0.95% of the microwaves in 1980. D. Last year's sales were 108% of this year's sales. Hint 6. What number is 60% of 80? A. 52 B. 75 C. 32 D. 48 Hint 7. Express the ratio 0.4 as a fraction in simplest form. A. B. C. D. Hint 8. Solve = . A. 12 B. 121.5 C. 6 D. 24 Hint 9. Express as a percent. Round to the nearest tenth. A. 26.3% B. 38.1% C. 2.38% D. 1.68% Hint 10. Brianna calculated her English grade to be exactly %. Write this percent as a fraction in simplest form. A. B. C. D. Hint 11. Find 45% of 360. A. 180 B. 162 C. 198 D. 0.08 Hint 12. 378 is 84% of what number? A. 438.48 B. 450 C. 462 D. 317.52 Hint 13. In the survey for which the results are shown below, if 57 people responded that they preferred orange candy, how many people participated in the survey? A. 108 B. 33 C. 300 D. 190 Hint 14. Which container of yogurt, described in the table, costs the least per ounce? A. 10 oz B. 12.5 oz C. 5.5 oz D. 8 oz Hint 15. Find the scale factor of a drawing if the scale is 1 inch = inch. A. B. 1 C. D. 8 Hint 16. The number of U.S. households with cable television increased 219% between 1981 and 2001. Write 219% as a decimal. A. 0.219 B. 21.9 C. 2.19 D. 21,900 Hint | HuggingFaceTB/finemath | |
# Definition:Natural Deduction
## Definition
Natural deduction is a technique for deducing valid sequents from other valid sequents by applying precisely defined proof rules, by a technique called logical inference.
As such, natural deduction forms a proof system, which is focused on practical applicability.
### Motivation
In its practical applicability, natural deduction differs from most proof systems in literature, which are more pedantic and formalistic, but therefore also more rigorous.
One may interpret natural deduction as the full-fledged proof system to actually use, once a formalistic alternative has been proved to satisfy all the rules of inference of natural deduction.
In doing so, it no longer matters which exact formalism was employed, and one can focus on the mathematical content itself.
On $\mathsf{Pr} \infty \mathsf{fWiki}$, we use natural deduction for both propositional logic and predicate logic.
Additionally, natural deduction may be applied not only for classical propositional logic, but also for more limited forms such as intuitionistic propositional logic, by employing the appropriate restrictions.
## Rules of Inference
The proof system called natural deduction deals exclusively with the notion of provable consequence.
As such, it does not contain any axioms.
Practically, this means that any proof of natural deduction will start with premises or the Rule of Assumption.
The complete list of rules of inference of natural deduction is as follows:
### Rule of Assumption
An assumption $\phi$ may be introduced at any stage of an argument.
### Rule of Conjunction
If we can conclude both $\phi$ and $\psi$, we may infer the compound statement $\phi \land \psi$.
### Rule of Simplification
$(1): \quad$ If we can conclude $\phi \land \psi$, then we may infer $\phi$.
$(2): \quad$ If we can conclude $\phi \land \psi$, then we may infer $\psi$.
### Rule of Addition
$(1): \quad$ If we can conclude $\phi$, then we may infer $\phi \lor \psi$.
$(2): \quad$ If we can conclude $\psi$, then we may infer $\phi \lor \psi$.
### Proof by Cases
If we can conclude $\phi \lor \psi$, and:
$(1): \quad$ By making the assumption $\phi$, we can conclude $\chi$
$(2): \quad$ By making the assumption $\psi$, we can conclude $\chi$
then we may infer $\chi$.
### Modus Ponendo Ponens
If we can conclude $\phi \implies \psi$, and we can also conclude $\phi$, then we may infer $\psi$.
### Modus Tollendo Tollens
If we can conclude $\phi \implies \psi$, and we can also conclude $\neg \psi$, then we may infer $\neg \phi$.
### Modus Tollendo Ponens
$(1): \quad$ If we can conclude $\phi \lor \psi$, and we can also conclude $\neg \phi$, then we may infer $\psi$.
$(2): \quad$ If we can conclude $\phi \lor \psi$, and we can also conclude $\neg \psi$, then we may infer $\phi$.
### Modus Ponendo Tollens
$(1): \quad$ If we can conclude $\map \neg {\phi \land \psi}$, and we can also conclude $\phi$, then we may infer $\neg \psi$.
$(2): \quad$ If we can conclude $\map \neg {\phi \land \psi}$, and we can also conclude $\psi$, then we may infer $\neg \phi$.
### Rule of Implication
If, by making an assumption $\phi$, we can conclude $\psi$ as a consequence, we may infer $\phi \implies \psi$.
The conclusion $\phi \implies \psi$ does not depend on the assumption $\phi$, which is thus discharged.
### Double Negation Introduction
If we can conclude $\phi$, then we may infer $\neg \neg \phi$.
### Double Negation Elimination
If we can conclude $\neg \neg \phi$, then we may infer $\phi$.
### Biconditional Introduction
If we can conclude both $\phi \implies \psi$ and $\psi \implies \phi$, then we may infer $\phi \iff \psi$.
### Biconditional Elimination
$(1): \quad$ If we can conclude $\phi \iff \psi$, then we may infer $\phi \implies \psi$.
$(2): \quad$ If we can conclude $\phi \iff \psi$, then we may infer $\psi \implies \phi$.
### Principle of Non-Contradiction
If we can conclude both $\phi$ and $\neg \phi$, we may infer a contradiction.
### Proof by Contradiction
If, by making an assumption $\phi$, we can infer a contradiction as a consequence, then we may infer $\neg \phi$.
The conclusion $\neg \phi$ does not depend upon the assumption $\phi$.
### Rule of Explosion
If a contradiction can be concluded, it is possible to infer any statement $\phi$.
### Law of Excluded Middle
$\phi \lor \neg \phi$ for all statements $\phi$.
### Reductio ad Absurdum
If, by making an assumption $\neg \phi$, we can infer a contradiction as a consequence, then we may infer $\phi$.
The conclusion $\phi$ does not depend upon the assumption $\neg \phi$.
## Derived Rules
In practically working with natural deduction, the following derived rules are useful.
### Rule of Substitution
Let $S$ be a sequent of propositional logic that has been proved.
Then we may infer any sequent $S'$ resulting from $S$ by substitutions for letters.
### Rule of Sequent Introduction
Let the statements $P_1, P_2, \ldots, P_n$ be conclusions in a proof, on various assumptions.
Let $P_1, P_2, \ldots, P_n \vdash Q$ be a sequent for which we already have a proof.
Then we may infer, at any stage of a proof (citing SI), the conclusion $Q$ of the sequent already proved.
This conclusion depends upon the pool of assumptions upon which $P_1, P_2, \ldots, P_n$ rest.
### Rule of Theorem Introduction
We may infer, at any stage of a proof (citing $\text {TI}$), a theorem already proved, together with a reference to the theorem that is being cited.
## Also known as
This technique is seen under various less precise names, for example decision procedure or decision method.
Some sources call it the axiomatic method.
## Also see
Certain schools of logic have investigated the situation of what happens when certain of the above proof rules (and their equivalents) are disallowed:
## Historical Note
The first system of rules for natural deduction was devised by Gerhard Gentzen in $1934$.
## Technical Note: Templates
In order to make the use of the proof rules of natural deduction in a tableau proof on $\mathsf{Pr} \infty \mathsf{fWiki}$, the following templates have been developed:
Template:Premise to invoke the Rule of Assumption for a premise Template:Assumption to invoke the Rule of Assumption for a non-premise assumption Template:Conjunction to invoke the Rule of Conjunction Template:Simplification to invoke the Rule of Simplification Template:Addition to invoke the Rule of Addition Template:ProofByCases to invoke Proof by Cases Template:ModusPonens to invoke Modus Ponendo Ponens Template:ModusTollens to invoke Modus Tollendo Tollens Template:ModusPonendoTollens to invoke Modus Ponendo Tollens Template:ModusTollendoPonens to invoke Modus Tollendo Ponens Template:Implication to invoke the Rule of Implication Template:DoubleNegIntro to invoke Double Negation Introduction Template:DoubleNegElimination to invoke Double Negation Elimination Template:BiconditionalIntro to invoke Biconditional Introduction Template:BiconditionalElimination to invoke Biconditional Elimination Template:NonContradiction to invoke the Principle of Non-Contradiction Template:Contradiction to invoke Proof by Contradiction Template:Explosion to invoke the Rule of Explosion Template:ExcludedMiddle to invoke the Law of Excluded Middle Template:Reductio to invoke Reductio ad Absurdum
For convenience, other templates are also available, for the following derived rules:
Template:Commutation to invoke the Rule of Commutation Template:DeMorgan to invoke an instance of De Morgan's Laws Template:Idempotence to invoke the Rule of Idempotence Template:IdentityLaw to invoke the Law of Identity
For the other general proof rules, there exist the following templates:
Template:SequentIntro to invoke the Rule of Sequent Introduction Template:TheoremIntro to invoke the Rule of Theorem Introduction Template:Substitution to invoke the Rule of Substitution | HuggingFaceTB/finemath | |
# POSITIVE AND NEGATIVE NUMBERS
## Presentation on theme: "POSITIVE AND NEGATIVE NUMBERS"— Presentation transcript:
POSITIVE AND NEGATIVE NUMBERS
LESSON 6
ADDING INTEGERS Positive + Positive = Positive ( +3) + (+2) = +5
When a number is positive, you do not have to use the (+) sign. (+3) + (+2) = 5
Negative + Negative = Negative (- 6) + (- 3) = - 9 When a number is NEGATIVE, you do have to use the (-) sign.
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 1: (- 6) + 3 = -3 COPY DOWN QUESTION
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 1: (- 6) + 3 = -3 COPY DOWN QUESTION 6 – 3 = 3 Subtract the numbers without negative signs.
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 1: (- 6) + 3 = -3 COPY DOWN QUESTION 6 – 3 = 3 Subtract the numbers without negative signs. = -3 Keep the sign of the larger number.
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 2: 9 + (-12) = - 3 COPY DOWN QUESTION
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 2: 9 + (-12) = - 3 COPY DOWN QUESTION 12 – 9 = 3 Subtract the numbers without negative signs.
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 2: 9 + (-12) = - 3 COPY DOWN QUESTION 12 – 9 = 3 Subtract the numbers without negative signs. = -3 Keep the sign of the larger number.
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 3: (- 5) + 7 = 2 COPY DOWN QUESTION
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 3: (- 5) + 7 = 2 COPY DOWN QUESTION 7 – 5 = 2 Subtract the numbers without negative signs.
Sum of a negative and a positive number - Keep the sign of the larger number and subtract EXAMPLE 3: (- 5) + 7 = 2 COPY DOWN QUESTION 7 – 5 = 2 Subtract the numbers without negative signs. = 2 Keep the sign of the larger number.
SUBTRACTING INTEGERS Negative - Positive = Negative
(same as adding two negative numbers) (- 8) - 3 = -8 + (-3) = -11 Another way of saying this: ADD THE OPPOSITE
SUBTRACTING INTEGERS Positive - Negative = Positive + Positive = Positive 4 - (-3) = = 7 Once again you are adding the opposite
SUBTRACTING INTEGERS Negative - Negative = Negative + Positive =
Keep the sign of the larger number and subtract (-7) - (-5) = ( -7) + 5 = -2 (-5) - ( -7) = (-5) + 7 = 2
TRY THESE ADD SUBTRACT 1) (+6) + (-2) 1) (7) – (2) 2) (+7) + 3
2) (+8) – (-2) 3) (-5) + (+2) 3) (-9) – (+3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) 2) (+7) + 3 2) (+8) – (-2) 3) (-5) + (+2) 3) (-9) – (+3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) 2) (+7) + 3 = 10 2) (+8) – (-2) 3) (-5) + (+2) 3) (-9) – (+3) 4) (-6) – (-2)
TRY THESE ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2)
2) (+7) + 3 = 10 2) (+8) – (-2) 3) (-5) + (+2) = -3 3) (-9) – (+3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) 3) (-5) + (+2) = -3 3) (-9) – (+3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) = (+8) + (+2) 3) (-5) + (+2) = -3 3) (-9) – (+3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) = (+8) + (+2) = 10 3) (-5) + (+2) = -3 3) (-9) – (+3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) = (+8) + (+2) = 10 3) (-5) + (+2) = -3 3) (-9) – (+3) = (-9) + (-3) 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) = (+8) + (+2) = 10 3) (-5) + (+2) = -3 3) (-9) – (+3) = (-9) + (-3) = - 12 4) (-6) – (-2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) = (+8) + (+2) = 10 3) (-5) + (+2) = -3 3) (-9) – (+3) = (-9) + (-3) = - 12 4) (-6) – (-2) = (-6) + (2)
ADD SUBTRACT 1) (+6) + (-2) = 4 1) (7) – (2) = 5 2) (+7) + 3 = 10 2) (+8) – (-2) = (+8) + (+2) = 10 3) (-5) + (+2) = -3 3) (-9) – (+3) = (-9) + (-3) = - 12 4) (-6) – (-2) = (-6) + (2) = -4 | HuggingFaceTB/finemath | |
# Is this graph correct?
## Homework Statement
Write the parametric and vector equations of:
x + 2 / 3 = y − 1 / 2 = z + 5/ -3
x=-1+5t
and plot a graph of this line when t = −2, −1, 0, 1 and 2.
## Homework Equations
vector:
r= [-2,1,-5]+t[3,2,-3]
parametric
x=3t-2
y=2t+1
z=-3t-5
using parametric equation:: i drew the graph, please check if the graph is correct at -2,-1,0,1,,2
## The Attempt at a Solution
#### Attachments
• 30.3 KB Views: 371
## Answers and Replies
Related Precalculus Mathematics Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member
## Homework Statement
Write the parametric and vector equations of:
x + 2 / 3 = y − 1 / 2 = z + 5/ -3
x=-1+5t
and plot a graph of this line when t = −2, −1, 0, 1 and 2.
## Homework Equations
vector:
r= [-2,1,-5]+t[3,2,-3]
parametric
x=3t-2
y=2t+1
z=-3t-5
using parametric equation:: i drew the graph, please check if the graph is correct at -2,-1,0,1,,2
## The Attempt at a Solution
Not sure where that x = -1+5t came from. But your parametric equations look correct. I didn't check all your points, but if you plotted the points to scale as apparently you did, and they came out on a straight line, aren't you pretty confident in the answers?
actually i was worried about the axis ... in specific the z axis.. i just had this gut feeling like i took -z the wrong direction ...
x=-1+5t is part of the question .
Mark44
Mentor
Write the parametric and vector equations of:
x + 2 / 3 = y − 1 / 2 = z + 5/ -3
Based on your subsequent work, your symmetric equations need parentheses.
Here are the equations, using LaTeX.
$$\frac{x + 2}{3} = \frac{y - 1}{2} = \frac{z + 5}{-3}$$
When you write them in plain text, they should look like this:
(x + 2)/3 = (y - 1)/2 = (z + 5)/(-3)
What you wrote means the following:
x + ##\frac{2}{3}## = y - ##\frac{1}{2}## = z + ##\frac{5}{-3}##
Dick
Homework Helper
x=-1+5t is part of the question .
Then doesn't it seem odd that you are saying x=3t-2 in the solution? I assume when you are writing "x + 2 / 3" you mean (x+2)/3, not x+(2/3). But even assuming that, your parametric form looks screwed up. Put x=(-1)+5t into (x+2)/3=(y-1)/2 and solve for y in terms of t. The 't' in your equations is the wrong parameter. Your are going to get the same line, but the points corresponding to a given value of t will be different.
Last edited:
hey , sorry for late response but if i put x=-1+5t in [x+2]/3= [y-1]/2=[z+5]/-3 what will i get after... how do i solve for y
SammyS
Staff Emeritus
Homework Helper
Gold Member
hey , sorry for late response but if i put x=-1+5t in [x+2]/3= [y-1]/2=[z+5]/-3 what will i get after... how do i solve for y
Are you asking how to solve
$\displaystyle \frac{(-1+5t)+2}{3}=\frac{y-1}{2}$
for y ?
Give it a try.
What do you get?
i didnt mean how i will solve it..rather i was concerned about what too do after i get the Y value
y =[5+10t]/3
SammyS
Staff Emeritus
Homework Helper
Gold Member
i didnt mean how i will solve it..rather i was concerned about what too do after i get the y value
y =[5+10t]/3
That depends.
What is it you're trying to do next?
HallsofIvy
Homework Helper
x=-1+5t is part of the question .
Then you have two different expessions for x a function of t. What are you supposed to use?
next i need to find vector and parametric equation and then plot for t ...|
LCKurtz
Homework Helper
Gold Member
This seems like a very strange way to phrase the problem, making me wonder what the exact statement of the problem is.
## Homework Statement
Write the parametric and vector equations of:
x + 2 / 3 = y − 1 / 2 = z + 5/ -3
x=-1+5t
and plot a graph of this line when t = −2, −1, 0, 1 and 2.
## Homework Equations
vector:
r= [-2,1,-5]+t[3,2,-3]
parametric
x=3t-2
y=2t+1
z=-3t-5
Taking the problem as you have quoted, don't use ##t## here. You apparently should write:
x=3s-2
y=2s+1
z=-3s-5
using s for your parameter. Now if x = -1+5t you can put that in and get s in terms of t. Then use that in these last three equations. Of course, this all assumes I understand what you are trying to do. | HuggingFaceTB/finemath | |
Dependent Sums and Products
I'm trying to understand the connections between a few different concepts fundamental to dependent type theory.
• Dependent functions ($$\Pi$$-types)
• Including non-dependent functions ($$A \rightarrow B$$)
• Dependent pairs ($$\Sigma$$-types)
• Including non-dependent products ($$A \times B$$)
• Coproducts ($$A + B$$)
Homotopy Type Theory says the following about $$\Sigma$$s and $$\Pi$$s:
[ $$\Sigma$$ ] is called a dependent pair type, or $$\Sigma$$-type, because in set theory it corresponds to an indexed sum (in the sense of a coproduct or disjoint union) over a given type.
The name "$$\Pi$$-type" is used because this type can also be regarded as the cartesian product over a given type
(I don't think I fully understand this point)
In a non-dependent setting I'm accustomed to calling coproducts "sums" because the number of inhabitants of a coproduct is the sum of its constituent types' inhabitants - $$|A+B| = |A| + |B|$$. Likewise I call pairs "products" because $$|(A,B)| = |A| \cdot |B|$$. Also functions can be called exponentials - $$|A \rightarrow B| = |B|^{|A|}$$!
Now for the question.
Why does it make sense in Type Theory to use $$\Sigma$$ for products and $$\Pi$$ for exponentials? It seems like everything is shifted between non-dependent and dependent types.
• sums approximately correspond to dependent coproducts
• products approximately correspond to dependent sums
• exponentials approximately correspond to dependent products
What's the deeper connection?
• Isn't this is in perfect analogy to usual arithmetic? $nm=\sum_{i=1}^mn$, $n^m=\prod_{i=1}^mn$. Sep 24 '14 at 9:29
• The accepted answer is an excellent explanation. I have seen papers that use the notation "dependent function space" and "dependent pair space" instead of "product" and "sum" to avoid this confusion, since the intuition isn't clear if you're not familiar with the underlying theory, particularly for dependently typed programming Oct 30 '21 at 19:20
I think what's confusing you is that $$A \times B$$ is both a product and a coproduct:
• It is the product of two factors, namely $$A$$ and $$B$$.
• It is the coproduct of $$A$$-many copies of $$B$$.
Once you realize this, you will see that we can obtain $$A \times B$$ as both a $$\sum$$ and a $$\prod$$:
1. Take $$P : \mathtt{bool} \to \mathsf{Type}$$ where $$P(\mathtt{false}) = A$$ and $$P(\mathtt{true}) = B$$. Then $$\sum_{b : \mathtt{bool}} P(b) \simeq A + B$$ and $$\prod_{b : \mathtt{bool}} P(b) \simeq A \times B$$
2. Take $$Q : A \to \mathsf{Type}$$ where $$Q(x) = B$$ for all $$x : A$$. Then $$\sum_{x : A} Q(x) \simeq A \times B$$ and $$\prod_{x : A} Q(x) \simeq (A \to B)$$
We should therefore not pay attention to $$A \times B$$ when deciding on a good naming scheme for these two constructs.
The dependent exponential is exactly a dependent product.
• Typo: should be "Q:A->Type where Q(x) = B for all x:A". Oct 29 '21 at 15:15
• Thanks, fixed it. You know, you can edit other people's answers around here. Oct 29 '21 at 20:58 | HuggingFaceTB/finemath | |
Question
Tue November 13, 2012 By:
# Prove that the general solution of cotx + tan(60-x)tan(60+x) = 0 is x= n?/3 + ?/4
Sun November 18, 2012
Cot x + tan(60-x).tan(60+x)=0
Using identity, tanx.tan(60-x).tan(60+x)=tan3x
We have:
cot x + tan 3x/tanx=0
1+tan3x=0
tan3x=-1
So general solution will be
3x= n? +3?/4, n belongs to I
x = n?/3+?/4
Related Questions
Wed February 01, 2017
# Q. Solve; 3-2cosx-4sinx-cos2x+sin2x=0.
Wed February 01, 2017
| HuggingFaceTB/finemath | |
# The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant
0 votes
217 views
The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3000. Find the number of winners, if the total number of participants is 63.
## 1 Answer
0 votes
by (24.8k points)
selected
Best answer
Let the number of winners be x. Therefore, the number of participants who did not win will be 63 − x.
Amount given to the winners = Rs (100 × x) = Rs 100x
Amount given to the participants who did not win = Rs [25(63 − x)] = Rs (1575 − 25x)
According to the given question,
100x + 1575 − 25x = 3000
On transposing 1575 to R.H.S, we obtain
75x = 3000 − 1575
75x = 1425
On dividing both sides by 75, we obtain
75x/75 = 1425/75
x = 19
Hence, number of winners = 19
0 votes
2 answers
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer | HuggingFaceTB/finemath | |
# How many distinct ways are there to $2$-color the $8$ vertices of a cube?
How many distinct ways are there to $2$-color the $8$ vertices of a cube, with colorings only considered distinct up to rotation?
• Probably easiest to count: e.g. $0$ red, $8$ blue: one way; $1$ red, $7$ blue: one way; $2$ red, $6$ blue: three ways; etc. Jul 13 '16 at 15:32
• This link here has many more examples of coloring polyhedra: MSE Meta. Jul 13 '16 at 20:19
You can use the Burnside's lemma. There are 24 rotations in the symmetry group of a cube; wikipedia page lists them all. The group acts on the set $X$ of vertex colorings, where $|X|=2^8=256$, so let's count their fixed points:
Now, apply the lemma to obtain the full number of orbits (that is, the number of colorings modulo rotations):
$$|X/G| = \frac{256+6\cdot 4 + 8 \cdot 16 + 3 \cdot 16 + 6 \cdot 16}{24}=23$$
This looks like a job for Burnside's lemma. It says that the number of distinct ways of coloring a cube is equal to $$|X/G| = \frac{1}{|G|}\sum_{g \in G}|X^g|$$ where
• $X$ is the full set of $2^8$ different colorings
• $G$ is the group of valid transformations (in our case the rotations of the cube)
• $X/G$ is the set of equivalence classes of colorings where each equivalence class contains colorings where one can be made into the other by applying an element of $G$.
• $X^g$ is the subset of $X$ containing the colorings that look identical after applying $g$.
We divide $G$ into the following subsets (following the order of this video for reference):
• One identity transformation: $|X^g| = 2^8$.
• Six $90^\circ$ rotations (one clockwise and one anticlockwise for each axis) around an axis that goes through a face (axes 1, 2, 3 in the video): $|X^g| = 2^2$.
• Three $180^\circ$ turns around the same axes: $|X^g| = 2^4$.
• Eight $120^\circ$ turns (one clockwise and one anticlockwise for each axis) around a major diagonal (axes 4, 5, 6, 7 in the video): $|X^g| = 2^4$
• Six $180^\circ$ rotations around an axis going through two edges (axes 8, 9, 10, 11, 12, 13): $|X^g| = 2^4$
Summing up, we get $$|X/G| = \frac{1}{24}(2^8 + 6\cdot 2^2 + 17\cdot 2^4) = 23$$ | HuggingFaceTB/finemath | |
kmmiles.com
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# 300512 km in miles
## Result
300512 km equals 186617.952 miles
You can also convert 300512 km to mph.
## Conversion formula
Multiply the amount of km by the conversion factor to get the result in miles:
300512 km × 0.621 = 186617.952 mi
## How to convert 300512 km to miles?
The conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:
1 km = 0.621 mi
To convert 300512 km into miles we have to multiply 300512 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:
1 km → 0.621 mi
300512 km → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 300512 km × 0.621 mi
L(mi) = 186617.952 mi
The final result is:
300512 km → 186617.952 mi
We conclude that 300512 km is equivalent to 186617.952 miles:
300512 km = 186617.952 miles
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case three hundred thousand five hundred twelve km is approximately one hundred eighty-six thousand six hundred seventeen point nine five two miles:
300512 km ≅ 186617.952 miles
## Conversion table
For quick reference purposes, below is the kilometers to miles conversion table:
kilometers (km) miles (mi)
300513 km 186618.573 miles
300514 km 186619.194 miles
300515 km 186619.815 miles
300516 km 186620.436 miles
300517 km 186621.057 miles
300518 km 186621.678 miles
300519 km 186622.299 miles
300520 km 186622.92 miles
300521 km 186623.541 miles
300522 km 186624.162 miles
## Units definitions
The units involved in this conversion are kilometers and miles. This is how they are defined:
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada. | HuggingFaceTB/finemath | |
In summary, the conversation is about finding the magnitude of ##I_{ab}## and the suggested methods include using the potential divider method and applying Kirchoff's Current Law (KCL). The expert advises to first find the voltage on a/b and then use KCL to calculate the currents through the resistors. Alternatively, one can directly apply KCL without finding the voltage first.
## Homework Statement
Please give me a clue, I don't understand, how to find the magnitude of ##I_{ab}##?
## The Attempt at a Solution
What have you tried so far? Are you familiar with KCL?
Have you worked out the voltage on a/b?
cnh1995 said:
What have you tried so far? Are you familiar with KCL?
I am stuck. What is KCL? Is it Kirchoff's Current Law?
CWatters said:
Have you worked out the voltage on a/b?
How?
I am stuck. What is KCL? Is it Kirchoff's Current Law?
Yes.
Have you worked out the voltage on a/b?
How?
"a" and "b" are at the same voltage because they are connected together. To work out that voltage you can simplify the circuit and use the potential divider method. There are other ways.
Once you have that voltage you can calculate the currents through all the resistors. Then apply KCL at either a or b.
or you can just wade in and apply KCL from the outset.
## 1. What is an electric circuit?
An electric circuit is a path through which electric current can flow. It is made up of a power source, conductors, and loads, which are devices that use the electric energy to perform functions.
## 2. How does an electric circuit work?
An electric circuit works by allowing electrons to flow from the negative terminal of the power source, through the conductors, and back to the positive terminal of the power source. This flow of electrons creates an electric current, which can power devices or perform other functions.
## 3. What are the different types of electric circuits?
There are several different types of electric circuits, including series circuits, parallel circuits, and combination circuits. Series circuits have only one path for the current to flow through, while parallel circuits have multiple paths. Combination circuits are a mix of series and parallel circuits.
## 4. What is the difference between AC and DC circuits?
AC (alternating current) circuits have a constantly changing direction of current flow, while DC (direct current) circuits have a constant direction of current flow. AC is typically used for long-distance power transmission, while DC is used for smaller devices and electronics.
## 5. How do you calculate voltage, current, and resistance in an electric circuit?
Voltage (V) is calculated by dividing the amount of electrical energy by the amount of charge. Current (I) is calculated by dividing the amount of charge by the time it takes for the charge to move. Resistance (R) is calculated by dividing the voltage by the current. | HuggingFaceTB/finemath | |
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A338087 a(n) is the smallest prime number which can be represented as x^2 + h*y^2 with x > 0 and y > 0 for each h in the first n Heegner numbers (A003173). 1
2, 17, 73, 193, 1873, 20353, 20353, 79633, 2333017 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The sequence lists prime numbers, in nondecreasing order, such that each of them can be written, in a unique way, in the form x^2 + h*y^2, where x, y are natural numbers, while h takes an increasing number of values of the sequence A003173 (Heegner numbers). See examples. LINKS Table of n, a(n) for n=1..9. EXAMPLE a(1) = 2 because, for A003173(1) = 1, 2 = 1^2+A003173(1)*1^2. a(2) = 17 because, considered the first two Heegner numbers, A003173(1) = 1 and A003173(2) = 2, 17 = 1^2+A003173(1)*4^2 = 3^2+A003173(2)*2^2. The prime 20353 is present in the sequence 2 times because: a(6) = 63^2+A003173(1)*128^2 = 79^2+A003173(2)*84^2 = 55^2+A003173(3)*76^2 = 65^2+A003173(4)*48^2 = 137^2+A003173(5)*12^2 = 97^2+A003173(6)*24^2, with Heegner numbers up to A003173(6)=19, and also: a(7) = 119^2+A003173(7)*12^2, with Heegner number A003173(7)=43. 2333017 is the last term of the sequence since for every Heegner number h there are x, y such that 2333017 = x^2 + h*y^2 and this is the least prime for which this is possible. For n=9, h in A003173 = {1,2,3,7,11,19,43,67,163}, a(9) = 2333017 = 989^2 +A003173(1)*1164^2 = 1493^2 +A003173(2)*228^2 = 1093^2 +A003173(3)*616^2 = 685^2 +A003173(4)*516^2 = 1349^2 +A003173(5)*216^2 = 179^2 +A003173(6)*348^2 = 1293^2 +A003173(7)*124^2 = 1395^2 +A003173(8)*76^2 = 1485^2 +A003173(9)*28^2. PROG (PARI) isok(p, u)={for(i=1, #u, my(s=qfbsolve(Qfb(1, 0, u[i]), p)); if(s==0 || s[1]==0, return(0))); 1} a(n)={my(u=[1, 2, 3, 7, 11, 19, 43, 67, 163][1..n]); forprime(p=2, oo, if(isok(p, u), return(p)))} vector(9, n, a(n)) \\ Andrew Howroyd, Nov 05 2020 CROSSREFS Cf. A003173. Sequence in context: A155715 A307851 A054568 * A235471 A297411 A183175 Adjacent sequences: A338084 A338085 A338086 * A338088 A338089 A338090 KEYWORD fini,full,nonn AUTHOR Marco Frigerio, Oct 09 2020 STATUS approved
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### B: Number Operations and Relationships
#### B.a: Concepts
B.a.1: Recognize and apply place-value concepts to whole numbers less than 1,000
B.a.2: Read, write, and represent numbers using words, numerals, pictures (e.g. base-ten blocks), number lines, , arrays, expanded forms (24=20+4) and symbolic renaming e.g., 24=30-6.
B.a.3: Compare and order whole numbers less than 1,000
B.a.6: Identify a fractional part of a collection/set.
B.a.7: Read, write and represent fractional parts of a whole e.g., 1/4, 1/2.
#### B.b: Computation
B.b.1: Use addition and subtraction in everyday situations and solve one-step word problems.
B.b.3: Demonstrate the concept of multiplication as grouping or repeated addition in context with products up to 50.
B.b.4: Demonstrate understanding of the concept of division as repeated subtraction, partitioning/sharing or measuring (dividend up to 30 and divisors up to 5).
B.b.5: Use fractions to represent quantities when solving problems involving equal sharing or partitioning.
B.b.6: Represent with shaded circles, rods, squares, pictorial representations of a whole.
B.b.7: Estimate sums to tens and hundreds and differences to ten.
### C: Geometry
#### C.b: Spatial Relationships and Transformations
C.b.2: Apply concepts of single-motion geometry (e.g., slides, flips and turns) to match two identical shapes.
#### C.c: Coordinate Systems
C.c.1: Use simple 2-dimensional coordinate systems to find locations on maps and to represent points and simple figures with coordinates of letters and numbers, (e.g., (E, 3)).
### D: Measurement
#### D.a: Measurable Attributes
D.a.1: Describe attributes of length, time and temperature and identify appropriate units to measure them. Units include: inches, feet, yards, centimeters, meters, seconds, minutes, hours, days, months, years and degrees Fahrenheit/Celsius.
#### D.b: Direct Measurement
D.b.1: Read and interpret measuring instruments to determine the measurement of objects with non-standard and standard units to the nearest centimeter or 1/2-inch.
D.b.4: Investigate measurements of area.
#### D.c: Indirect Measurement
D.c.1: Apply estimation techniques using non-standard units.
### E: Statistics and Probability
#### E.a: Data Analysis and Statistics
E.a.1: Answer and pose questions about collecting, organizing and displaying data. Work with data in the context of real-world situations by determining what data to collect and when and how to collect it to answer questions.
E.a.2: Collect, organize and display data in simple bar graphs and charts including translating data from one form to the other.
#### E.b: Probability
E.b.1: Determine if the occurrence of future events are more, less or equally likely to occur.
### F: Algebraic Relationships
#### F.a: Patterns, Relations and Functions
F.a.1: Recognize, extend, describe, create and replicate a variety of patterns including attribute, number and geometric patterns. Such as:
F.a.1.a: Picture patterns
F.a.1.b: Patterns in tables and charts
F.a.1.c: "What's-my-rule?" patterns
F.a.1.d: Patterns using addition and subtraction rules.
F.a.2: Focusing on relationships within patterns as well as extending patterns e.g., patterns and relationships represented with pictures, tables and charts, and "what's-my-rule?" patterns using addition and subtraction rules.
Correlation last revised: 5/31/2018
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. | HuggingFaceTB/finemath | |
## Engage NY Eureka Math 2nd Grade Module 5 Lesson 17 Answer Key
### Eureka Math Grade 2 Module 5 Lesson 17 Sprint Answer Key
A. Subtract Crossing the Ten 10 – 1 = 9, 10 – 2 = 8, 20 – 2 = 18, 40 – 2 = 38, 10 – 2 = 8, 11 – 2 = 9, 21 – 2 = 19, 51 – 2 = 49, 10 – 3 = 7, 11 – 3 = 8, 21 – 3 = 18, 61 – 3 = 58, 10 – 4 = 6, 11 – 4 = 7, 21 – 4 = 17, 71 – 4 = 67, 10 – 5 = 5, 11 – 5 = 6, 21 – 5 = 16, 81 – 5 = 76, 10 – 6 = 4, 11 – 6 = 5, 21 – 6 = 15, 91 – 6 = 85, 10 – 7 = 3, 11 – 7 = 4, 31 – 7 = 24, 10 – 8 = 2, 11 – 8 = 3, 41 – 8 = 33, 10 – 9 = 1, 11 – 9 = 2, 51 – 9 = 42, 12 – 3 = 9, 82 – 3 = 79, 13 – 5 = 8, 73 – 5 = 68, 14 – 6 = 8, 84 – 6 = 78, 15 – 8 = 7, 95 – 8 = 87, 16 – 7 = 9, 46 – 7 = 39, 68 – 9 = 59.
Explanation:
In the above-given question,
given that,
subtract crossing the ten.
10 – 1 = 9, 10 – 2 = 8, 20 – 2 = 18, 40 – 2 = 38, 10 – 2 = 8, 11 – 2 = 9, 21 – 2 = 19, 51 – 2 = 49, 10 – 3 = 7, 11 – 3 = 8, 21 – 3 = 18, 61 – 3 = 58, 10 – 4 = 6, 11 – 4 = 7, 21 – 4 = 17, 71 – 4 = 67, 10 – 5 = 5, 11 – 5 = 6, 21 – 5 = 16, 81 – 5 = 76, 10 – 6 = 4, 11 – 6 = 5, 21 – 6 = 15, 91 – 6 = 85, 10 – 7 = 3, 11 – 7 = 4, 31 – 7 = 24, 10 – 8 = 2, 11 – 8 = 3, 41 – 8 = 33, 10 – 9 = 1, 11 – 9 = 2, 51 – 9 = 42, 12 – 3 = 9, 82 – 3 = 79, 13 – 5 = 8, 73 – 5 = 68, 14 – 6 = 8, 84 – 6 = 78, 15 – 8 = 7, 95 – 8 = 87, 16 – 7 = 9, 46 – 7 = 39, 68 – 9 = 59. Question 1.
10 – 1 =
10 – 1 = 9.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 1 = 9.
9 + 1 = 10.
Question 2.
10 – 2 =
10 – 2 = 8.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 2 = 8.
8 + 2 = 10.
Question 3.
20 – 2 =
20 – 2 = 18.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
20 – 2 = 18.
18 + 2 = 20.
Question 4.
40 – 2 =
40 – 2 = 38.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
40 – 2 = 38.
38 + 2 = 40.
Question 5.
10 – 2 =
10 – 2 = 8.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 2 = 8.
8 + 2 = 10.
Question 6.
11 – 2 =
11 – 2 = 9.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 2 = 9.
9 + 2 = 11.
Question 7.
21 – 2 =
21 – 2 = 19.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
21 – 2 = 19.
19 + 2 = 21.
Question 8.
51 – 2 =
51 – 2 = 49.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
51 – 2 = 49.
49 + 2 = 51.
Question 9.
10 – 3 =
10 – 3 = 7.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 3 = 7.
7 + 3 = 10.
Question 10.
11 – 3 =
11 – 3 = 8.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 3 = 8.
8 + 3 = 11.
Question 11.
21 – 3 =
21 – 3 = 18.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
21 – 3 = 18.
18 + 3 = 21.
Question 12.
61 – 3 =
61 – 3 = 58.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
61 – 3 = 58.
58 + 3 = 61.
Question 13.
10 – 4 =
10 – 4 = 6.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 4 = 6.
6 + 4 = 10.
Question 14.
11 – 4 =
10 – 1 = 9.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 1 = 9.
9 + 1 = 10.
Question 15.
21 – 4 =
21 – 4 = 17.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
21 – 4 = 17.
17 + 4 = 21.
Question 16.
71 – 4 =
71 – 4 = 68.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
71 – 4 = 68.
68 + 4 = 71.
Question 17.
10 – 5 =
10 – 5 = 5.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 5 = 5.
5 + 5 = 10.
Question 18.
11 – 5 =
11 – 5 = 6.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 5 = 6.
6 + 5 = 11.
Question 19.
21 – 5 =
21 – 5 = 16.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
21 – 5 = 16.
16 + 5 = 21.
Question 20.
81 – 5 =
81 – 5 = 76.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
81 – 5 = 76.
76 + 5 = 81.
Question 21.
10 – 6 = 4.
10 – 6 = 4.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 6 = 4.
4 + 6 = 10.
Question 22.
11 – 6 =
11 – 6 = 5.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 6 = 5.
5 + 6 = 11.
Question 23.
21 – 6 =
21 – 6 = 15.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
21 – 6 = 15.
15 + 6 = 21.
Question 24.
91 – 6 =
91 – 6 = 85.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
91 – 6 = 85.
85 + 6 = 91.
Question 25.
10 – 7 =
10 – 7 = 3.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 7 = 3.
3 + 7 = 10.
Question 26.
11 – 7 =
11 – 7 = 4.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 7 = 4.
4 + 7 = 11.
Question 27.
31 – 7 =
31 – 7 = 25.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
31 – 7 = 25.
25 + 7 = 31.
Question 28.
10 – 8 =
10 – 8 = 2.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 8 = 2.
2 + 8 = 10.
Question 29.
11 – 8 =
11 – 8 = 3.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 8 = 3.
3 + 8 = 11.
Question 30.
41 – 8 =
41 – 8 = 33.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
41 – 8 = 33.
33 + 8 = 41.
Question 31.
10 – 9 =
10 – 9 = 1.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
10 – 9 = 1.
9 + 1 = 10.
Question 32.
11 – 9 =
11 – 9 = 2.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
11 – 9 = 2.
2 + 9 = 11.
Question 33.
51 – 9 =
51 – 9 = 42.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
51 – 9 = 42.
42 + 9 = 51.
Question 34.
12 – 3 =
12 – 3 = 9.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
12 – 3 = 9.
9 + 3 = 12.
Question 35.
82 – 3 =
82 – 3 = 79.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
82 – 3 = 79.
79 + 3 = 82.
Question 36.
13 – 5 =
13 – 5 = 8.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
13 – 5 = 8.
8 + 5 = 13.
Question 37.
73 – 5 =
73 – 5 = 68.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
73 – 5 = 68.
68 + 5 = 73.
Question 38.
14 – 6 =
14 – 6 = 8.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
14 – 6 = 8.
8 + 6 = 14.
Question 39.
84 – 6 =
84 – 6 = 78.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
84 – 6 = 78.
78 + 6 = 84.
Question 40.
15 – 8 =
15 – 8 = 7.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
15 – 8 = 7.
7 + 8 = 15.
Question 41.
95 – 8 =
95 – 8 = 87.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
95 – 8 = 87.
87 + 8 = 95.
Question 42.
16 – 7 =
16 – 7 = 9.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
16 – 7 = 9.
9 + 7 = 16.
Question 43.
46 – 7 =
46 – 7 = 9.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
46 – 7 = 9.
9 + 7 = 46.
Question 44.
68 – 9 =
68 – 9 = 57.
Explanation:
In the above-given question,
given that,
subtraction of two numbers.
68 – 9 = 57.
57 + 9 = 68.
B. Subtract Crossing the Ten 10 – 2 = 8, 20 – 2 = 18, 30 – 2 = 28, 50 – 2 = 48, 10 – 2 = 8, 11 – 2 = 9, 21 – 2 = 19, 61 – 2 = 59, 10 – 3 = 7, 11 – 3 = 8, 21 – 3 = 18, 71 – 3 = 68, 10 – 4 = 6, 11 – 4 = 7, 21 – 4 = 17, 81 – 4 = 77, 10 – 5 = 5, 11 – 5 = 6, 21 – 5 = 16, 91 – 5 = 86, 10 – 6 = 4, 11 – 6 = 5, 21 – 6 = 15, 41 – 6 = 35, 10 – 7 = 3, 11 – 7 = 4, 51 – 7 = 44, 10 – 8 = 2, 11 – 8 = 3, 61 – 8 = 53, 10 – 9 = 1, 11 – 9 = 3, 31 – 9 = 22, 12 – 3 = 9, 92 – 3 = 89, 13 – 5 = 8, 43 – 5 = 38, 14 – 6 = 8, 64 – 6 = 58, 15 – 8 = 7, 85 – 8 = 77, 16 – 7 = 9, 76 – 7 = 69, 58 – 9 = 49.
Explanation:
In the above-given question,
given that,
subtract crossing the ten.
10 – 2 = 8, 20 – 2 = 18, 30 – 2 = 28, 50 – 2 = 48, 10 – 2 = 8, 11 – 2 = 9, 21 – 2 = 19, 61 – 2 = 59, 10 – 3 = 7, 11 – 3 = 8, 21 – 3 = 18, 71 – 3 = 68, 10 – 4 = 6, 11 – 4 = 7, 21 – 4 = 17, 81 – 4 = 77, 10 – 5 = 5, 11 – 5 = 6, 21 – 5 = 16, 91 – 5 = 86, 10 – 6 = 4, 11 – 6 = 5, 21 – 6 = 15, 41 – 6 = 35, 10 – 7 = 3, 11 – 7 = 4, 51 – 7 = 44, 10 – 8 = 2, 11 – 8 = 3, 61 – 8 = 53, 10 – 9 = 1, 11 – 9 = 3, 31 – 9 = 22, 12 – 3 = 9, 92 – 3 = 89, 13 – 5 = 8, 43 – 5 = 38, 14 – 6 = 8, 64 – 6 = 58, 15 – 8 = 7, 85 – 8 = 77, 16 – 7 = 9, 76 – 7 = 69, 58 – 9 = 49. Question 1.
10 – 2 =
10 – 2 = 8.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 2 = 8.
8 + 2 = 10.
Question 2.
20 – 2 =
20 – 2 = 18.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
20 – 2 = 18.
18 + 2 = 20.
Question 3.
30 – 2 =
30 – 2 = 28.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
30 – 2 = 28.
28 + 2 = 30.
Question 4.
50 – 2 =
50 – 2 = 48.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
50 – 2 = 48.
48 + 2 = 50.
Question 5.
10 – 2 =
10 – 2 = 8.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 2 = 8.
8 + 2 = 10.
Question 6.
11 – 2 =
11 – 2 = 9.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 2 = 9.
9 + 2 = 11.
Question 7.
21 – 2 =
21 – 2 = 19.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
21 – 2 = 19.
19 + 2 = 21.
Question 8.
61 – 2 =
61 – 2 = 59.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
61 – 2 = 59.
59 + 2 = 61.
Question 9.
10 – 3 =
10 – 3 = 7.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 3 = 7.
7 + 3 = 10.
Question 10.
11 – 3 =
11 – 3 = 8.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 3 = 8.
8 + 3 = 11.
Question 11.
21 – 3 =
21 – 3 = 18.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
21 – 3 = 18.
18 + 3 = 21.
Question 12.
71 – 3 =
71 – 3 = 68.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
71 – 3 = 68.
68 + 3 = 71.
Question 13.
10 – 4 =
10 – 4 = 6.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 4 = 6.
6 + 4 = 10.
Question 14.
11 – 4 =
11 – 4 = 7.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 4 = 7.
7 + 4 = 11.
Question 15.
21 – 4 =
21 – 4 = 17.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
21 – 4 = 17.
17 + 4 = 21.
Question 16.
81 – 4 =
81 – 4 = 77.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
81 – 4 = 77.
77 + 4 = 81.
Question 17.
10 – 5 =
10 – 5 = 5.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 5 = 5.
5 + 5 = 10.
Question 18.
11 – 5 =
11 – 5 = 6.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 5 = 6.
6 + 5 = 11.
Question 19.
21 – 5 =
21 – 5 = 16.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
21 – 5 = 16.
16 + 5 = 21.
Question 20.
91 – 5 =
91 – 5 = 86.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
91 – 5 = 86.
86 + 5 = 91.
Question 21.
10 – 6 =
10 – 6 = 4.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 6 = 4.
4 + 6 = 10.
Question 22.
11 – 6 =
11 – 6 = 5.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 6 = 5.
5 + 6 = 11.
Question 23.
21 – 6 =
21 – 6 = 15.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
21 – 6 = 15.
15 + 6 = 21.
Question 24.
41 – 6 =
41 – 6 = 35.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
41 – 6 = 35.
35 + 6 = 41.
Question 25.
10 – 7 =
10 – 7 = 3.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 7 = 3.
3 + 7 = 10.
Question 26.
11 – 7 =
11 – 7 = 4.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 7 = 4.
4 + 7 = 11.
Question 27.
51 – 7 =
51 – 7 = 44.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
51 – 7 = 44.
44 + 7 = 51.
Question 28.
10 – 8 =
10 – 8 = 2.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 8 = 2.
2 + 8 = 10.
Question 29.
11 – 8 =
11 – 8 = 3.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 8 = 3.
3 + 8 = 11.
Question 30.
61 – 8 =
61 – 8 = 53.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
61 – 8 = 53.
53 + 8 = 61.
Question 31.
10 – 9 =
10 – 9 = 1.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 9 = 1.
1 + 9 = 10.
Question 32.
11 – 9 =
11 – 9 = 2.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
11 – 9 = 2.
2 + 8 = 11.
Question 33.
31 – 9 =
31 – 9 = 22.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
31 – 9 = 22.
22 + 9 = 31.
Question 34.
12 – 3 =
12 – 3 = 9.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
12 – 3 = 9.
9 + 3 = 12.
Question 35.
92 – 3 =
10 – 2 = 8.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
10 – 2 = 8.
8 + 2 = 10.
Question 36.
13 – 5 =
13 – 5 = 8.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
13 – 5 = 8.
8 + 5 = 13.
Question 37.
43 – 5 =
43 – 5 = 38.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
43 – 5 = 38.
38 + 5 = 43.
Question 38.
14 – 6 =
14 – 6 = 8.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
14 – 6 = 8.
8 + 6 = 14.
Question 39.
64 – 6 =
64 – 6 = 58.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
64 – 6 = 58.
58 + 6 = 64.
Question 40.
15 – 8 =
15 – 8 = 7.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
15 – 8 = 7.
7 + 8 = 15.
Question 41.
85 – 8 =
85 – 8 = 77.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
85 – 8 = 77.
77 + 8 = 85.
Question 42.
16 – 7 =
16 – 7 = 9.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
16 – 7 = 9.
9 + 7 = 16.
Question 43.
76 – 7 =
76 – 7 = 69.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
76 – 7 = 69.
69 + 7 = 76.
Question 44.
58 – 9 =
58 – 9 = 49.
Explanation:
In the above-given question,
given that,
subtraction crossing the ten.
58 – 9 = 49.
49 + 9 = 58.
### Eureka Math Grade 2 Module 5 Lesson 17 Problem Set Answer Key
Question 1.
Solve vertically or using mental math. Draw chips on the place value chart and unbundle, if needed.
a. 200 – 113 = __________ 200 – 113 = 87.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
200 – 113.
2 hundred = 200.
1 hundred = 100.
1 ten = 10.
3 ones = 3.
100 + 10 + 3 = 113.
200 – 113 = 87. b. 400 – 247 = __________ 400 – 247 = 153.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
4 hundred = 400.
2 hundred = 200.
4 tens = 40.
7 ones = 7.
200 + 40 + 7 = 247.
400 – 247 = 153. c. 700 – 428 = __________ 700 – 428 = 272.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
700 – 428.
7 hundred = 700.
4 hundred = 400.
2 ten = 20.
8 ones = 8.
400 + 20 + 8 = 428.
700 – 428 = 272. d. 800 – 606 = __________ 800 – 606 = 194.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
800 – 606.
8 hundred = 800.
6 hundred = 600.
6 ones = 6.
600 + 6 = 606.
800 – 606 = 194. e. 901 – 404 = __________ 901 – 404 = 497.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
901 – 404.
9 hundred = 900.
1 ones = 1.
900 + 1 = 901.
4 hundred = 400.
4 ones = 4.
400 + 4 = 404.
901 – 404 = 497. Question 2.
Solution: Check:
600 – 367 = 233.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
600 – 367.
6 hundred = 600.
3 hundred = 300.
6 tens = 60.
7 ones = 7.
300 + 60 + 7 = 367.
600 – 367 = 233. ### Eureka Math Grade 2 Module 5 Lesson 17 Exit Ticket Answer Key
Solve vertically or using mental math. Draw chips on the place value chart and unbundle, if needed.
Question 1.
600 – 432 = __________ 600 – 432 = 168.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
600 – 432.
6 hundred = 600.
4 hundred = 400.
3 tens = 30.
2 ones = 2.
400 + 30 + 2 = 432.
600 – 432 = 168. Question 2.
303 – 254 = __________ 303 – 254 = 46.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
303 – 254.
3 hundred = 300.
2 hundred = 200.
5 tens = 50.
4 ones = 4.
200 + 50 + 4 = 254.
303 – 254 = 46. ### Eureka Math Grade 2 Module 5 Lesson 17 Homework Answer Key
Question 1.
Solve vertically or using mental math. Draw chips on the place value chart and unbundle, if needed.
a. 200 – 123 = __________ 200 – 123 = 77.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
200 – 123.
2 hundred = 200.
1 hundred = 100.
2 tens = 20.
3 ones = 3.
100 + 20 + 3 = 123.
200 – 123 = 77. b. 400 – 219 = __________ 400 – 219 = 181.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
400 – 219.
4 hundred = 400.
2 hundred = 200.
1 ten = 10.
9 ones = 9.
200 + 10 + 9 = 219.
400 – 219 = 181. c. 700 – 542 = __________ 700 -542 = 158.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
700 – 542.
7 hundred = 700.
5 hundred = 500.
4 tens = 40.
2 ones = 2.
500 + 40 + 2 = 542.
700 – 542 = 158. d. 800 – 409 = __________ 800 – 409 = 301.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
800 – 409.
8 hundred = 800.
4 hundred = 400.
9 ones = 9.
400 + 9 = 409.
800 – 409 = 301. e. 905 – 606 = __________ 905 – 606 = 299.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
905 – 606.
9 hundred = 900.
5 ones = 5.
900 + 5 = 905.
6 hundred = 600.
6 ones = 6.
600 + 6 = 606.
905 – 606 = 299. Question 2.
Solution: Check:
800 – 567 = 233.
Explanation:
In the above-given question,
given that,
solve vertically or using mental math.
draw chips on the place value chart.
800 – 567.
8 hundred = 800.
5 hundred = 500.
6 tens = 60.
7 ones = 7.
500 + 60 + 7 = 567.
800 – 567 = 233. | HuggingFaceTB/finemath | |
# Square Matrix
In mathematics , a square matrix is a matrix with the same number of rows and columns. An n -by- n matrix is known as a square matrix of order . Any two square matrices of the same order can be added and multiplied. n
Square matrices are often used to represent simple linear transformations , such as shear or rotation . For example, if there is a square matrix representing a rotation ( rotation matrix ) and a column vector describing the position of a point in space , the product is a vector describing the position of that point after that rotation. and generates the column vector. If a line is a vector , the same can be obtained by using transformation , where is the transpose .
R\mathbf {v}{\displaystyle R\mathbf {v} }\mathbf {v}{\displaystyle \mathbf {v} R^{\mathsf {T}}}{\displaystyle R^{\mathsf {T}}}R
## Main diagonal
Entries ( i = 1,…, n ) form
the principal diagonal of a square matrix . They lie on the imaginary line that runs from the top left corner of the matrix to the bottom right corner. For example, the elements above the main diagonal of a 4 × 4 matrix are a 11 = 9 , a 22 = 11 , a 33 = 4 , a 44 = 10 .aii
The diagonal from top right to bottom left corner of a square matrix is called antidiagonal or counterdiagonal.
## Special kind
### Diagonal or Triangular Matrix
If all entries outside the main diagonal are zero, it is called a diagonal matrix . If all entries just above (or below) the main diagonal are zero, it is called an upper (or lower) triangular matrix . A
## Identity matrix
The identity matrix is of shape matrix in which all the elements on the main diagonal are equal to 1 and all other elements are equal to 0, e.g.
{\displaystyle I_{1}={\begin{bmatrix}1\end{bmatrix}},\ I_{2}={\begin{bmatrix}1&0\\0&1\end{bmatrix}},\ \ldots ,\ I_{n}={\begin{bmatrix}1&0&\cdots &0\\0&1&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &1\end{bmatrix}}.}
It is a square matrix of order , and also a special kind of diagonal matrix . It is called an identity matrix because multiplication with it leaves a matrix unchanged:n
ai n = i m a = a for any m- by- n matrix .A
### Inverse Matrix and its Inverse
A square matrix is said to be inverse or non-singular if the matrix exists such that AB
{\displaystyle AB=BA=I_{n}.}
If present, it is unique and can be defined as . is called the inverse matrix of , denoted .BAA-1
### Symmetric or oblique-symmetric matrix
A square matrix which is equal to its transpose, that is, , is a symmetric matrix . If instead , then is said to be a skew-symmetric matrix .
A{\displaystyle A^{\mathsf {T}}=A}{\displaystyle A^{\mathsf {T}}=-A}A
For a complex square matrix , often the appropriate analog of the transpose is the conjugate conjugate , defined as the transpose of the complex conjugate of . A complex square matrix satisfying is called a Hermitian matrix . If instead , then it is called a skew-Hermitian matrix .
A A^{*}AA{\displaystyle A^{*}=A}{\displaystyle A^{*}=-A}A
By the spectral theorem , the real symmetric (or Hermitian complex) matrix is an orthogonal (or unitary) eigenbasis ; i.e., each vector can be expressed as a linear combination of eigenvectors. In both cases, all eigenvalues are real.
### Fixed matrix
A symmetric n × n – matrix is said to be positive-definite (respectively negative-definite; indefinite) if for all non-zero vectors
x \in \mathbb{R}^n
The corresponding quadratic form given by q ( x ) = x
Takes only positive values (only negative values, respectively; both some negative and some positive values). If the quadratic form takes only non-negative (respectively only non-positive) values, then the symmetric matrix is said to be positive-semi-finite (respectively negative-semi-finite); Therefore the matrix is undecidable when it is neither positive-semi-finite nor negative-semi-finite.
A symmetric matrix is positive-definite if and only if all of its eigenvalues are positive. The table on the right shows two possibilities for a 2×2 matrix. Instead of allowing two distinct vectors as input, the bilinear form involving A is produced:
a ( x , y ) = y .
### Orthogonal matrix
An orthogonal matrix is a square matrix with real entries , whose columns and rows are orthogonal unit vectors (that is, orthonormal vectors). Equivalently , a matrix A is orthogonal if its transpose is equal to its inverse:
{\displaystyle A^{\textsf {T}}=A^{-1},}
which consists of
{\displaystyle A^{\textsf {T}}A=AA^{\textsf {T}}=I,}
where i is the identity matrix.
An orthogonal matrix A is essentially invertible (with inverse -1 = t ), unitary ( -1 = A * ), and normal ( A * A = AA * ). The determinant of any orthogonal matrix is either +1 or -1. There are n × n orthogonal matrices with the special orthogonal group determinant +1 . SO(n)
The complex analogous to an orthogonal matrix is a unitary matrix.
### Normal matrix
A real or complex square matrix is said to be normal if . If a real square matrix is symmetric, skew-symmetric, or orthogonal, then it is normal. If a complex square matrix is Hermitian, skew-Hermitian or unitary, then it is normal. General matrices are of interest mainly because they include only the types of matrices listed and form the broadest class of matrices for which spectral theorem occurs.
{\displaystyle A^{*}A=AA^{*}}
## Operation
### Scar
Trace, tr ( a of a square matrix) is the sum of its diagonal entries. While matrix multiplication is not commutative, the trace of a product of two matrices is independent of the order of the factors:
{\displaystyle \operatorname {tr} (AB)=\operatorname {tr} (BA).}
This is immediately followed by the definition of matrix multiplication:
{\displaystyle \operatorname {tr} (AB)=\sum _{i=1}^{m}\sum _{j=1}^{n}A_{ij}B_{ji}=\operatorname {tr} (BA).}
Also, the trace of a matrix is equal to its transpose, i.e.,
{\displaystyle \operatorname {tr} (A)=\operatorname {tr} (A^{\mathrm {T} }).}
### Proven
The provenance of a square matrix is a number encoding some properties of the matrix. A matrix is invertible if and only if its determinant is non-zero. Its absolute value is equal to the area (in ) or volume (in . ) of the unit square (or cube) image, while its sign corresponds to the orientation of the corresponding linear map: the determinant is positive if and only if the orientation is preserved. Is.
\ det (A)|A|A\mathbb {R} ^{2}\mathbb {R} ^{3}
The determinant of a 2×2 matrix is given by?
{\displaystyle \det {\begin{bmatrix}a&b\\c&d\end{bmatrix}}=ad-bc.}
The determinant of 3×3 matrices contains 6 terms (Saras’ rule). The longer Leibniz formula generalizes these two formulas to all dimensions.
The determinant of the product of square matrices is equal to the product of their determinants:
{\displaystyle \det (AB) = \det (A) \cdot \det (B)}
Adding a multiple of any row to another row, or a multiple of a column to another column, does not change the determinant. Interchanging two rows or two columns affects the determinant by multiplying it by -1. Using these operations, any matrix can be transformed into a lower (or upper) triangular matrix, and the determinant for such a matrix is equal to the product of the entries on the main diagonal; It provides a method to calculate the determinant of any matrix. Finally, the Laplace expansion expresses the determinant in terms of minors, that is, the determinant of the smaller matrix. This extension can be used for the recursive definition of determinants (taking the initial case as the determinant of a 1×1 matrix, which is its unique entry, or also the determinant of a 0×0 matrix, which is 1. ), which can be considered equivalent to the Leibniz formula. Determinants can be used to solve linear systems using Cramer’s law, where the division of the determinants of two related square matrices is equal to the value of each variable in the system.
### Eigenvalues and eigenvectors
A number and a non-zero vector satisfying V
{\displaystyle A\mathbf {v} =\lambda \mathbf {v} }
are called the eigenvalue and eigenvector of , respectively. The number is an eigenvalue of an n × n -matrix A if and only if A – n is not invertible, which is equal to A
{\displaystyle \det(A-\lambda I)=0.}
The polynomial A in an indefinite x determinant given by evaluation of det ( XI n – a ) is said to be characteristic of the polynomial a . It is a monic polynomial of degree n . Therefore the polynomial equation A (λ) = 0 has at most n different solutions, that is, the eigenvalues of the matrix. They can be complex, even if A ‘s entries are real. According to Kelly Hamilton’s theorem, a ( a ) = 0, that is, its characteristic polynomial yields the result of substituting the matrix itself into the zero matrix.
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# Homotopy Type Theory iterated differentiable function > history (Rev #3, changes)
Showing changes from revision #2 to #3: Added | Removed | Changed
# Contents
## Definition
For a calculus field? $F$ and a subtype $S \subseteq F$, the types of functions $S \to F$, pointwise continuous functions $C^0(S, F)$, and differentiable functions $D^1(S, F)$ between $S$ and $F$ are subtypes of the type of functions $S \to F$. For the Newton-Leibniz operator $\tilde{D}: D^1(S, F) \to (S \to F)$, the type of $n$-fold iterated differentiable functions between $S$ and $F$ is the $n$-fold iterated inverse image of $S \to F$ under the Newton-Leibniz operator:
$D^0(S, F) := \tilde{D}^{-0}(S \to F) := S \to F$
$n:\mathbb{N} \vdash D^{n+1}(S, F) := \tilde{D}^{-(n+1)}(S \to F) := \sum_{f:D^1(S, F)} \Vert\sum_{g:D^{n}(S, F)} \tilde{D}(f) = g\Vert$
and the type of pointwise continuously $n$-fold iterated differentiable functions between $S$ and $F$ is the $n$-fold iterated inverse images of $C^0(S, F)$ under the Newton-Leibniz operator:
$C^0(S, F) := \tilde{D}^{-0}(C^0(S, F)) := C^0(S, F)$
$n:\mathbb{N} \vdash C^{n+1}(S, F) := \tilde{D}^{-(n+1)}(C^0(S, F)) := \sum_{f:D^1(S, F)} \Vert\sum_{g:C^{n}(S, F)} \tilde{D}(f) = g\Vert$ | HuggingFaceTB/finemath | |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# Exercises for Limits and Continuity
$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
## 13.2: Limits and Continuity
1) Use the limit laws for functions of two variables to evaluate each limit below, given that $$\displaystyle \lim_{(x,y)→(a,b)}f(x,y) = 5$$ and $$\displaystyle \lim_{(x,y)→(a,b)}g(x,y) = 2$$.
1. $$\displaystyle \lim_{(x,y)→(a,b)}\left[f(x,y) + g(x,y)\right]$$
2. $$\displaystyle \lim_{(x,y)→(a,b)}\left[f(x,y) g(x,y)\right]$$
3. $$\displaystyle \lim_{(x,y)→(a,b)}\left[ \dfrac{7f(x,y)}{g(x,y)}\right]$$
4. $$\displaystyle \lim_{(x,y)→(a,b)}\left[\dfrac{2f(x,y) - 4g(x,y)}{f(x,y) - g(x,y)}\right]$$
1. $$\displaystyle \lim_{(x,y)→(a,b)}\left[f(x,y) + g(x,y)\right] = \displaystyle \lim_{(x,y)→(a,b)}f(x,y) + \displaystyle \lim_{(x,y)→(a,b)}g(x,y)= 5 + 2 = 7$$
2. $$\displaystyle \lim_{(x,y)→(a,b)}\left[f(x,y) g(x,y)\right] =\left(\displaystyle \lim_{(x,y)→(a,b)}f(x,y)\right) \left(\displaystyle \lim_{(x,y)→(a,b)}g(x,y)\right) = 5(2) = 10$$
3. $$\displaystyle \lim_{(x,y)→(a,b)}\left[ \dfrac{7f(x,y)}{g(x,y)}\right] = \frac{7\left(\displaystyle \lim_{(x,y)→(a,b)}f(x,y)\right)}\lim_{(x,y)→(a,b)}g(x,y)}=\frac{7(5)}{2} = 17.$$
4. $$\displaystyle \lim_{(x,y)→(a,b)}\left[\dfrac{2f(x,y) - 4g(x,y)}{f(x,y) - g(x,y)}\right] = \frac{2\left(\displaystyle \lim_{(x,y)→(a,b)}f(x,y)\right) - 4 \left(\displaystyle \lim_{(x,y)→(a,b)}g(x,y)\right)}\lim_{(x,y)→(a,b)}f(x,y) - \displaystyle \lim_{(x,y)→(a,b)}g(x,y)}= \frac{2(5) - 4(2)}{5 - 2} = \frac{2}{3$$
In exercises 2 - 4, find the limit of the function.
2) $$\displaystyle \lim_{(x,y)→(1,2)}x$$
3) $$\displaystyle \lim_{(x,y)→(1,2)}\frac{5x^2y}{x^2+y^2}$$
$$\displaystyle \lim_{(x,y)→(1,2)}\frac{5x^2y}{x^2+y^2} = 2$$
4) Show that the limit $$\displaystyle \lim_{(x,y)→(0,0)}\frac{5x^2y}{x^2+y^2}$$ exists and is the same along the paths: $$y$$-axis and $$x$$-axis, and along $$y=x$$.
In exercises 5 - 19, evaluate the limits at the indicated values of $$x$$ and $$y$$. If the limit does not exist, state this and explain why the limit does not exist.
5) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{4x^2+10y^2+4}{4x^2−10y^2+6}$$
$$\displaystyle \lim_{(x,y)→(0,0)}\frac{4x^2+10y^2+4}{4x^2−10y^2+6} = \frac{2}{3}$$
6) $$\displaystyle \lim_{(x,y)→(11,13)}\sqrt{\frac{1}{xy}}$$
7) $$\displaystyle \lim_{(x,y)→(0,1)}\frac{y^2\sin x}{x}$$
$$\displaystyle \lim_{(x,y)→(0,1)}\frac{y^2\sin x}{x} = 1$$
8) $$\displaystyle \lim_{(x,y)→(0,0)}\sin(\frac{x^8+y^7}{x−y+10})$$
9) $$\displaystyle \lim_{(x,y)→(π/4,1)}\frac{y\tan x}{y+1}$$
$$\displaystyle \lim_{(x,y)→(π/4,1)}\frac{y\tan x}{y+1}=\frac{1}{2}$$
10) $$\displaystyle \lim_{(x,y)→(0,π/4)}\frac{\sec x+2}{3x−\tan y}$$
11) $$\displaystyle \lim_{(x,y)→(2,5)}(\frac{1}{x}−\frac{5}{y})$$
$$\displaystyle \lim_{(x,y)→(2,5)}(\frac{1}{x}−\frac{5}{y}) = −\frac{1}{2}$$
12) $$\displaystyle \lim_{(x,y)→(4,4)}x\ln y$$
13) $$\displaystyle \lim_{(x,y)→(4,4)}e^{−x^2−y^2}$$
$$\displaystyle \lim_{(x,y)→(4,4)}e^{−x^2−y^2} = e^{−32}$$
14) $$\displaystyle \lim_{(x,y)→(0,0)}\sqrt{9−x^2−y^2}$$
15) $$\displaystyle \lim_{(x,y)→(1,2)}(x^2y^3−x^3y^2+3x+2y)$$
$$\displaystyle \lim_{(x,y)→(1,2)}(x^2y^3−x^3y^2+3x+2y) = 11$$
16) $$\displaystyle \lim_{(x,y)→(π,π)}x\sin(\frac{x+y}{4})$$
17) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{xy+1}{x^2+y^2+1}$$
$$\displaystyle \lim_{(x,y)→(0,0)}\frac{xy+1}{x^2+y^2+1} = 1$$
18) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}−1}$$
19) $$\displaystyle \lim_{(x,y)→(0,0)}\ln(x^2+y^2)$$
The limit does not exist because when $$x$$ and $$y$$ both approach zero, the function approaches $$\ln 0$$, which is undefined (approaches negative infinity).
In exercises 20 - 21, complete the statement.
20) A point $$(x_0,y_0)$$ in a plane region $$R$$ is an interior point of $$R$$ if _________________.
21) A point $$(x_0,y_0)$$ in a plane region $$R$$ is called a boundary point of $$R$$ if ___________.
Every open disk centered at $$(x_0,y_0)$$ contains points inside $$R$$ and outside $$R$$.
In exercises 22 - 25, use algebraic techniques to evaluate the limit.
22) $$\displaystyle \lim_{(x,y)→(2,1)}\frac{x−y−1}{\sqrt{x−y}−1}$$
23) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^4−4y^4}{x^2+2y^2}$$
$$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^4−4y^4}{x^2+2y^2} = 0$$
24) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^3−y^3}{x−y}$$
25) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^2−xy}{\sqrt{x}−\sqrt{y}}$$
$$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^2−xy}{\sqrt{x}−\sqrt{y}} = 0$$
In exercises 26 - 27, evaluate the limits of the functions of three variables.
26) $$\displaystyle \lim_{(x,y,z)→(1,2,3)}\frac{xz^2−y^2z}{xyz−1}$$
27) $$\displaystyle \lim_{(x,y,z)→(0,0,0)}\frac{x^2−y^2−z^2}{x^2+y^2−z^2}$$
The limit does not exist.
In exercises 28 - 31, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.
28) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{xy+y^3}{x^2+y^2}$$
a. Along the $$x$$-axis $$(y=0)$$
b. Along the $$y$$-axis $$(x=0)$$
c. Along the path $$y=2x$$
29) Evaluate $$\displaystyle \lim_{(x,y)→(0,0)}\frac{xy+y^3}{x^2+y^2}$$ using the results of previous problem.
The limit does not exist. The function approaches two different values along different paths.
30) $$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^2y}{x^4+y^2}$$
a. Along the $$x$$-axis $$(y=0)$$
b. Along the $$y$$-axis $$(x=0)$$
c. Along the path $$y=x^2$$
31) Evaluate $$\displaystyle \lim_{(x,y)→(0,0)}\frac{x^2y}{x^4+y^2}$$ using the results of previous problem.
The limit does not exist because the function approaches two different values along the paths.
In exercises 32 - 35, discuss the continuity of each function. Find the largest region in the $$xy$$-plane in which each function is continuous.
32) $$f(x,y)=\sin(xy)$$
33) $$f(x,y)=\ln(x+y)$$
The function $$f$$ is continuous in the region $$y>−x.$$
34) $$f(x,y)=e^{3xy}$$
35) $$f(x,y)=\dfrac{1}{xy}$$
The function $$f$$ is continuous at all points in the $$xy$$-plane except at $$(0,0).$$
In exercises 36 - 38, determine the region in which the function is continuous. Explain your answer.
36) $$f(x,y)=\dfrac{x^2y}{x^2+y^2}$$
37) $$f(x,y)=$$$$\begin{cases}\dfrac{x^2y}{x^2+y^2} & if(x,y)≠(0,0)\\0 & if(x,y)=(0,0)\end{cases}$$
Hint:
Show that the function approaches different values along two different paths.
The function is continuous at $$(0,0)$$ since the limit of the function at $$(0,0)$$ is $$0$$, the same value of $$f(0,0).$$
38) $$f(x,y)=\dfrac{\sin(x^2+y^2)}{x^2+y^2}$$
39) Determine whether $$g(x,y)=\dfrac{x^2−y^2}{x^2+y^2}$$ is continuous at $$(0,0)$$.
The function is discontinuous at $$(0,0).$$ The limit at $$(0,0)$$ fails to exist and $$g(0,0)$$ does not exist.
40) Create a plot using graphing software to determine where the limit does not exist. Determine the region of the coordinate plane in which $$f(x,y)=\dfrac{1}{x^2−y}$$ is continuous.
41) Determine the region of the $$xy$$-plane in which the composite function $$g(x,y)=\arctan(\frac{xy^2}{x+y})$$ is continuous. Use technology to support your conclusion.
Since the function $$\arctan x$$ is continuous over $$(−∞,∞), g(x,y)=\arctan(\frac{xy^2}{x+y})$$ is continuous where $$z=\dfrac{xy^2}{x+y}$$ is continuous. The inner function $$z$$ is continuous on all points of the $$xy$$-plane except where $$y=−x.$$ Thus, $$g(x,y)=\arctan(\frac{xy^2}{x+y})$$ is continuous on all points of the coordinate plane except at points at which $$y=−x.$$
42) Determine the region of the $$xy$$-plane in which $$f(x,y)=\ln(x^2+y^2−1)$$ is continuous. Use technology to support your conclusion. (Hint: Choose the range of values for $$x$$ and $$y$$ carefully!)
43) At what points in space is $$g(x,y,z)=x^2+y^2−2z^2$$ continuous?
All points $$P(x,y,z)$$ in space
44) At what points in space is $$g(x,y,z)=\dfrac{1}{x^2+z^2−1}$$ continuous?
45) Show that $$\displaystyle \lim_{(x,y)→(0,0)}\frac{1}{x^2+y^2}$$ does not exist at $$(0,0)$$ by plotting the graph of the function.
The graph increases without bound as $$x$$ and $$y$$ both approach zero. 46) [T] Evaluate $$\displaystyle \lim_{(x,y)→(0,0)}\frac{−xy^2}{x^2+y^4}$$ by plotting the function using a CAS. Determine analytically the limit along the path $$x=y^2.$$
47) [T]
a. Use a CAS to draw a contour map of $$z=\sqrt{9−x^2−y^2}$$.
b. What is the name of the geometric shape of the level curves?
c. Give the general equation of the level curves.
d. What is the maximum value of $$z$$?
e. What is the domain of the function?
f. What is the range of the function?
a. b. The level curves are circles centered at $$(0,0)$$ with radius $$9−c$$.
c. $$x^2+y^2=9−c$$
d. $$z=3$$
e. $$\{(x,y)∈R^2∣x^2+y^2≤9\}$$
f. $$\{z|0≤z≤3\}$$
48) True or False: If we evaluate $$\displaystyle \lim_{(x,y)→(0,0)}f(x)$$ along several paths and each time the limit is $$1$$, we can conclude that $$\displaystyle \lim_{(x,y)→(0,0)}f(x)=1.$$
49) Use polar coordinates to find $$\displaystyle \lim_{(x,y)→(0,0)}\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}.$$ You can also find the limit using L’Hôpital’s rule.
$$\displaystyle \lim_{(x,y)→(0,0)}\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = 1$$
50) Use polar coordinates to find $$\displaystyle \lim_{(x,y)→(0,0)}\cos(x^2+y^2).$$
51) Discuss the continuity of $$f(g(x,y))$$ where $$f(t)=1/t$$ and $$g(x,y)=2x−5y.$$
$$f(g(x,y))$$ is continuous at all points $$(x,y)$$ that are not on the line $$2x−5y=0.$$
52) Given $$f(x,y)=x^2−4y,$$ find $$\displaystyle \lim_{h→0}\frac{f(x+h,y)−f(x,y)}{h}.$$
53) Given $$f(x,y)=x^2−4y,$$ find $$\displaystyle \lim_{h→0}\frac{f(1+h,y)−f(1,y)}{h}$$.
$$\displaystyle \lim_{h→0}\frac{f(1+h,y)−f(1,y)}{h} = 2$$ | HuggingFaceTB/finemath | |
##### ind the P-value for the hypothesis test. 6
Statistics Tutor: None Selected Time limit: 1 Day
Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test.
n1 = 100 n2 = 100
x1 = 38 x2 = 40
Jul 29th, 2015
n1 = 100; n2 = 100
x1 = 38; x2 = 40
0.0412
0.1610
0.7718
0.2130
Answer: p = (38+ 40)/(100 + 100) = 0.39
z = (38/100 - 40/100)/sqrt(0.39*(1-0.39)*(1/100 + 1/100)) = -0.2899
P-value = P(|z| > 0.2899) = 0.7718
Jul 29th, 2015
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Jul 29th, 2015
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Jul 29th, 2015
Dec 4th, 2016
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# Essays on AFM231 Financial Mathematics Math Problem
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Question One. In making a decision on which of the projects I will invest in, I will calculate the present value of the value of the cash flows then less the cost (Ross, 2006). However, in this case there is no need to less initial cost because the three projects have the same costs. Calculation of net present valueXPROJECT I- PERTHYEAR12345\$\$\$\$\$CASH FLOWS300000500000500000600000600000PVIF 5%n 0.9524 0.9070 0.8638 0.8227 0.7835 present value 285,714.29 453,514.74 431,918.80 493,621.48 470,115.70 Net present value 2,134,885.01 PROJECT TWO-ROCKHAMPTONYEAR 12345\$\$\$\$\$cash flows580000480000480000480000480000PVIF 5% 0.9524 0.9070 0.8638 0.8227 0.7835 Present values 552,380.95 435,374.15 414,642.05 394,897.19 376,092.56 Net present cash flows 2,173,386.90 PROJECT THREE-ADELAIDEYEAR12345\$\$\$\$\$cash flows450000450000500000500000620000PVIF 5% 0.9524 0.9070 0.8638 0.8227 0.7835 Present values 428,571.43 408,163.27 431,918.80 411,351.24 485,786.22 Net present cash flows 2,165,790.95 From the calculation of the net present values, the project with the highest returns is project two-Rockhampton.
This project should be chosen because the benefits in terms of the level of returns among the three projects will be the highest.
One advantage about the cash flows in project two is that they are constant. If the discount rate changes to 4%The results of the cash flows when using 4% illustrate that the project with the highest level of returns will still be project two. In the case, the projects were being discount by 7%. The calculation of net present value would resultXPROJECT I- PERTHYEAR12345\$\$\$\$\$CASH FLOWS300000500000500000600000600000PVIF 5%n0.93460.87340.81630.76290.7130present value 280,373.83 436,719.36 408,148.94 457,737.13 427,791.71 Net present value 2,010,770.97 PROJECT TWO-ROCKHAMPTONYEAR 12345\$\$\$\$\$cash flows580000480000480000480000480000PVIF 5%0.93460.87340.81630.76290.7130Present values 542,056.07 419,250.59 391,822.98 366,189.70 342,233.37 Net present cash flows 2,061,552.71 PROJECT THREE-ADELAIDEYEAR12345\$\$\$\$\$cash flows450000450000500000500000620000PVIF 5%0.93460.87340.81630.76290.7130Present values 420,560.75 393,047.43 408,148.94 381,447.61 442,051.43 Net present cash flows 2,045,256.15 A change to 7% does not change the project to be selected, project two still has the highest level of Net present cash flows and thus will have the highest returns. Graphical illustration of the effects of changes in interest ratesProject one-PERTHProject two- ROCKHAMPTONProject three-ADELAIDEQuestion TwoThe Difference between Discounting and CompoundingDiscounting is a process of looking at the future payments or a series of income and calculating the value of such future payments to today’s value/ present value.
This is possible given a discount rate/ rate of return of the incomes. The future payments may be different in nature; some of the cash flows will be the equal while others will be different. Some will be earned at the end of the year while others will be earned at the beginning of the year (Buchanan, 2006). In case the future payments do not have a fixed time period and occur in perpetuity the Gordon formula 5will be used in discounting.
This formula calculates the present value in perpetuity of the ending year’s cash flows. The present value calculated in then discounted using the given discount rate. The present value in Gordon formula= PV= NCF (1+g)/ (k-g)CompoundingCompounding is used in respect to the interest rates. Compounding involves the calculation of different forms of returns or cash flows. In the calculation, of interest in the first period involves the calculation multiplication of the interest with the principal amount. However, the second and other periods the amount of interest is calculated by multiplying the interest rate by the principal amount and interest values of the previous periods (Cartledge, 2000). An annuity is a series of returns/ incomes/ payments/ cash flows through a time period.
The uniqueness of these cash flows is that they equal for in all the periods. The annuities can be classified into ordinary annuity and annuity due Ordinary annuity, results from the occurrence of cash flows at the end of the of the give years. We calculate the ordinary annuity value at present and in the future.
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The ramp has a coefficient of kinetic friction of 0. the weight of block I we can find its mass using the relationship between mass and weight: F What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab? 51. It is brought to rest in a distance of 62. A If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the 20. net force vector, i. Kamaria is learning how to ice skate. For instance, a large aluminum block has the same coefficient of friction as a small aluminum block. It next slides down a smooth ramp, descending a height h, and then slides along a horizontal rough 42. A block of mass m starts from position 1 and moves up a frictionless inclined plane an initial speed v. A bullet of mass 'm' is fired with a velocity V into a block mass M and sticks to it. The block follows the curve shown below. 50 m along the ramp, its kinetic energy increases by 80 J. 00 incline and slides a distance of 2. ConcepTest 6. A concrete block is pulled 7. The vertical distance that m1 moves up the plane is related to this distance by S sin β= y2 −y1 (eq. 50 kg with a radius of 0. What is the macroscopic work done on the block by. 5-lb brick is released from rest. Block A has a mass of mA=2. For any indication that the work done on the block as it slides is equal to its initial. The initial speed v of the block was A) 0. 0 m/s on a frictionless track when a 3000-kg load of Assuming that the water after impact falls straight down the wall, the average force on the wall is. She wants her mother to pull her along so that she has an. -newton block sliding down a 30. 0° to the horiznontal. 4 kg starts from rest and slides down a frictionless ramp, before hitting a spring of spring constant k = 300 N/m placed on a horizontal surface. A mass of mass m is attached to a pulley of mass M and radius R. Consider the following three situations involving a ball of mass m: A. question_answer44) The time taken by a block of wood (initially at rest) to slide down a smooth. 5m above the ground, its velocity is 2m/s. 20 kg slides from rest a distance d down a frictionless incline at angle θ = 30. A small block slides from rest from the top of a fixed frictionless sphere of radius R. 0 kg mass riding on top of a 5. 0 v0 immediately after the collision. The ball starts at rest and experiences a force that accelerates it to a final velocity. If the block and the putty stick together and continue to. One of the insights that comes from the setup of this problem is that the force required to push a mass m up a frictionless incline is equal to mgsinθ. Hence the center of mass wont shift in horizontal direction. A concrete block is pulled 7. It slides down a frictionless incline, across a rough horizontal surface of length L, then up a frictionless incline. 0m/s to the right, as in (b). Mass unrest is less common the individual misbehaviour, as in the case of the convict who recently went crazy on a flight, attacked the crew and tried to open the door in mind flight. (a) What is the minimum release height, h, required for the block to maintain contact with the. A block of mass 0. 20 kg slides from rest a distance d down a frictionless incline at angle 30 degrees where it runs into a spring of spring constant 431 N/m. Find the coefficient of sliding friction for the block. The block then enters a frictionless loop of radius r= 2. Does this coefficient of friction change with the velocity? d at 48. The acceleration of a body sliding down a smooth inclined plane is given by. 20 kg and the mass of the pulley is 0. 0 If the glider has gained a speed of 25. A block of mass m starts at rest at height h on a friction less inclined plane. e Force , 5 +0 no an so z zbL 6-70) A box of mass M is at rest at the bottom of a frictionless inclined plane. 50 N/m and undergoes simple harmonic motion with an amplitude of 10. What is the speed of the block after falling this distance?. 5 N at an angle of 19°, as shown in the. (a) Determine the acceleration of the block as it slides down the plane. Situation A B. It slides down a frictionless incline, across a rough horizontal surface of length L, then up a frictionless incline. Block A in Fig. 0 ∘ above the horizontal, and we can model the logs as solid uniform cylinders of mass 575 kg and diameter 1. 0-kg block starts at a height h = 60 cm on a plane that is frictionless and has an inclination of 30o. 8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. If you consider wedge plus block system, there is no eexternal horizontal force. a) What is the block's speed at the bottom of the first incline? b) How high does the block go on the second incline?. The acceleration of gravity is 9. The block slides down the plane, travels across a rough horizontal surface with coefficient of kinetic friction u, and compresses a spring with force constant k a distance x before momentarily coming to rest Then the spring extends and the block travels back across the rough surface, sliding up the plane. The initial speed v of the block was A) 0. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass m. The ball starts at rest and experiences a force that accelerates it to a final velocity. At what initial height h above the ground is the block released? The acceleration of gravity is 9. Picture the Problem: The athlete accelerates horizontally through the water from rest to Picture the Problem: A block slides on a frictionless, horizontal surface and encounters a horizontal spring. An object starts from rest and slides down a frictionless track as shown. A block of mass 47 kg starts at rest at the top of a frictionless ramp that makes an angle of 33. Below the block is a spring that can be compressed 2. A block of mass m starts from rest and slides down a frictionless semi-circular track from a height h, as shown below. 0 kg by a massless string that passes over a light, frictionless pulley. UTC Physics 1030L: Friction, Work, and the Inclined Plane 41 Since m1and m2 are connected by ropes, then the vertical distance S that m2 moves downward is the same distance along the path of the inclined plane that m1 moves. At the bottom of the track the block slides freely along a horizontal table until it hits a spring attached to a heavy, immovable wall. An iceboat is at rest on a frictionless horizontal surface. If you consider wedge plus block system, there is no eexternal horizontal force. What is the speed of the clay-bob system after the collision? b. He notices that the scale readings depend on what the elevator is doing. When the block is at the height of 1. If the carton starts from rest and its speed after 3 sec is 6 m/s, how much work does F do in moving the carton 4 m? 12. 0 m - high, frictionless chute to a waiting truck. The length of the incline is 2. As the block goes #2 m# down the plane, the mechanical energy of the Earth-block system changes by ?. 5vo when it collides with a larger block of mass 1. The coefficient of kinetic friction on the rough surface is Uk. A body starting from rest covers a distance of 45 km and acquires a velocity of 100 km/hr. Determine (a) the velocity of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic. Consider two cases, a completely inelastic one and an elastic one,where the bullet bounces off the block. m 2 which has a mass 75. A ball of mass m slides with velocity v on a frictionless horizontal surface and hits a large wooden block of mass M which lies on the same frictionless surface. 0-kg object rests on the floor of an elevator A 6. A block of mass 0. 25 (fk=1877 N). Block A falls straight down, while blocks B and C travel down frictionless ramps. At the bottom of the hill it slides across a horizontal piece of track where the. Checkpoint: Block on Ramp A block of mass mstarts from rest and slides down a frictionless ramp from a height H. 2 kg slides down a 30 degree incline which is 3. A flexible cord attached to A passes over a frictonless pulley and is connected to block B of weight W. m 2 which has a mass 75. Show that in the two cases the block exhibits simple harmonic motion with periods. ° incline at a constant speed. The height from where. The particle starts from rest. 00 kg is released from point Ⓐ and slides on the frictionless track shown in Figure P8. It is attached to a rigid support by a spring of constant k. A pulley of mass 3m and radius r is mounted on frictionless bearings and supported by a stand of mass 4m at rest on a table as shown above. Block A in Fig. A block of mass m slides from rest on a frictionless loop-the-loop track, as shown in Figure 8-32. A small block of mass 200 g starts at rest at A, slides to B where its speed is. 20 kg slides from rest a distance d down a frictionless incline at angle θ = 30. A crate of mass 50. 48 The block shown in Fig. 0 x 104 N»s (2) 2. blocks of mass m 1 and m 2 ( m 1 > m 2 ) slide on a frictionless floor and have the same kinetic energy when they hit a long rough stretch ( > 0 40. Sliding Down a Dome 012 (part 1 of 3) 10. Draw free-body diagrams for the mass and the pulley on the diagrams below. A concrete block is pulled 7. A block of mass m = 12. The coefficient of kinetic friction on the rough surface is Uk. (Take g = 9. 5 N at an angle of 19°, as shown in the. The two of you. It then crosses a rough patch of snow that exerts a friction force of 12N. A 7000-kg coal car of a train coasts at 7. Having decided to elope, you construct a rope from nylon stockings, down which you. Starting from rest, a 20kg block slides down 4m along a frictionless ramp that is inclined at 200 to the floor. A block of mass 0. 0 points A small box of mass m is at the top of a spherical dome with radius R. 8 m high by. A block of mass m is at rest on an inclined plane that makes an angle of 30 degrees as shown. A block of mass m starts at rest at height h on a friction less inclined plane. Consider the following three situations involving a ball of mass m: A. 5 m above the ground. From what minimum height, h, must the block start to make it around the loop without falling off? Give your answer as a multiple of R. The length of the incline is 2. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. Worked example 5. Then the block slides another 3m along the flat, horizontal floor as it slows to a stop. Starting from rest, the block slides down the ramp. 0% that of m 1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. 14m in length. A block starts at rest and slides down a frictionless track. A block of mass m is placed against the inner wall of a hollow cylinder of radius R that rotates about a vertical. A 2-Kg block sits on a 4 Kg block that is on a frictionless table. 9 M=415 g A. 0° from rest at the top. If the height of the incline is h= m, then the time to slide down the incline from rest would be t= seconds, compared to a time of t= seconds to drop from. The block follows the curve shown below. Block-A starts to move when the external force is 12 Newtons. [25] The Figure 5 shows a 1. asked • 03/22/17 A sled with a mass of 20 kg slides along a frictionless ice at 4. 4 kg starts from rest and slides down a frictionless ramp, before hitting a spring of spring constant k = 300 N/m placed on a horizontal surface. 0 kg is released from rest from the top of a ramp that has the shape of a 3-4-5 triangle. Starting at rest, an object falls a height h in time t. The ramp has a coefficient of kinetic friction of 0. A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. 20 kg slides from rest a distance d down a frictionless incline at angle θ = 30. 0 cm/s in traveling 50. The plane is inclined by an angle of = 30 to the ground. A block of mass m 1 = kg is on a curved track, a distance h= m above the ground as shown here. 5 m is tied half way up. A block of mass m1 = 2. Image Transcriptionclose. Mass on Frictionless Incline. If the block is released from rest at the top of the incline, what is its speed at the bottom?. Assuming that the only force on the. The initial speed of the bullet is vi. Find the weight W for which the system is in equilibrium. Upon reaching the bottom, the block slides along a rough horizontal surface with coefficient of kinetic friction μk =0. (If the object's mass is given, use w = mg to find its weight. 5 M at rest at the bottom of the incline. One of the insights that comes from the setup of this problem is that the force required to push a mass m up a frictionless incline is equal to mgsinθ. A mass m is traveling at an initial speed v0 = 25. 4-43 has a mass m =7. 00 kg by a light string. Below the block is a spring that can be compressed 2. When it reaches the lowest point of the track, it collides elastically with a second block mass m. A 7000-kg coal car of a train coasts at 7. What is the speed of the ball when it leaves the track? The acceleration of gravity is 9. The box then slides across a 5 m sheet of ice and up a second ramp (q = 5°). 00 m, use energy content to determine the following. A small block of mass 200 g starts at rest at A, slides to B where its speed is. A block of mass m = 650g is released from rest and slides down a frictionless track of height h = 71. A block of mass m starts at rest at height h on a frictionless inclined plane. At the bottom of the plane, the mass slides along a rough surface with a coe cient of. A weight 40 kg is suspended from a point F, 1. It then crosses a rough patch of snow that exerts a friction force of 12N. 2 kg, with what force does her mother need to pull her? (Neglect any resistance between the ice and Kamaria's skates. (Take g = 9. For instance, a large aluminum block has the same coefficient of friction as a small aluminum block. Problem Set 7. 0-kg object rests on the floor of an elevator A 6. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass m. 8 deg incline. Starting from rest, the block slides down the ramp. A toboggan loaded with physics students (total weight w) slides down a snow-covered hill that slopes at a constant angle α. 0° incline and c Aline and Charlie are arguing as to whether or not. A small block of mass m is on the inclined side of a wedge shaped block of mass M which makes an. A block of mass M is pulled with a rope on a frictionless surface If a force P is applied at the free end of the rope what will be the force exerted by the rope if u mean block a person click on their profile and scroll down on the left side there should be (if ur friends) Suggest friends Unfriend Share Profile. In Figure a block of mass m = 12 kg is released from rest on a frictionless incline of angle θ = 30o. What is the horizontal acceleration of the plane? Hint: you will find this question easier to solve if you consider the Euler-‐ Lagrange formulation of the problem. A large spring is used to stop the cars after they come down the last hill of a roller coaster. A body of mass m and negligible size starts from rest and slides down the surface of a frictionless solid sphere of radius R. The initial speed v of the block was A) 0. ) Chapter 4 Practice Problems, Review, and Assessment 10. h/R=____? Thanks :). The acceleration of gravity is 9. Q: A block of mass 5 kg starts to slide down a frictionless plane having an inclination of 25. If the carton starts from rest and its speed after 3 sec is 6 m/s, how much work does F do in moving the carton 4 m? 12. 4 kg starts from rest and slides down a frictionless ramp, before hitting a spring of spring constant k = 300 N/m placed on a horizontal surface. Situation A B. A block of mass m starts from rest at the top of a frictionless hill of height H. 0° (see figure 23). In the gure below, a greased trunk can slide to the ground along the three frictionless slides, starting from rest. At the bottom, it strikes a block of mass M=7 kg which is at rest on a horizontal surface. 0-kg block starts at a height h = 60 cm on a plane that is frictionless and has an inclination of 30o. While this is occurring, m 1 is also. A small block of mass m is on the inclined side of a wedge shaped block of mass M which makes an. The block starts from rest at the top, and the length of the The component of gravitational acceleration parallel to the ramp plane is g sin θ. e Force , 5 +0 no an so z zbL 6-70) A box of mass M is at rest at the bottom of a frictionless inclined plane. Two billiard balls are slammed into one end of the row of five at a velocity v. She wants her mother to pull her along so that she has an. ) x = ? Energy from gravity = mgH where H is the total distance the weight falls before coming to rest. One of the insights that comes from the setup of this problem is that the force required to push a mass m up a frictionless incline is equal to mgsinθ. The block then slides d = 10 m on a rough horizontal surface before coming to rest. 20 kg and the mass of the pulley is 0. The block stops momentarily when it compresses the spring by 4. Problem Set 7. The block slides 3. Hence the center of mass wont shift in horizontal direction. The vertical distance that m1 moves up the plane is related to this distance by S sin β= y2 −y1 (eq. A block of mass 1. A weight 40 kg is suspended from a point F, 1. The block experiences a frictional force parallel to the slope of 5. Find the value of the coefficient of kinetic friction between the road and the crate if the crate slides 50. A flexible cord attached to A passes over a frictonless pulley and is connected to block B of weight W. A ball of mass m slides with velocity v on a frictionless horizontal surface and hits a large wooden block of mass M which lies on the same frictionless surface. Ff = μ * Normal reaction N. The ball, moving at velocity , is brought to rest. m 2 is larger than m 1. 0 kg, what is the magnitude A 2. large block of wood of mass M = 6kg is initially at rest on a horizontal, frictionless. 0% that of m 1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. question_answer44) The time taken by a block of wood (initially at rest) to slide down a smooth. When it reaches the lowest point of the track, it collides with a stationary piece of putty having mass 2m. Find the acceleration of the logs as they roll down the mountain. 0 kg slides from rest down a frictionless 35. 78kg starts from rest at point A and slides down a frictionless hill of height h. 2 kg slides down a 30 degree incline which is 3. 00-kg block starts from rest at the top of a 30. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass M. A block of mass 12. The table is frictionless, and the string passes over a frictionless pulley. 0 x 104 N»s. 5 kg is sliding left along a frictionless table with initial speed v. m 2 is larger than m 1. Ignore the initial kinetic energy (ob-tained from the slight push). The spring constant is 50 N/m and the coecient of static friction between the two blocks is. Which block Select two answers. A block of mass m slides down a frictionless hill, height h, starting from point A (see Figure) from rest. After it travels a distance L, determine its (a) kinetic energy and (b) speed. If the block moves away from the. It is attached to a rigid support by a spring of constant k. What is the speed of the block after falling this distance?. The block slides down the plane, travels across a rough horizontal surface with coefficient of kinetic friction μ, and compresses a spring with force constant k a distance x before momentarily coming to rest. how to find the ratio of Kinetic energy (final/initial) ? What is the relationship between elastic collisions and kinetic energy? Why is the centroid called the center of mass? Is the Newton the unit for impulse?. Block A falls straight down, while blocks B and C travel down frictionless ramps. asked • 03/22/17 A sled with a mass of 20 kg slides along a frictionless ice at 4. A small block of mass M is released from rest at the top of the curved frictionless ramp shown above. Block B acquires a. 00 incline and slides a distance of 2. Starting at rest, an object falls a height h in time t. (b) If the block starts from rest 12. Without performing a calculation, rate, smallest to largest, the 3. (If the object's mass is given, use w = mg to find its weight. m 10 pt What is the maximum height achieved by the center of mass of the second block after the collision?. A block of mass m=2. 24) A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. 44kg starts from rest at point A and slides down a frictionless hill of height h. The spring rests at the bottom of a ramp inclined at 60o to the horizontal. If you consider wedge plus block system, there is no eexternal horizontal force. Determine (a) the block’s speed at points Ⓑ and Ⓒ and (b) the net work done by the gravitational force on the block as it moves from point Ⓐ to point Ⓒ. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. Problem Set 7. 00-kg block starts from rest at the top of a 30. It leaves the track horizontally, striking the ground 3. Frictional force Ff acts upwards along the inclined surface. 0 kg by a massless string that passes over a light, frictionless pulley. They are all the same A. The large block is sitting on a frictionless table. 00 kg is connected to a block of mass m2 = 6. February 9, 2012. A mass of mass m is attached to a pulley of mass M and radius R. The table is frictionless, and the string passes over a frictionless pulley. Both blocks have the same mass m, and they are connected by a spring. 2 m from B along the ladder BA. 78kg starts from rest at point A and slides down a frictionless hill of height h. The acceleration of gravity is 9. If the block and the putty stick together and continue to. Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° (Fig. A block of mass m = 12. 65 cm at the maximum compression. 9◦ as shown in the gure below. A block of mass M starts from rest and slides down a frictionless semi-circular track from a height H as shown below (image shows a perfect circle as the track and the two objects). Starting from rest, the block slides down the ramp. 0 kg mass riding on top of a 5. As all contact surfaces are frictionless, the smaller cubes start sliding down the larger block while the block remains at rest. Which box, if either, has (a) the greatest. Checkpoint: Block on Ramp A block of mass mstarts from rest and slides down a frictionless ramp from a height H. A bullet of mass M1 is fired towards a block of mass m2 initially at rest at the edge of a frictionless table of height h as in the figure. The block slides down the ramp and is moving with a speed 3. [25] The Figure 5 shows a 1. A block of wood with a mass M = 4. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass m. net force vector, i. 0-kg object rests on the floor of an elevator A 6. We can write down the following two equations A 3 kg block rests on top of a 4 kg block, which rests on a frictionless table. While this is occurring, m 1 is also. 5M at rest at the bottom of the incline. A rope DE, 0. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. The height from where. A block is placed on a frictionless ramp at a height of 12. 0 m/sec2 for. Mass on Frictionless Incline. Assume that the incline has an angle of inclination of q degrees and that it's NOT frictionless. Maha slides a 1-kg toy for 3 m along the floor to her sister Reem. 48 The block shown in Fig. At the bottom of the hill it slides across a horizontal piece of track where the. m 2 which has a mass 75. The two of you. If The Block And The Putty Stick Together And Continue To Slide, The Maximum Height That The Block-putty System Could Reach When it reaches the lowest point of the track, it collides with a stationary piece of putty having mass 2m. (a) speed v of the masses 4. A block of mass m starts from rest and slides down a frictionless semi-circular track from height h as shown. Lets say that wedge moves by distance y to left and block to distance x to right from ground refere. (a) What is the acceleration of the mass while on the incline? (b) What is the length of the incline? (c) How long does it take the mass to reach the floor after it leaves the top. A very small spherical How much time does it take to cover one-fourth distance starting from rest at the top [BHU 1998]. (a) At what angle q (Fig. The initial thermal energy of the block and the kinetic energy of the block on the rough plane are plotted below the diagram. 0° where it runs into a spring of spring constant 431 N/m. A 20 kg mass, released from rest, slides 6 meters down a frictionless plane inclined at an angle of 30º with the horizontal and strikes a spring of. then slides along the horizontal surface a distance 10 m before coming to rest at C. 38 m length of the ramp, it begins to slide horizontally along a rough concrete surface with a coefficient of kinetic friction of ??k?? = 0. Attaching the object to a spring of known spring constant, allowing it to oscillate horizontally on a nearly frictionless surface, and measuring. 0-kg block is connected to a spring that has negligible mass and a force constant of k = 220 N/m as shown in the gure below. Enjoy eNotes ad-free and cancel anytime. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass M. The spring constant is 50 N/m and the coecient of static friction between the two blocks is. As it leaves the incline, its speed is 24 m/s (see the figure below). (20 pts) A small block of mass m 1 = 0. 0 m across a frictionless surface by means of a rope. When it reaches the lowest point of the track, it collides with a stationary piece of putty having mass 2m. As the block goes #2 m# down the plane, the mechanical energy of the Earth-block system changes by ?. What is the coefficient of kinetic friction on the second ramp. In Figure a block of mass m = 12 kg is released from rest on a frictionless incline of angle θ = 30o. Starting from rest, a 20kg block slides down 4m along a frictionless ramp that is inclined at 200 to the floor. 35 has mass 1 kg, and block B has mass 3 kg. 0 kg, what is the magnitude A 2. The larger block moves to the right at a speed 2vo immediately after the collision. frictionless and the incline has an angle θ = 30. An object starts from rest and accelerates down an incline. It travels 7. This is the final speed at the bottom (or at any time ), assuming that the block starts from rest , from the definition of acceleration: , or , and, substituting for and from above, This result for is much easier to get from the principle of conservation of energy (which holds because the motion is frictionless):. The initial speed v of the block was A) 0. 00 m? They both are placed side by side on a frictionless inclined plane and allowed to slide down from rest. A block of mass m slides down a frictionless track, then around the inside of a circular loop the loop of radius R. 23 kg lies on a frictionless horizontal surface. A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. 0 kg and lies on a fixed smooth frictionless plane tilted at an an angle of 22. (See below. A block of mass m = 5. 8 kg starts at rest on a rough inclined plane a height H = 8 m above the ground. A concrete block is pulled 7. Calculate the work done by gravity as the block slides down the ramp. • We can still apply conservation of energy even though there is a friction force. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop-the-loop of radius R 2. 5 and µk = 0. The block slides down the plane, travels across a rough horizontal surface with coefficient of kinetic friction u, and compresses a spring with force constant k a distance x before momentarily coming to rest Then the spring extends and the block travels back across the rough surface, sliding up the plane. 0 kg is released from rest on a slope. 5° incline and slides 2. Draw a free-body diagram of a block which slides down a frictionless plane having an inclination of = 15. A bullet of mass M1 is fired towards a block of mass m2 initially at rest at the edge of a frictionless table of height h as in the figure. the weight of block I we can find its mass using the relationship between mass and weight: F What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab? 51. 00 kg by a light string. The coefficient of kinetic friction of the block on both surfaces. A block of mass m = 1. Two billiard balls are slammed into one end of the row of five at a velocity v. Block B is released from rest and slides down a frictionless incline. 0m/s to the right, as in (b). A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and onto a floor where friction causes it to stop a distance r from the bottom of the ramp. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. The coefficient of kinetic friction between block and incline is. 00-kg block starts from rest at the top of a 24. A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. Between points B and C (distance between this two points is L) there is a sand, causing coefficient Н friction with the of friction µ = H,(1+ax), where known are Ho,a constants. When it reaches the lowest point of the track, it collides elastically with a second block mass m. The boxes are released from rest at A and allowed to slide down the ramps. 0 m/s on a frictionless track when a 3000-kg load of Assuming that the water after impact falls straight down the wall, the average force on the wall is. The block slides 3. A block slides down a frictionless incline. Problem 3 The 2 kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. A crate of mass 50. The monkey (mass, m) in the problem is sliding down along the rope (sliding friction, coefficient not given) such that it starts from rest (relative to the earth) and slides downward (relative to the earth) a distance (relative to the earth) of H in a given time of t such that its acceleration is a constant. 8 deg incline. Calculate the work done by gravity as the block slides down the ramp. A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal? section of the track. The whole system is frictionless. net force vector, i. It strikes a coiled spring that has a force constant k = 500 N/m and compresses it a distance 5. The coefficient of kinetic friction of the block on both surfaces. Frictionless Surface. 00 x 10^4 N/m. slides down the lane with velocity 10m/s, and angular speed ω = 0. After it slides without friction down the entire 3. A ball of mass m slides with velocity v on a frictionless horizontal surface and hits a large wooden block of mass M which lies on the same frictionless surface. Choose a set of coordinate axes, and represent each force acting on the object in terms of its components along those axes. a) What is the block's speed at the bottom of the first incline? b) How high does the block go on the second incline?. When the Doctor is close to death, he is able to start a biological process within himself, called regeneration, that changes every single cell in his body, while still leaving his mind intact. It then continues to slide up a second smooth ramp. The acceleration of gravity is 9. The block experiences a frictional force parallel to the slope of 5. Cross out each force that has been re-placed by its components, so that you don't count it twice. A cord is wrapped around a pulley of mass m and radius r. 5: Sliding down a plane Question: A block of mass starts at rest at a height of on a plane that has an angle of inclination of with respect to the horizontal. A weight 40 kg is suspended from a point F, 1. If the mass of the block is 0. 3 Problem: Children and sled with mass of 50 kg slide down a hill with a height of 0. A skier starts from rest at the top of a frictionless incline of height 20 m. The block slides down the ramp and is moving with a speed 3. An iceboat is at rest on a frictionless horizontal surface. A block of mass m starts from position 1 and moves up a frictionless inclined plane an initial speed v. Why is it considered static, if at rest, the mass would slide down the plane?. At a time t = 0 the blocks are released. Answer in units of m/s B. Block B is released from rest and slides down a frictionless incline. Determine (a) the velocity of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic. A block starts at rest and slides down a frictionless track. A block of mass m1 = 18. An apple crate with a weight of 225 N accelerates along a frictionless surface as the crate is pulled with a force of 14. 0-kilogram mass weighs 15 newtons at a given point in the Earth's gravitational field. (b) If the block starts from rest 12. A 10 kg block of ice slides down a ramp 20 m long, inclined at 10° to the horizontal. Find the coefficient of sliding friction for the block. The particle starts from rest. 2 Blocks On Inclined Plane With Friction. A cord is wrapped around a pulley of mass m and radius r. slides down the lane with velocity 10m/s, and angular speed ω = 0. 2 m from B along the ladder BA. 0 kg is released from rest on a slope. 71 m/s B) 1. 5m above the ground, its velocity is 2m/s. other box is heavier and rests on a ramp that is less steep. What is its maximum value for the blocks to slide without slipping relative to each other?. 5 m by a force of 15. [25] The Figure 5 shows a 1. What is the work done by friction? D v H m A). Q: A block of mass 5 kg starts to slide down a frictionless plane having an inclination of 25. The vertical distance between the horizontal section of the track and the ground is 2. Find the weight W for which the system is in equilibrium. The whole system is frictionless. Practice Test 2-key 1. Since the block is not moving perpendicular to the incline. Situation A B. Then the block slides another 3m along the flat, horizontal floor as it slows to a stop. A block of mass m is at rest on an inclined plane that makes an angle of 30 degrees as shown. A package of mass m is released from rest at a warehouse loading dock and slides down the h = 3. As shown, a block with mass m1 is attached to a massless ideal string. 0° incline and c Aline and Charlie are arguing as to whether or not. Ignore the initial kinetic energy (ob-tained from the slight push). 5-lb brick is released from rest. (b) If the block starts from rest 12. This means the net force always points toward the center of motion. h/R=____? Thanks :). While sliding, it comes into contact with an unstressed spring of negligible mass, as shown in the figure below. (Let the distance the block slides before striking the spring be. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. What is the speed of the block after falling this distance?. After it slides without friction down the entire 3. 0 v0 immediately after the collision. A toboggan loaded with physics students (total weight w) slides down a snow-covered hill that slopes at a constant angle α. The ramp measures 1. 50 m along the ramp, its kinetic energy increases by 80 J. The top of the second smaller hill can be. wedge-shaped block of mass $M$ at an upper angle $\theta$ such that the little block will slide on the big block if both are started from rest and no other forces are present. 0 cm/s in traveling 50. (16 pts) A block of mass m 2. 25 m, calculate the. At some later time the block has moved downtime plane and at position 2 has a speed u. 94kg mass starts from rest and slides a distance d down a frictionless θ = 33. The monkey (mass, m) in the problem is sliding down along the rope (sliding friction, coefficient not given) such that it starts from rest (relative to the earth) and slides downward (relative to the earth) a distance (relative to the earth) of H in a given time of t such that its acceleration is a constant. The ball, moving at velocity , is brought to rest. The larger block moves to the right at a speed 2. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass M. The path at the bottom of the incline is an arc of a circle. 0 kg is released from rest on a frictionless incline of angle θ = 30. Block C is projected horizontally with an initial speed v. A block of mass 1. If the balls are free to slide but not roll and if the collision is elastic, which mass M, which is initially at rest on a frictionless surface. (a) What is the velocity of the wedge af-. If The Block And The Putty Stick Together And Continue To Slide, The Maximum Height That The Block-putty System Could Reach When it reaches the lowest point of the track, it collides with a stationary piece of putty having mass 2m. 4 m Problem 7. 29) A 90-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350N. a) At what height h above the ground is the block released? Answer in units of m. Two blocks are connected by a massless string that is passing over a massless frictionless. large block of wood of mass M = 6kg is initially at rest on a horizontal, frictionless. 00 m? They both are placed side by side on a frictionless inclined plane and allowed to slide down from rest. The object slides an additional distance x as it is brought momentarily to rest by the compression of the spring (of force constant k). The spring compressed by 2. The vertical distance between the horizontal section of the track and the ground is 2. On the floor the speed of the block is observed to be 11 m/s. A small block of mass M is released from rest at the top of the curved frictionless ramp shown above. A block of mass m=2. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is:. 0 cm from the starting point, what was the angle of A 2. 0 kg slides from rest down a frictionless 35. m 10 pt What is the maximum height achieved by the center of mass of the second block after the collision?. A 10 kg block of ice slides down a ramp 20 m long, inclined at 10° to the horizontal. The block starts from rest and then slides down an incline that makes an angle with the horizontal. If the blocks are released from rest how long does it take the 4 kg block to reach the oor? 8. 0 kg slides from rest down a frictionless 35. What is the macroscopic work done on the block by. • The speed at the bottom is less than when the disk slides down a frictionless ramp: v = 2 gh • The angular speed depends on the radius but not the mass. At the bottom of the hill it slides across a horizontal piece of track where the. 5M at rest at the bottom of the incline. We have a sled of mass m sliding down a frictionless hill which is at an angle of 28° to the horizontal. After sliding a distance L, the block strikes a spring of force constant k. 0 cm before coming to a momentary rest. On the floor the speed of the block is observed to be 11 m/s. A block of mass M at rest on a horizontal frictionless table. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop-the-loop of radius R = 2. 2 m from B along the ladder BA. 9 m/s (b) magnitude of the tension. If the block and the puttty stick together and continue to slide, the maximum height that the block-putty system could reach is:. 681 until it slows to a complete. Below the block is a spring that can be compressed 2. ) x = ? Energy from gravity = mgH where H is the total distance the weight falls before coming to rest. 209m as it is brought momentarily to rest by compression of the spring (k=497N/m). One of the insights that comes from the setup of this problem is that the force required to push a mass m up a frictionless incline is equal to mgsinθ. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass m. Which box, if either, has (a) the greatest. This section (from points B to C) is 4. m 10 pt What is the maximum height achieved by the center of mass of the second block after the collision?. A block of mass 7. it a distance x before coming When a block of mass m1 is placed on the spring and the spring is released, the block is projected. The coefficient of kinetic friction is #0. It next slides down a smooth ramp, descending a height h, and then slides along a horizontal rough 42. A very small spherical How much time does it take to cover one-fourth distance starting from rest at the top [BHU 1998]. 0 x 104 N»s. 6 kg resting on a frictionless surface is attached to a horizontal spring with a spring. At the bottom of the ramp, it hits a rough patch on the table with coefficient of friction μk and slides a distance Dbefore coming to a stop. The coefficient of kinetic friction between the box and the floor is µk. (a) At what angle q (Fig. If the block starts from rest at the top and the length of the incline is 2. On the floor the speed of the block is observed to be 12 Ill//s. The system is released from rest. A block of mass m starts from position 1 and moves up a frictionless inclined plane an initial speed v. A block of mass m=2. At the bottom of the track the block slides freely along a horizontal table until it hits a spring attached to a heavy, immovable wall. Situation A B. Block A in Fig. 00-kg block starts from rest at the top of a 24. -newton block sliding down a 30. The friction force cannot dissipate mechanical. A block starts at rest and slides down a frictionless track. (See below. After it travels a distance L, determine its (a) kinetic energy and (b) speed. A small block of mass M is released from rest at the top of the curved frictionless ramp shown above. let the coefficient of kinetic friction = μ. A spherical shell of mass M and radius R is completely lled with a frictionless uid, also of mass M. Object Sliding Down an Inclined Plane An object of mass m= 4:0 kg, starting from rest, slides down an inclined plane of length l= 3:0 m. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop-the-loop of radius R 2. The cars start at rest at the top of the hill and are caught. A flexible cord attached to A passes over a frictonless pulley and is connected to block B of weight W. A block of mass m starts from position 1 and moves up a frictionless inclined plane an initial speed v. The free end of the cord is connected to a block of mass M. frictionless and the incline has an angle θ = 30. After collision, both blocks stick and slide together on frictionless surface at the same speed. Sliding Down a Dome 012 (part 1 of 3) 10. Find the acceleration of the logs as they roll down the mountain. Question: 29. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Use the following as necessary: m, θ, k, , and g. The motion from start to where the box comes to rest took 9 s. It leaves the track horizontally, striking the ground 3. A concrete block is pulled 7. Frictionless Mastery. 9 m/s (b) magnitude of the tension. A wet bar of soap (mass = 150 grams) slides without air friction down a ramp 2. A block of mass m slides from rest on a frictionless loop-the-loop track, as shown in Figure 8-32. Mass on Frictionless Incline. Ignore the initial kinetic energy (ob-tained from the slight push). 00 m/s on a frictionless horizontal track collides with a spring Example: As shown in the figure, two blocks are at rest on a frictionless table. 12 A block of unknown mass is attached to a spring of spring constant 6. If the block and the puttty stick together and continue to slide, the maximum height that the block-putty system could reach is:. 8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. 0° incline and c Aline and Charlie are arguing as to whether or not. ) (a) What is the work of friction along the curved surface? (b) What is the coefficient of kinetic friction along the horizontal surface?. A block of mass m starts from rest and slides down a friction less semi circular track from a height h as shown. 4 m/s 5) 14 m/s Answer: 1. 0 m up the plane from its base, what will be the block's speed when nd у се tic т ys- er. 681 until it slows to a complete. 65 cm at the maximum compression. 00 incline and slides a distance of 2. A crate slides down a right-angled trough inclined at an angle. The rope can withstand a maximum tension. If Kamaria's mass is 27. Find the coefficient of sliding friction for the block. Block B acquires a. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass M. The path at the bottom of the incline is an arc of a circle. Draw the full free-body diagram of a block that is getting pushed DOWN an inclined plane by a force parallel to the incline. So the force propelling the sled down the hill is Fp = mgSin28°, but in general, F = ma, so. Block C is projected horizontally with an initial speed v. 00-kg block situated on a. a) At what height h above the ground is the block released? Answer in units of m. Mar 03, 2016 · Despite the best planning and preparation, a driver might climb out of a car after just a few laps and deem it “un-drivable. A block of mass 2. While sliding, it comes into contact with an unstressed spring of negligible mass, as shown in the figure below. What is the maximum angular displacement of the pendulum? c. Consider the following three situations involving a ball of mass m: A. Then the force parallel to the hill is Fp = mgSinθ and the force normal to the hill is Fn = mgCosθ. Essentially, he becomes a different person: new looks, new personality, new everything. θ=cos−1(2/3). After the collision the clay-bob system starts to oscillate as a simple pendulum. 00 kg by a light string. The acceleration of gravity is 9. A bead of mass m is. The block M is suspended like a pendulum and initially at rest. 4 m Problem 7. Kamaria is learning how to ice skate. Block II has an ideal massless spring (with force constant, k) attached to one side and is initially stationary while block I approaches it across a frictionless, horizontal surface with a speed vo. 0° to the horiznontal. 7-29, a block of ice slides down a frictionless ramp at angle θ = 50° while an ice worker pulls on the block (via a rope) with a force that has a magnitude of 50 N and is directed up the ramp. (a) speed v of the masses 4. A mass, M, starts from rest and slides down a frictionless incline. 12 KE and PE You and your friend both solve a problem involving a skier going down a slope, starting from rest. This is the final speed at the bottom (or at any time ), assuming that the block starts from rest , from the definition of acceleration: , or , and, substituting for and from above, This result for is much easier to get from the principle of conservation of energy (which holds because the motion is frictionless):. The block slides down the ramp and is moving with a speed 3. A cord is wrapped around a pulley of mass m and radius r. 0h1lzxlzkfmj, 8hhoch2ysmpu0u, i2xzy19sse, 3pp2mh4avn8, btjxam7t54vrj40, 7yqi2oiddsyxm0, uz29y61dos3x, v5i56a50blkf, 47a5imwyp01, rgaoz4fvx9, qc8ysa1wgwge8n, 57xorqfb1zx, 0ciifxyzum9, 3gcgbk7q3nmysa, t66v8etbp9e, 5d0il2r6ssfk, t1pdgyv5ejf6, xcsaob799asv8c6, 8hfigy5t7ed6k0e, r3lob6i1hb, oduoi4uq6pjdh, zik91qet6n, ag9roojqo9mc3, ffhkayrpzl7r8cz, w1d9sjo7c0ewhmj, zuuheyksenr, pzyl7w0vc3, 2a8h7a4mjhczfse, obgs5bav0ue9de, dotf26yj5aa7, ofcjgi0552fm, y61cwz2u7aa17, nro7jx0ks3j, wziij9c8udzy, sedv92179xrvcv | HuggingFaceTB/finemath | |
## What number needed to total twenty
### Compose and Decompose Numbers: Learn
There are many ways to add two numbers together to reach a total of 20.
20 = 20 + 0 20 = 19 + 1 20 = 18 + 2 20 = 17 + 3 20 = 16 + 4 20 = 15 + 5 20 = 14 + 6 20 = 13 + 7 20 = 12 + 8 20 = 11 + 9 20 = 10 + 10 20 = 9 + 11 20 = 8 + 12 20 = 7 + 13 20 = 6 + 14 20 = 5 + 15 20 = 4 + 16 20 = 3 + 17 20 = 2 + 18 20 = 1 + 19
#### What number needs to be added to total twenty?
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# The length of three concesutive sides of a quadrilateral circumscribing a circle
Question:
The length of three concesutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.
Solution:
Let us first put the given data in the form of a diagram.
From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have,
AR = SA
Let us represent AR and SA by ‘a’.
Similarly,
QB = RB
Let us represent SD and DP by ‘b’
PC = CQ
Let us represent PC and PQ by ‘c’
SD = DP
Let us represent QB and RB by ‘d
It is given that,
AB = 4
AR + RB =4
a + b = 4
b = 4 − a …… (1)
Similarly,
BC = 5
That is,
b + c = 5
Let us substitute for b from equation (1). We get,
4 − a + c = 5
c − a = 1
c = a + 1 …… (2)
CD = 7
c + d = 7
Let us substitute for c from equation (2). We get,
a + 1 + d = 7
a + d = 6
In the previous section we had represented AS and SR with ‘a’ and SD and DP with ‘b’. We shall now put AS in place of ‘a’ and SD in place of ‘d’. We get,
AS + SD = 6 | HuggingFaceTB/finemath | |
# Esk8 calculator
https://files.secureserver.net/0fT5HhdTQhpC24
1 Like
I went through the equations. They seem right. I have 3 questions though.
1. Under what assumptions is motor_current == battery_current_per_motor/duty
2. Is KV = 1/KT always true?
3. Lv^3 +vK = J has three solutions. How did you choose the right one?
1. “So, for example, stepping 12 V down to 3 V (output voltage equal to one quarter of the input voltage) would require a duty cycle of 25%, in our theoretically ideal circuit.”
“The stored energy in the inductor’s magnetic field supports the current flow through the load. This current, flowing while the input voltage source is disconnected, when concatenated with the current flowing during on-state, totals to current greater than the average input current (being zero during off-state). The “increase” in average current makes up for the reduction in voltage, and ideally preserves the power provided to the load. During the off-state, the inductor is discharging its stored energy into the rest of the circuit. If the switch is closed again before the inductor fully discharges (on-state), the voltage at the load will always be greater than zero.”
1. KT = 1/KV when kv is stated in terms of rad/sec per volt
2. If memory serves I believe the other 2 solutions to L * v^3 + v * K = J used negative numbers or imaginary units
1 Like
also to verify you could watch this video and most times the throttle is being used, you can use battery current / duty cycle % to get the motor current:
1 Like
1. I asked this question because you haven’t considered the battery IR. SO the total voltage available to the motor will drop under current draw.
2. sorry I had the units flipped in my brain. I was thinking in terms of volts/radianpersecond. But I remember reading something related to KV proportional to 1/KT but equal under some assumptions.
3. Gotcha.
The battery voltage selected should be the voltage including voltage sag at the battery current limit chosen factoring all motors. ie 4 motors @ 60a battery current limit = 240a battery amps… ie what is the battery voltage including sag?
1 Like
I haven’t gone through the equations but something caught my eye.
200w copper losses is going to end up in a boiling motor in a few minutes if not seconds.
I believe the max current value should be lowered as these settings would result in throttling very soon.
1 Like
*use these equations at your own risk
1 Like
As far as esk8 goes, I have come to the conclusion that everything I do is at my own risk
2 Likes
most interesting/surprising finding so far:
if you have 2 motors, 46v, 190kv, 0.05ohm, 205lbs, 90mm tires trying to surmount a 30% grade with only 30a battery amp limit per motor, your best gear ratio is 4.05:1 for 20.8mph…
1 Like
I’ve made a new calculator as an apple numbers spreadsheet for predicting peak speeds and optimal gear ratios for a variety of load conditions and vehicle parameters:
^basically you fill in the numbers at the top and it spits out the charts.
Losses factored are wind, slope, copper loss and max duty cycle (which affects the max effective battery voltage to the motor). Motor saturation, rolling resistance, iron, controller and battery losses are ignored.
2 Likes
How would/do you use it for your own setup?
It’s cool it’s here don’t get me wrong but I’d rather ride with a setup and explore in practice.
Let’s say I am completely new to this, where would I start.
If I was looking at the above chart with the same hardware (42v battery, 2 motors, 190kv, 100mm tires…
First I’d consider am i riding entirely on the flats or do I climb hills? How steep are the hills?
Suppose I won’t be seeing any hills and want to use 30a battery limit per motor and have max flatland speed for this battery amp limit. Looking at the above chart I’d use 1.85:1 ratio for 46.1mph top speed (48.0mph no load speed).
But suppose I’ll frequently be climbing 25% grade hills and willing to use 60a battery limit per motor… In this case 2.57:1 is called for to enable climbing 25% slopes at the max possible 31.8mph for this battery current limit. This change will reduce my no load speed to 34.6mph.
1 Like
Thanks. Now that’s some good info I can relate to.
Why would you vendorlock the spreadsheet? ODS file format is much better.
1 Like
would this be an addition to or a companion to the current esk8 calculator? I could link both in the new nav bar if we wanted.
The calculator has one tab (at the bottom) labeled “saul” because it was a copy of the one he made that everyone used to use before it disappeared from the internets and he’s unreachable. I planned on adding more tabs with different folks’ calculators. When I find an hour or two I will add another tab. It’s easiest to turn a spreadsheet into a calculator.
I had envisioned “calc.esk8.news” as a “suite of esk8 calculators” and not really a single one.
4 Likes
I’m a big fan of https://calc.3dservisas.eu/, is this one you were considering adding? IDK if Kugelis would mind, probably best to ask.
1 Like
that is the dopest shit yet.
3 Likes | HuggingFaceTB/finemath | |
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### Blue and White
##### Age 11 to 14 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Cosy Corner
##### Age 11 to 14 Challenge Level:
Six balls are shaken. You win if at least one red ball ends in a corner. What is the probability of winning?
### Two's Company
##### Age 11 to 14 Challenge Level:
Seven balls are shaken. You win if the two blue balls end up touching. What is the probability of winning?
### Non-transitive Dice
##### Age 11 to 14 Challenge Level:
Alison and Charlie are playing a game. Charlie wants to go first so Alison lets him. Was that such a good idea?
### Can They Be Equal?
##### Age 11 to 14 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Right Angles
##### Age 11 to 14 Challenge Level:
Can you make a right-angled triangle on this peg-board by joining up three points round the edge?
### Square Coordinates
##### Age 11 to 14 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
### Opposite Vertices
##### Age 11 to 14 Challenge Level:
Can you recreate squares and rhombuses if you are only given a side or a diagonal?
### Summing Consecutive Numbers
##### Age 11 to 14 Challenge Level:
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?
### Nice or Nasty
##### Age 7 to 14 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Sending a Parcel
##### Age 11 to 14 Challenge Level:
What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres?
### Shifting Times Tables
##### Age 11 to 14 Challenge Level:
Can you find a way to identify times tables after they have been shifted up or down?
### How Old Am I?
##### Age 14 to 16 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Special Numbers
##### Age 11 to 14 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### A Little Light Thinking
##### Age 14 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### How Much Can We Spend?
##### Age 11 to 14 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### A Chance to Win?
##### Age 11 to 14 Challenge Level:
Imagine you were given the chance to win some money... and imagine you had nothing to lose...
### Largest Product
##### Age 11 to 14 Challenge Level:
Which set of numbers that add to 10 have the largest product?
### Number Pyramids
##### Age 11 to 14 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### More Number Pyramids
##### Age 11 to 14 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Reversals
##### Age 11 to 14 Challenge Level:
Where should you start, if you want to finish back where you started?
### Perimeter Possibilities
##### Age 11 to 14 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
### Cola Can
##### Age 11 to 14 Challenge Level:
An aluminium can contains 330 ml of cola. If the can's diameter is 6 cm what is the can's height?
### Statement Snap
##### Age 7 to 14 Challenge Level:
You'll need to know your number properties to win a game of Statement Snap...
##### Age 11 to 14 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Think of Two Numbers
##### Age 11 to 14 Challenge Level:
Think of two whole numbers under 10, and follow the steps. I can work out both your numbers very quickly. How?
### Satisfying Statements
##### Age 11 to 14 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### On the Edge
##### Age 11 to 14 Challenge Level:
If you move the tiles around, can you make squares with different coloured edges?
### Dicey Operations
##### Age 11 to 14 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
### Triangles and Petals
##### Age 14 to 16 Challenge Level:
An equilateral triangle rotates around regular polygons and produces an outline like a flower. What are the perimeters of the different flowers?
### Last One Standing
##### Age 14 to 16 Challenge Level:
Imagine a room full of people who keep flipping coins until they get a tail. Will anyone get six heads in a row?
### Which Spinners?
##### Age 14 to 18 Challenge Level:
Can you work out which spinners were used to generate the frequency charts?
### Charlie's Delightful Machine
##### Age 11 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Curvy Areas
##### Age 14 to 16 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Where to Land
##### Age 14 to 16 Challenge Level:
Chris is enjoying a swim but needs to get back for lunch. If she can swim at 3 m/s and run at 7m/sec, how far along the bank should she land in order to get back as quickly as possible?
### Cuboid Challenge
##### Age 11 to 16 Challenge Level:
What's the largest volume of box you can make from a square of paper?
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Pick's Theorem
##### Age 14 to 16 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Beelines
##### Age 14 to 16 Challenge Level:
Is there a relationship between the coordinates of the endpoints of a line and the number of grid squares it crosses? | HuggingFaceTB/finemath | |
6.3 Rate Equations - Kinetics
Kinetics6.3 Rate Equations
A summary of the differential and integrated rate equations for the different reaction orders, along with their derived half-lives is given in Table 1.11.
6.3.1 Differential Rate Equation
For the reaction
The rate law of the form
can be expressed as
(6.5)
which constitutes the differential form of the rate equation.
The exponent in the rate law, m, determines the order of the reaction; it is said to proceed with the m-th order. The differential rate equations for the different reaction orders are:
(6.6)
(6.7)
(6.8)
(6.9)
6.3.2 Integrated Rate Expression for the First Order Rate Law
In order to derive an expression for a given rate law that allows calculation of numerical values for the concentration of a reactant over time, the differential rate expression needs to be integrated:
To integrate the differential 1st-order rate law
(6.7)
one first identifies the two variables that are of interest for the integration; here, these are c(A) and t. The equation is then re-arranged such that the two variables get isolated on opposite sides:
The integration has to be carried out with respect to the two differential variables, i.e. on the left side and dt on the right side. The factors and k are constants with respect to the integration and thus can be taken outside the integral. In order to calculate determined integrals, one also needs to specify the boundaries within which the integration is supposed to happen. We start at time t 0 where the concentration of A is the initial concentration c 0(A), and proceed up to time t, where the concentration of A is c t (A):
(6.10)
The integral on the left side, , corresponds to the type ∫x −1dx and is resolved to lnx. On the right hand side, ∫dt is of the type ∫dx which resolves to x (see Table A.2). Equation 6.10 with determined integrals thus yields:
(6.11)
The expressions of the resolved integrals need to be calculated as per:
Equation 6.11 then yields:
With t 0 = 0 one obtains:
In order to calculate a numerical value for the only unknown, c t (A), we isolate the expression for c t (A) on one side:
(6.12)
(6.13)
Equation 6.13 shows that if we measure the concentration of A, c t (A), at various time points t, and plot ln c t (A) versus t, we obtain a line with a negative slope, (−νA·k), and a y-intercept of ln c 0(A) (Fig. 6.2). Linear relationships in xy plots are a convenient way of analysing data, and historically (before the advent of the computer) the only way to determine numerical values of fitting constants.
Fig. 6.2
Analysis of 1st-order kinetics by means of linear regression
With modern fitting software, it is, of course, possible to fit non-linear relationships. If Eq. 6.12 is resolved for c t (A), one obtains:
(6.14)
(6.15)
A closer look at Eqs. 6.13 and 6.15 reveals that the experimentally determined rate constant is indeed (νA·k) and includes the stoichiometric coefficient! This needs to be taken into account when calculating the rate constant k from experimental data.
6.3.3 Half-lives and Time Constants
Half-lives and time constants are secondary parameters derived from the rate equations and can be calculated for reactions of any order.
The half-life t 1/2 describes the time required to reduce the concentration of a reactant to 1/2 of its initial value. The time constant τ describes the time required to reduce the concentration of a reactant to of its initial value.
For 1st-order chemical reactions, the half-life and time constant are particularly useful indicators for the rate of the reaction, since with 1st-order reactions the two parameters do not depend on the initial concentration of reactants is the half-life t 1/2 of a reactant.
In order to derive an expression for half-life of a 1st-order reaction, we consider its definition:
Using this relationship in the 1st-order rate Eq. 4.13 yields:
Making use of a logarithm rule that multiplies (−1) into the argument of the logarithm yields:
and thus
Irrespective of the initial concentration, the time it takes in a 1st-order reaction to reduce the initial concentration to half its value is always .
Earlier, we have introduced the time constant τ of a reaction as the time it takes for the concentration of a reactant to fall to (e ≈ 2.71828) of its initial value. As above in the case of the half-life, we use this relationship and apply it to the 1st-order rate law 4.13 which yields:
(6.17)
6.3.4 Integrated Rate Expression for the Second Order Rate Law
The differential 2nd-order rate law is given by Eq. 6.8:
(6.8)
We re-arrange the equation such that the two variables c(A) and t get isolated on opposite sides:
and integrate from time t0 where the concentration of A is c 0 (A) to time t, where the concentration of A is c t (A):
(6.18)
The integral on the left side, corresponds to the type ∫x −2dx and is resolved to . On the right hand side, ∫dt is of the type ∫dx which resolves to x (see Table A.2). Equation 6.18 with determined integrals thus yields:
With t 0 = 0, this yields:
(6.19)
or, when resolving for c t (A):
(6.20)
Equation 6.19 shows that for the second order rate law, a linear correlation is obtained when plotting versus t, where the slope is νAk and the y-intercept is (Fig. 6.3).
Fig. 6.3
Analysis of 2nd-order kinetics by means of linear regression
6.3.5 Integrated Rate Expression for the Third Order Rate Law
The differential 3rd-order rate law is given by Eq. 6.9:
(6.9)
For integration, we again isolate the two variables, c(A) and t, on opposite sides:
and integrate from [t 0 , c 0 (A)] to [t, c t (A)]:
(6.21)
The integral on the left side, corresponds to the type ∫x −3dx and is resolved to . On the right hand side, is of the type ∫dx which resolves to x (see Table A.2). Equation 6.21 with determined integrals thus yields:
(6.22)
Therefore, a plot of against t will yield a line with slope 2 ⋅ νAk and intercept (Fig. 6.4).
Fig. 6.4
Analysis of 3rd-order kinetics by means of linear regression
6.3.6 Integrated Rate Expression for the Zero-th Order Rate Law
In order to integrate the differential rate law of zero-th order
(6.6)
the two variables, c(A) and t, are isolated on opposite sides of the equation:
We then integrate from [t 0 ,c 0 (A)] to [t, c t (A)]:
(6.23)
The integrals on both sides side, and , correspond to the type ∫dx which resolves to x (see Table A.2). Equation 6.23 with determined integrals thus yields:
which results in
With t 0 = 0 this yields:
(6.24)
Therefore, a plot of c t (A) against t will yield a line with slope −νAk and intercept c 0(A) (Fig. 6.5).
Fig. 6.5
Analysis of 0th-order kinetics by means of linear regression
| HuggingFaceTB/finemath | |
# Work Input And Work Output And Efficiency Problems Pdf
On Sunday, April 4, 2021 5:36:23 AM
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Published: 04.04.2021
A few years ago, a major manufacturing-based conglomerate asked a gifted mathematician to join its corporate staff. He devoted many months to the assignment and also tapped the knowledge of […]. Some productivity indexes boast technical elegance and statistical precision—but have little to do with daily management decision making, or even, for that matter, the bottom line.
Most curricular materials in TeachEngineering are hierarchically organized; i. Some activities or lessons, however, were developed to stand alone, and hence, they might not conform to this strict hierarchy. Related Curriculum shows how the document you are currently viewing fits into this hierarchy of curricular materials. Figure 1. A lever, one of the six simple machines.
## Efficiency of Machines
The efficiency of a machine indicates how well its input energy is converted to useful output energy or work. It is a major factor in the usefulness of a machine and is the fraction or percentage of the output divided by the input. According to the Law of Conservation of Energy , the total output energy or work must equal the total input energy. However, some of the input energy does not contribute to the output work and is lost to such things as friction and heat. This lesson will answer those questions. Useful tool: Units Conversion.
Section Basic and Modified Brayton Cycles. The compressor pressure ratio is The turbine inlet temperature is o C. Determine a the thermal efficiency, b net power output and c back work ratio. Use the PG model for air.
## Efficiency of Machines
Simple machines make work easier, but they do not decrease the amount of work you have to do. Recall that in closed systems the total amount of energy is conserved. A machine cannot increase the amount of energy you put into it. So, why is a simple machine useful? Although it cannot change the amount of work you do, a simple machine can change the amount of force you must apply to an object, and the distance over which you apply the force. In most cases, a simple machine is used to reduce the amount of force you must exert to do work.
## Efficiency of Machines
Efficiency is the often measurable ability to avoid wasting materials, energy, efforts, money, and time in doing something or in producing a desired result. In a more general sense, it is the ability to do things well, successfully, and without waste. It is not something we want for its own sake, but rather because it helps us attain more of the things we value". It often specifically comprises the capability of a specific application of effort to produce a specific outcome with a minimum amount or quantity of waste, expense, or unnecessary effort.
Mechanical efficiency is a dimensionless number that measures the effectiveness of a machine in transforming the power input to the device to power output. A machine is a mechanical linkage in which force is applied at one point, and the force does work moving a load at another point. At any instant the power input to a machine is equal to the input force multiplied by the velocity of the input point, similarly the power output is equal to the force exerted on the load multiplied by the velocity of the load. Since a machine does not contain a source of energy, nor can it store energy, from conservation of energy the power output of a machine can never be greater than its input, so the efficiency can never be greater than 1. All real machines lose energy to friction ; the energy is dissipated as heat.
Mechanical advantage is defined as the ratio of resistance overcomes to the effort applied. The simple machine requires force to do work. The resistive force to be overcome is called load and the force applied to overcome the load is called effort.
Когда церковь получит все останки этого великого человека, она причислит его к лику святых и разместит отдельные части его тела в разных соборах, чтобы все могли проникнуться их величием. - А у вас здесь… - Беккер не сдержал смешка. - Да. Это очень важная часть! - заявил лейтенант.
Голос его звучал спокойно и чуточку игриво. - Откроем пачку тофу. - Нет, спасибо. - Сьюзан шумно выдохнула и повернулась к .
Если бы я сумел слегка модифицировать этот код, - продолжал Стратмор, - до его выхода в свет… - Он посмотрел на нее с хитрой улыбкой. Сьюзан потребовалось всего мгновение. Стратмор сразу заметил изумление, мелькнувшее в ее глазах, и взволнованно изложил свой план: - Если бы я получил ключ, то смог бы взломать наш экземпляр Цифровой крепости и внести нужные изменения… - Черный ход, - сказала Сьюзан, мгновенно забыв о том, что Стратмор ей лгал. Она все поняла. - Вроде Попрыгунчика.
Действительно хорошая новость. ГЛАВА 54 - Пусти. А потом раздался нечеловеческий крик.
Сьюзан в ужасе смотрела на экран.
Розы, шампанское, широченная кровать с балдахином. Росио нигде не. Дверь, ведущая в ванную, закрыта.
Единственной проблемой оставался Хейл. Чатрукьян посмотрел на комнату Третьего узла - не следит ли за ним криптограф. - Какого черта, - промычал он себе под нос.
Если Танкадо убьют, этот человек опубликует пароль. - Его партнер опубликует ключ? - недоуменно переспросила Сьюзан. Стратмор кивнул: - Он разместит его в Интернете, напечатает в газетах, на рекламных щитах.
for pdf and pdf
### Introduction to accounting and finance geoff black pdf
23.12.2020 at 22:38
### Probability questions and solutions pdf
31.12.2020 at 08:49
### Introduction to microbiology and microorganism pdf
17.05.2021 at 18:47 | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Combinatorics similar to Online:
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### There are 40 results
Broad Topics > Decision Mathematics and Combinatorics > Combinatorics
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Greetings
##### Stage: 3 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
##### Stage: 3 Challenge Level:
Is it possible to use all 28 dominoes arranging them in squares of four? What patterns can you see in the solution(s)?
### Cube Paths
##### Stage: 3 Challenge Level:
Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B?
### Tri-colour
##### Stage: 3 Challenge Level:
Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
### Flagging
##### Stage: 3 Challenge Level:
How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours?
### Euromaths
##### Stage: 3 Challenge Level:
How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Bell Ringing
##### Stage: 3 Challenge Level:
Suppose you are a bellringer. Can you find the changes so that, starting and ending with a round, all the 24 possible permutations are rung once each and only once?
### Magic Caterpillars
##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Olympic Magic
##### Stage: 4 Challenge Level:
in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?
### Painting Cubes
##### Stage: 3 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Master Minding
##### Stage: 3 Challenge Level:
Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions?
### Ways of Summing Odd Numbers
##### Stage: 3
Sanjay Joshi, age 17, The Perse Boys School, Cambridge followed up the Madrass College class 2YP article with more thoughts on the problem of the number of ways of expressing an integer as the sum. . . .
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### An Investigation Based on Score
##### Stage: 3
Class 2YP from Madras College was inspired by the problem in NRICH to work out in how many ways the number 1999 could be expressed as the sum of 3 odd numbers, and this is their solution.
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Russian Cubes
##### Stage: 4 Challenge Level:
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
### Doodles
##### Stage: 4 Challenge Level:
Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?
### Penta Colour
##### Stage: 4 Challenge Level:
In how many different ways can I colour the five edges of a pentagon red, blue and green so that no two adjacent edges are the same colour?
### Snowman
##### Stage: 4 Challenge Level:
All the words in the Snowman language consist of exactly seven letters formed from the letters {s, no, wm, an). How many words are there in the Snowman language?
### Paving Paths
##### Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### Tangles
##### Stage: 3 and 4
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Deep Roots
##### Stage: 4 Challenge Level:
Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
### Lost in Space
##### Stage: 4 Challenge Level:
How many ways are there to count 1 - 2 - 3 in the array of triangular numbers? What happens with larger arrays? Can you predict for any size array?
### Links and Knots
##### Stage: 4 and 5
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots, prime knots, crossing numbers and knot arithmetic.
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### Permute It
##### Stage: 3 Challenge Level:
Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers.
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
##### Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
### One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
### Euler's Officers
##### Stage: 4 Challenge Level:
How many different ways can you arrange the officers in a square?
### Counting Binary Ops
##### Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
### Scratch Cards
##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### In a Box
##### Stage: 4 Challenge Level:
Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair? | HuggingFaceTB/finemath | |
Computer Science
Convert the following binary numbers to decimal (6 points):
11111
101011
1101011
Convert the following decimal numbers to binary (6 points):
49
367
1023
Convert the following hex numbers to decimal (4 points):
ACE
800
Convert the following decimal numbers to hex (4 points):
64,206
16,383
If b is a binary number, what number is represented by b0? That is, what number do you get when you place a 0 at the end of a binary number? What happens if you place two 0's at the end? Generalize your answer: what happens if you place n 0's at the end? (3 points)
Determine the binary representation of the following integers using 8 bits (6 points):
-15
-64
-127
1. 👍 0
2. 👎 0
3. 👁 99
1. 1). binary to decimal
11111 = 2^4 2^3 2^2 2^1 2^0 = 16+8+4+2+1 = 31
Ans. 31
101011 = 43
1101011 = 107
2). decimal to binary
49 = 00110001
367 = 0000000101101111
1023 = 0000001111111111
3). Hex to decimal
ACE = 2766
800 = 2048
4). Decimal to hex
64,206 = FACE
16,383 = 3FFF
5). Binary using octal
-15 = -1101
-64 = -110100
-127 = -1010111
1. 👍 0
2. 👎 0
posted by Excel
Similar Questions
1. Math
please convert decimal numbers to binary 32.75 10th power
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2. COMPUTER DATA NETWORKS
A firm is assigned the network part 128.171. It selects a 10-bit subnet part. a) Draw the bits for the four octets of the IP address of the first host on the first subnet. (Hint: Use Windows Calculator.) b) Convert this answer
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3. computers networking
A firm is assigned the network part 128.171. It selects a 10-bit subnet part. a) Draw the bits for the four octets of the IP address of the first host on the first subnet. (Hint: Use Windows Calculator.) b) Convert this answer
asked by Anonymous on May 5, 2013
4. Networking
firm is assigned the network part 128.171. It selects a 10-bit subnet part. a) Draw the bits for the four octets of the IP address of the first host on the first subnet. (Hint: Use Windows Calculator.) b) Convert this answer into
asked by Steve on April 22, 2014
5. computers networking
A firm is assigned the network part 128.171. It selects a 8-bit subnet part. a) Draw the bits for the four octets of the IP address of the first host on the first subnet. (Hint: Use Windows Calculator.) b) Convert this answer into
asked by Nguyen on May 8, 2009
6. maths
Convert binary num into decimal 1100
asked by highgc on January 17, 2015
7. Computers
Convert from binary to decimal (1000)2 - (0011)2 = (?)10
asked by Mike on December 11, 2017
8. math
1:how do you order 3/4 , 7/8 ,and 13/6 from least to greatest. 2:how do you convert 2 2/4 ito an improper fractin. 3:how do you convert 3% into a decimal and how do you convert 4.56 into a percent ... i am in 6th grade by the way
asked by makiya on February 18, 2013
9. math help
emaily says that she can convert 18/25 to a decimal by using equivalent fractions instead of dividing 18 by 25. Use emily's method to convert 18/25 to a decimal. I am confused because there is not just 1 number equivalent to both
asked by reece on September 11, 2014
10. Converting decimal(floating point) to octal/hex
Hi everybody, Can anybody show me how to convert 0.59375(decimal) to Octal and Hexadecimal? go to wikpediaa and look up Hexadecimal and Octal It will give you an idea Hi can anyone convert 5.4 * 10 ^-79 to IBM hex floating point
asked by David on August 10, 2006
More Similar Questions | HuggingFaceTB/finemath | |
# Since the z-transform of the zero order hold is 1, why bother including it in analysis or simulation?
The $z$-transform of the zero order hold is 1. So why should it even be considered in the discrete time analysis or simulation of discrete time control systems?
$$Z\left(\frac{1-e^{-sT}}s\right)=\frac{z}{z-1}-z^{-1}\frac{z}{z-1}=1$$
• There is no z transform for ZOH, since it is not a discrete-time system, it is continuous time system meant to model reconstruction of discrete time signal into continuous time signal. – Eugene Sh. Feb 10 '15 at 20:37
• @EugeneSh. Thanks - you actually solved my main problem. My system is purely discrete, so no need for a ZOH. The interesting thing is it is used as a device to 'bridge' discrete and continuous components, but taken in isolation its Z-transform is indeed 1. Or did I make a mathematical error? – docscience Feb 10 '15 at 21:16
• The point is that when sampled, the continuous function f(t) and ZOH(f(t)) will yield the same data points, so when performing the Z-transform on both, you will get the same result, so this is the meaning of your 1 result, I guess. – Eugene Sh. Feb 10 '15 at 21:19
• @EugeneSh. To further add to the mystery of the ZOH you will find that some text books include $T$ as a factor in the denominator, which makes more sense to me. There's actually been heated discussions on whether $T$ should be included or not. – docscience Feb 10 '15 at 21:22
The ZOH TF above is a link between continuous and discrete domains in hybrid systems. This is the most convenient mechanism for representing a hybrid system in transfer function form. There is not, of course, a one-to-one relationship between $s$ and $z$ domains, hence it's a mathematical convenience. In the relationship above, the exponential term should be negative (not positive as given), giving a $z$-equivalent of $1-\exp(-sT)$ as $(z -1)/z$ to be included with purely discrete blocks (filters, etc) and the $1/s$ part of the ZOH should be included with the other continuous $s$-blocks. The $s$-TF of the continuous elements is then transformed into the $z$-domain, giving an overall $z$-TF.
If anybody cares, it's interesting to do the z-transform of the ZOH at a DIFFERENT rate. In other words, say you have a digital signal $X_{in}$ that is digitized at 50 Hz, so $T_1 = 0.02 \mathrm{s}$, each sample of the signal is 20 ms apart. Say you have a computer program that is operating on this signal at 100 Hz (let $T_2 = 0.01 \mathrm{s}$). What is happening in the computer is that a variable $X_{in}$ that changes every 20 ms is being sampled by your a program every 10 ms and some operation is performed producing an output $X_{out}$. Since the value of $X_{in}$ simply holds its value until it changes every 20 ms, you can consider this to be a ZOH:
$(1 - e^{-sT_1})/s$
Now, you have to redigitize at the $1/T_2$ rate. The numerator simply becomes
$(1-z_1^{-1})$
and the numerator (through s-to-Z transform):
$\frac{1}{(1-z_2^{-1})}$
resulting in:
$\frac{(1-z_1^{-1})}{(1-z_2^{-1})}$
Let me go back a bit and define $z_1$ and $z_2$.
Since, in general, $z = e^{-sT}$, define $z_1 = e^{-sT_1}$, and $z_2 = e^{-sT_2}$
Noting that $T_1 = 2 T_2$, $z_1 = e^{-sT_1} = e^{-s2T_2} = e^{2(-sT_2)} = z_2^2$
substitute $z_2^2$ for $z_1$ :
$\frac{1-z_1^{-1}}{1-z_2^{-1}} = \frac{1-z_2^{-2}}{1-z_2^{-1}} = \frac{(1-z_2^{-1})(1+z_2^{-1})}{1-z_2^{-1}} = 1+z_2^{-1}$
So, the final result is actually very intuitive when you think about it. The program that is upsampling $X_{in}$ from 50 to 100 Hz is simply taking the original copy $X_{in}$ (the '1' part of the result) and adding to it a copy of the same sample delayed by one $T_2$ sample step (the $z_2^{-1}$ part). A simple example:
Say $X_{in} = [1;2;3;4]$, with sample spacing of $T = 20\mathrm{ms}$ Then $X_{in}$ upsampled with the ZOH would look like
$X_{inupzoh} = [1;1;2;2;3;3;4;4]$ with sample spacing of $T = 10\mathrm{ms}$
NOTE, that WITHOUT the ZOH, (say $X_{in}$ was cleared every time it was read), the upsampled $X_{in}$ would be:
$X_{inup} = [1;0;2;0;3;0;4;0]$
Now for the fun part (if you stuck with me): If you have access to a tool (like Matlab) to do ffts and bode plots:
Do the fft of $X_{in}=[1;2;3;4]$, as well as $X_{inup}$ and $X_{inupzoh}$. You can look at the real and imag parts of the fft results or convert the fft results to magnitude and phase and see the surprising results.
What you should see is that upsampling without a ZOH simply moves the Nyquist frequency. You have to remember that the spectrum of a digital signal is periodic and infinite, that means that even though we usually only look at the spectrum from $0$ to $\frac{T_s}{2}$, that spectrum is actually repeating every $\frac{kT_s}{2}$ on the freq axis. Upsampling without the ZOH didn't actually change the signal (it only inserted zeros), but it DID move the Nyquist freq. Then you can see that ADDING another copy of the signal delayed by one sample point has an interesting gain AND phase effect. At low frequencies the gain is actually doubled, and the gain doesn't get to unity until $\frac{2}{3}\times 50\mathrm{Hz}$ and drops off after that. This is a very real effect that CAN cause problems for some control systems that are gain and/or phase sensitive.
Downsampling is a bit more complicated mathematically... I've typed too much already....
• Zero stuffing followed by a low pass filter with a gain of 2 is actually a good way to double the sampling rate in a streaming DSP chain. – Simon Richter Apr 1 '16 at 23:14 | HuggingFaceTB/finemath | |
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# What is the remainder when x^1999 is divided by (x^2)-1?
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Solution
## (x)= x^1999 factors of x^2-1=(x+1)(x-1) we know, dividend=divisor*quotient+remainder since p(x) is a polynomial of degree 1999 so it will leave a linear remainder in form of ax+b where a and b are constants. x^1999=divisor*(x+1)(x-1)+ax+b put x=1 so, 1=a+b ..................(i) put x=-1 so, -1=-a+b .................(ii) solving equations we get, a=1 and b=0 so remainder=ax+b=1*x+0=x so when x^1999 is divided by x^2-1 then it leaves remainder as x.
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# What is the volume occupied by a 91.8*g mass of carbon dioxide that is heated to 517.2*K, and subjected to a pressure of 287*"Torr"?
Jun 20, 2017
$V \cong 235 \cdot L$
#### Explanation:
Given the Ideal Gas equation.....$P V = n R T$
So $V = \frac{n R T}{P}$, and we know that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$
V=((91.8*g)/(44.01*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx517.2*K)/((287*"Torr")/(760*"Torr"*atm^-1))=??L.
The gas constant conveniently gives us the volume in $\text{litres}$.
Note here that I have (i) converted the pressure in $\text{Torr}$, and (ii) converted the temperature to $\text{degrees Kelvin}$, which are typically used in these reactions. | HuggingFaceTB/finemath | |
Question: Which Is Greater CP Or CV?
What is the difference between a CP and a CV?
CV: CV is the amount of heat energy that a substance absorbs or releases (per unit mass) with the change in temperature where a volume change does not occur.
CP: CP is the amount of heat energy that a substance absorbs or release (per unit mass) with the change in temperature where a pressure change does not occur..
Can CP ever be less than CV if so give an example?
Cp is always greater than Cv for the same temperature difference because: For Cp, i.e., heat capacity for constant pressure process, take for example piston cylinder arrangement. … But for Cv, i.e., heat capacity for constant volume process, take for example insulated rigid container.
What is the CP CV ratio?
In thermal physics and thermodynamics, the heat capacity ratio, also known as the adiabatic index, the ratio of specific heats, or Laplace’s coefficient, is the ratio of the heat capacity at constant pressure (CP) to heat capacity at constant volume (CV). … The pressure inside is equal to atmospheric pressure.
What is CP of steam?
Steam Cp=1.8723 kJ/kg. K Cv=1.4108 kJ/kg. K – see steam tables. At IUPAC standard temperature and pressure (0 °C and 101.325 kPa), dry air has a density of 1.2754 kg/m3.
How do you get CP and CV?
From the ideal gas law, P V = nRT, we get for constant pressure d(P V ) = P dV + V dP = P dV = nRdT . Substituting this in the previous equation gives Cp dT = CV dT + nRdT . Dividing dT out, we get CP = CV + nR .
How do you calculate CP CV for gas mixture?
Helium (He) is a monoatomic gas and dioxygen (O2) is a diatomic gas.For a monoatomic gas and a diatomic gas, value of Cv are (3/2)R and (5/2)R respectively.Hence for the given mixture, (Cv)mix = [2. … Now since Cp- Cv = R hence Cp = Cv + R = (17) R / 6.So we get, Cp / Cv = (17 / 11).More items…
Can CP be less than CV?
Cp is greater than the molar specific heat at constant volume Cv because energy must now be supplied not only to raise the temperature of the gas but also for the gas to do work. … More heat would be required at constant pressure to cause the same temperature rise and Cp will be greater than Cv.
Does CP depend on pressure?
Cp is (dH over dT) at constant pressure. Let’s start from enthalpy as a function of temperature and pressure. Then, the total differential of enthalpy is like this. … So the temperature dependence of this function, dH over dP, gives pressure dependence of Cp.
Is CP a CV?
So, Cp represents the molar heat capacity, C when pressure is constant. … Cv represents the molar heat capacity C when volume is constant. Under a constant volume, the volume of a substance does not change, so the change in volume is zero.
What is CV equal to?
molar specific heatCv is the molar specific heat at constant Volume means the amount of heat transfer required to raise temperature of one mole of a gas by 1 K at constant Pressure.
What is the value of CP?
The nominal values used for air at 300 K are CP = 1.00 kJ/kg. K, Cv = 0.718 kJ/kg. K,, and k = 1.4.
What is CV for an ideal gas?
The molar specific heat capacity of a gas at constant volume (Cv) is the amount of heat required to raise the temperature of 1 mol of the gas by 1 °C at the constant volume. … Its value for monatomic ideal gas is 5R/2 and the value for diatomic ideal gas is 7R/2. | HuggingFaceTB/finemath | |
# physics
A lamp hangs vertically from a cord in a descending elevator that declerates at 2.4m/s^2 a) if the tension in the cord is 89N what is the lamp's mass? b) what is the cord's tension when the elevator is ascends with an upward acceleration of 2.4m/s^2?
I know T=w where w=mg. Would I use T-mg=ma? I should be getting 31.3 kN for a) and 24.3kN for part b but I can't figure out my problem.
Yes, use
T - mg = ma.
T does NOT equal mg when the elevator is accelerating or decelerating.
In the first case,
89 N = m (g + a)
m = 89/(9.8 + 2.4) = 7.3 kg
I used a plus sign for "a" since it is decelerating and descending
The (a) problem asks for a mass, so I don't see how the answer can be in Newtons.
1. 👍 0
2. 👎 0
3. 👁 132
1. Acceleration used should be -2.4.
If you used acceleration as +2.4, the tension should be negative according to its free diagram
mg-t=ma , which yields to 12.02 kg
1. 👍 0
2. 👎 0
posted by P6
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CLASS_11_MATHS_SOLUTIONS_NCERT
Class XI Chapter 14 –
Class XI Chapter 14 – Mathematical Reasoning Maths ______________________________________________________________________________ Exercise 14.4 Question 1: Rewrite the following statement with “ if-then †in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd. Solution 1: The given statements can be written in five different ways as follows. (i)A natural number is odd implies that its square is odd. (ii)A natural number is odd only if its square is odd. (iii)For a natural number to be odd, it is necessary that its square is odd. (iv)For the square of a natural number to be odd, it is sufficient that the number is odd. (v)If the square of a natural number is not odd, then the natural number is not odd. Question 2: Write the contrapositive and converse of the following statements. (i) If x is a prime number, then x is odd. (ii) It the two lines are parallel, then they do not intersect in the same plane. (iii) Something is cold implies that it has low temperature. (iv) You cannot comprehend geometry if you do not know how to reason deductively. (v) x is an even number implies that x is divisible by 4 Solution 2: (i) The contrapositive is as follows. If a number x is not odd, then x is not a prime number. The converse is as follows. If a number x is odd, then it is a prime number. (ii) The contrapositive is as follows. If two lines intersect in the same plane, then they are not parallel. The converse is as follows. If two lines do not intersect in the same plane, then they are parallel. (iii) The contrapositive is as follows. If something does not have low temperature, then it is not cold. The converse is as follows. If something is at low temperature, then it is cold. (iv) The contrapositive is as follows. If you know how to reason deductively, then you can comprehend geometry. The converse is as follows. If you do not know how to reason deductively, then you cannot comprehend geometry. Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 14 – Mathematical Reasoning Maths ______________________________________________________________________________ (v) The given statement can be written as follows. If x is an even number, then x is divisible by 4. The contrapositive is as follows. If x is not divisible by 4, then x is not an even number. The converse is as follows. If x is divisible by 4, then x is an even number. Question 3: Write each of the following statement in the form “ if-then â€. (i) You get a job implies that your credentials are good. (ii) The Banana trees will bloom if it stays warm for a month. (iii) A quadrilateral is a parallelogram if its diagonals bisect each other. (iv) To get A+ in the class, it is necessary that you do the exercises of the book. Solution 3: (i)If you get a job, then you credentials are good. (ii)If the Banana tree stays warm for a month, then it will bloom. (iii)If the diagonals of a quadrilateral bisect each other, then it is parallelogram. (iv)If you want to get an A + in the class, then you do all the exercise of the book. Question 4: Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other. (a) If you live in Delhi, then you have winter clothes. (i) If you do not have winter clothes, then you do not live in Delhi. (ii) If you have winter clothes, then you live in Delhi. (b) If a quadrilateral is a parallelogram, then its diagonals bisect each other. (i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram. (ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Solution 4: (a) (i) This is the contrapositive of the given statement (a). (ii) This is the converse of the given statement (a). (b) (i) This is the contrapositive of the given statement (b). (ii) This is the converse of the given statement (b). Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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Home | Math | 10 Examples of Vertical Angles
# 10 Examples of Vertical Angles
October 14, 2023
written by Rida Mirza
In mathematics, vertical angles are a important concept. Vertical angles are formed when two lines intersect, and their opposing angles are congruent.
In this article, we will discuss ten examples of vertical angles in mathematics.
## Examples of Vertical Angles
These are 10 examples of vertical angles.
### 1: Open Scissors
When you open a pair of scissors, the two blades form an “X” shape. The angles formed at the intersection of the blades are vertical angles.
### 2: Book Pages
When you open a book, the pages often form vertical angles at the center crease.
The angle formed by each page with the center crease is a pair of vertical angles.
### 3: Carpenter’s Square
Carpenter’s squares, commonly used in woodworking and construction, have two arms that meet at a right angle.
The angles formed when the arms intersect are vertical angles.
### 4: Street Signs
Poles that support street signs often intersect with the horizontal and vertical bars, forming vertical angles.
### 5: Picture Frames
Picture frames with corners create vertical angles at the corners.
### 6: Traffic Lights
Bars holding traffic lights at intersections meet at specific angles. These angles are vertical angles.
### 7: Fences and Gates
Fences and gates often have vertical posts intersected by horizontal beams or bars. The angles at the intersection points are vertical angles.
### 8: Computer Screens
In dual-monitor setups, the screens are often aligned at an angle to meet the viewer’s eye level.
The angle at which the screens intersect is a pair of vertical angles.
Railroad tracks on a straight stretch of railway form pairs of vertical angles where they intersect.
### 10: Sports Field Goalposts
In sports like football and soccer, the goalposts are supported by vertical posts. The angles at which the crossbar intersects with the vertical posts are vertical angles.
File Under: | HuggingFaceTB/finemath | |
Associated Topics || Dr. Math Home || Search Dr. Math
### Handshakes and Polygon Diagonals
```
Date: 09/12/2001 at 20:42:42
From: Brendan
Subject: If a polygon has # sides, how many diagonals will it have?
Dear Dr. Math,
I can't figure this one out!
If a polygon has 42 sides, how many diagonals does it have?
```
```
Date: 09/13/2001 at 05:40:04
From: Doctor Pete
Subject: Re: If a polygon has # sides, how many diagonals will it
have?
Hi,
Let's look at a hexagon, with 6 sides. Pick a vertex. How many
diagonals come from that vertex? (Three.) How many vertices in a
hexagon? (Six.)
So there should be 3*6 = 18 diagonals, but in fact if you count them
there are only half as many. Why? Well, since a single diagonal joins
exactly two vertices, for each vertex we counted the same diagonal
twice.
So look at a 42-gon.
How many diagonals come from a single vertex?
How many vertices are there in a 42-gon?
How many diagonals are in a 42-gon?
Let's look at something similar to this. If you have 100 people at a
party, and everyone shakes hands with everyone else, how many
handshakes take place? Well, pick a person. He shakes hands with 99
others. In fact, each person shakes hands with 99 others, so there
should be 100*99 = 9900 handshakes; but again, it takes two people to
shake hands, so this is double the correct amount, which is 4950
handshakes.
By now the pattern should be clear. If there are n people at a party,
the number of handshakes is n(n-1)/2.
Now, there is a different way to count these handshakes. The first
person shakes hands with 99 others. The second person shakes hands
with 98 others, not including the first person, whom we have already
counted. The third person shakes hands with 97 others, not including
the first two. And so on, until the 99th person, who shakes hands with
the 100th person. So there are
99 + 98 + 97 + ... + 2 + 1
handshakes. But we have seen that this is equal to 100*99/2, and this
suggests a formula. In general, if there are n people at a party, then
counting the number of handshakes in two different ways gives the
formula
1 + 2 + ... + (n-2) + (n-1) = n(n-1)/2.
Finally, we can prove this summation in yet another way. Write
S = 1 + 2 + ... + (n-2) + (n-1)
S = (n-1) + (n-2) + ... + 2 + 1
----------------------------------------
2S = n + n + ... + n + n,
and since there are n-1 "n"'s, then we have 2S = n(n-1), or
S = n(n-1)/2.
However, the handshake analogy isn't exact. At a party, each person gets to
shake hands with everyone but himself. But in a polygon, each vertex
doesn't get to form a diagonal with himself... or with the two vertices
nearest him. (He's already connected to them by edges.) So for a
polygon, we have to subtract the number of edges from the total for
the handshake problem:
S = n(n-1)/2 - n
= [n(n-1) - 2n]/2
= [n(n - 1 - 2)]/2
= n(n-3)/2
- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Permutations and Combinations
High School Triangles and Other Polygons
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | HuggingFaceTB/finemath | |
# 139 imperial cups in UK tablespoons
## Conversion
139 imperial cups is equivalent to 2632.94379166667 UK tablespoons.[1]
## Conversion formula How to convert 139 imperial cups to UK tablespoons?
We know (by definition) that: $1\mathrm{brcup}\approx 18.9420416666667\mathrm{brtablespoon}$
We can set up a proportion to solve for the number of UK tablespoons.
$1 brcup 139 brcup ≈ 18.9420416666667 brtablespoon x brtablespoon$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{brtablespoon}\approx \frac{139\mathrm{brcup}}{1\mathrm{brcup}}*18.9420416666667\mathrm{brtablespoon}\to x\mathrm{brtablespoon}\approx 2632.943791666671\mathrm{brtablespoon}$
Conclusion: $139 brcup ≈ 2632.943791666671 brtablespoon$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 UK tablespoon is equal to 0.000379803018645907 times 139 imperial cups.
It can also be expressed as: 139 imperial cups is equal to $\frac{1}{\mathrm{0.000379803018645907}}$ UK tablespoons.
## Approximation
An approximate numerical result would be: one hundred and thirty-nine imperial cups is about two thousand, six hundred and thirty-two point nine four UK tablespoons, or alternatively, a UK tablespoon is about zero times one hundred and thirty-nine imperial cups.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | HuggingFaceTB/finemath | |
Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago
# physics help?
Two children are playing on a balanced seesaw. One child with a mass of 43 kg sits 1.6 m from the center. Where on the other side must the second child sit if her mass is 32 kg?
Relevance
• 43*1.6=68.8kg m/32=21.5m from center.
• Moments clockwise must equal moments anticlockwise for balacing to occur
i.e. 43 x 1.6 = 32 x d
so, d = 68.8/32 => 2.15 metres from centre
:)>
• 43*1.6 = 32*d
d = 43*1.6/32 = 2.1500000 m
Now you must round answer to 2 sigfigs, never answer with more digits than given!
2.2 m is correct
• balance implies :
43*1.6 = 32*d
d = 1.6*43/32 = 43/20 = 4.3/2 = 2.15 m | HuggingFaceTB/finemath | |
# Is Jensen's inequality an iff condition on convex functions?
According to wikipedia this is Jensen's inequality:
If X is a random variable and φ is a convex function, then:
$$\varphi\left(\mathbb{E}[X]\right) \leq \mathbb{E}\left[\varphi(X)\right].$$
Which stated as an implication reads as follows:
If Convex $\implies$ $\varphi\left(\mathbb{E}[X]\right) \leq \mathbb{E}\left[\varphi(X)\right].$
However, I was wondering if the converse was also true.
$\varphi\left(\mathbb{E}[X]\right) \leq \mathbb{E}\left[\varphi(X)\right]$ $\implies$ Convex.
I thought it would be an iff because of the definition of convexity. Recall what convexity means (and let me explain why I thought iff for jensen's):
f is called convex if: $$\forall x_1, x_2 \in X, \forall t \in [0, 1]: \qquad f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2).$$
Since its a definition, the the word convex and its definition statement are logically equivalent. Therefore, its an iff, i.e.
f is called convex $\Leftrightarrow \forall x_1, x_2 \in X, \forall t \in [0, 1]: \qquad f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2).$
right? Does that logic generalize to Jensen's inequality or am I wrong for both?
Context:
The reason that I care about this is because I wanted to proof that a multivariable function f is convex. My plan of attack was to show Jensen's inequality "worked" on it and since Jensen's inequality and convexity are logically equivalent, then I can conclude that f is convex. But that only works if its an iif. Also, I absolutely wanted to avoid taking the Hessian or derivatives to prove it was convex, I strictly wanted to stick with the "original" definition of convexity to prove this (i.e. the inequality definition).
• If the inequality holds for all random variables (with expectation), then $\varphi$ is convex. – Daniel Fischer Nov 17 '14 at 15:57
• @DanielFischer how is that not a contradiction to Seyhmus Güngören answer bellow? It seems he said that Jensen's cannot imply convexity but you say it does. I am confused right now. – Pinocchio Nov 17 '14 at 16:15
• That answer says that the inequality $\psi(\mathbb{E}[X]) \leqslant \mathbb{E}[\psi(X)]$ can hold for some random variables $X$ even if $\psi$ is not convex. [Note that in that answer, the chosen $\psi$ is convex on the range of the chosen $U$.] – Daniel Fischer Nov 17 '14 at 16:17
• Is the random variable for that answer the continuous random variable that takes values in $U$ with any probability distribution? – Pinocchio Nov 17 '14 at 16:23
The convexity of $f$ is equivalent to Jensen's inequality holding for all random variables $X$. (If it holds for all $X$, then in particular it holds for the random variable with $P(X=x_1)=t$ and $P(X=x_2)=1-t$, where Jensen reduces to the definition of convexity.) So if somehow it's easier to prove Jensen for all $X$ (perhaps because the generality of that statement clears away distractions), then sure, your method is sound.
That is not true take $U$ uniform in $[0,1]$ and consider the function $f(x)=x^3$. Then one can see that $E[\psi(X)]\geq \psi(E[X])$ but $f$ is non-convex.
In general convexity is a much stronger argument than Jensen's inequality.
• Sorry if this is a very easy question, but what does it mean that "convexity is a much stronger argument/statement". Are you using the word stronger in a precise way or is it just intuitive? Or both. Either way not sure what it means. Btw, thanx for your response. – Pinocchio Nov 17 '14 at 15:58
• Stronger in the sense that in order to imply covexity one needs to do a little bit more that Jensens inequality. More formally, it means that Jensens inequality is not strong enough to imply convexity. – Seyhmus Güngören Nov 17 '14 at 16:01
• Seyhmus, I apologize but a comment on my question confused me. Daniel Fischer in the comments section said: "If the inequality holds for all random variables (with expectation), then φ is convex.", however, you said that jensen's inequality is not strong enough to imply convexity, but he said it can. Currently, I am confused, so can it or can it not imply convexity? Thanks for your time :) – Pinocchio Nov 17 '14 at 16:18
• there is no contradiction. Hear what he is saying. – Seyhmus Güngören Nov 17 '14 at 16:24
• Oh ok, I think I get it, so its an iff only when its true for all r.vs X. Right? Thanks for your patience :) – Pinocchio Nov 17 '14 at 16:27
It depends on the underlying probability space.
For example, if the measure is $\delta_{1\over2}$ on $[0,1]$, then $E X= X({1 \over 2})$, $E \phi(X) = \phi(X({1 \over 2}))$, so Jensen's inequality holds whether or not $\phi$ is convex.
For all $t\in [0,1]$ you need to be able to find a set $A$ such that $pA = t$.
If you can do this, let $X=x_1 \cdot 1_A + x_2 \cdot 1_{X \setminus A}$. Then $EX = t x_1 + (1-t) x_2$ and $E \phi(X) = t \phi(x_1) + (1-t) \phi(x_2)$, which implies that $\phi$ is convex if Jensen's inequality holds. | HuggingFaceTB/finemath | |
# Quadratic Formula | Picture, Making use of & Instances
Quadratic Formula | Picture, Making use of & Instances
The fresh new Quadratic Algorithm is actually a milestone over the way to totally skills algebra. Knowing they, to help you worthy of it, also to utilize it correctly, you should know a tiny bit of their records, upcoming enjoy the title inside it.
## Desk Of Material
Polynomials (phrases with many terms) can have linear, rectangular, and you will cubic viewpoints. Frustration enters whenever we go through the term “quadratic” since it indicates four away from anything, particularly an effective quadrilateral. Nevertheless the supply of your word mode “while making square,” such as length times thickness. Inside mathematics, the definition out-of square are an enthusiastic exponent to your second-degree. These types of exponents are energies away from dos :
• x 2
• 2 dos
• b dos
## Quadratic Formula during the Fundamental Form
As long as a beneficial ? 0 , you need to be capable factor the brand new quadratic picture. Both, whether or maybe not, so it gets complicated otherwise dirty, or you dont factor it.
You can use that it formula to resolve quadratic equations. Otherwise, if for example the equation factored, you might make use of the quadratic algorithm to evaluate in case your options of your own quadratic equation is actually correct.
The definition of b 2 – 4 a-c , which is under the (sqrt) in the quadratic algorithm is known as the fresh discriminant. The discriminant is used to choose how many choices new quadratic picture enjoys.
• we f b dos – 4 a c = 0 > step 1 s o l you t we o n
• i f b 2 – 4 a-c > 0 > dos s o l you t i o n s
• i f b 2 – 4 a c https://datingranking.net/cs/cuddli-recenze/ n o r e a l s o l u t i o n
## Utilizing The fresh Quadratic Algorithm
Into Quadratic Algorithm to apply, the equation you’re untangling should be from the means one sets most of the variables on one hand of one’s translates to sign and 0 on the other:
Our very own quadratic picture have a tendency to foundation, so it is good starting point. Basic we foundation the fresh picture. We have been seeking to a few amounts you to definitely proliferate so you’re able to 6 and you may incorporate to help you 5 :
We could notice that both term translates to 0 (as multiplying it times others term production 0 ). We could put per phrase comparable to 0 and solve to possess x:
Comparing our analogy, x 2 + 5 x + six = 0 , toward important types of the quadratic formula (that will along with just be titled “the fresh quadratic”), we have these types of opinions:
• b = 5
• c = six
Now we can fool around with those in the fresh new Quadratic Algorithm and look, since i already know our very own email address details are – dos and you can – step 3 :
## Get a hold of X-Intercepts
Inside the a picture eg an excellent x dos + b x + c = y , lay y = 0 and make the fresh new equation. The latest you’ll be able to x-philosophy may be the x-intercepts; where you range crosses this new x-axis.
Graphing calculators will probably not be equal to the precision of the Quadratic Formula. Use the calculator to verify the rounded results, but expect them to be slightly different.
For example, assume you may have a response on Quadratic Algorithm within they. Get-off as is, in lieu of composing it a decimal comparable ( step 3.16227766 ) , having higher accuracy.
Inside resolving quadratics, your help your self of the once you understand numerous a means to resolve people equation. Initiate fixing an effective quadratic from the watching in the event it tend to basis (what a few affairs proliferate giving c that will along with sum to provide b ?).
Explore any of these steps, and graphing, to evaluate a reply derived using other means. Make use of the Quadratic Algorithm to evaluate factoring, for-instance.
For individuals who then plotted this quadratic function with the a graphing calculator, your own parabola could have an effective vertex away from ( 1.twenty-five , – ) which have x-intercepts out-of – step 1 and you will step three.5 .
Here is a quadratic that will not factor: x 2 – 7 x – 3 = 0
Zero products of – step three enhance – 7 , and that means you usually do not have fun with factoring. What direction to go? If the anything you knew is actually factoring, you would certainly be stuck. However you see to try the latest Quadratic Algorithm, with the help of our viewpoints:
## When to Make use of the Quadratic Formula
Quadratic equations are usually used each and every day. They may be used to determine portion, formulate the rate regarding an object, as well as to choose a beneficial item’s earnings. It is important that you know how to get choice to own quadratic equations using the Quadratic Algorithm.
When using the Quadratic Formula, you must be attentive to the smallest details. For example, placing the entire numerator over 2 a is not optional. Everything, from – b to the square root, is over 2 a .
That pesky b right at the beginning is tricky, too, since the Quadratic Formula makes you use – b . Try not to think of – b as ” n e g a t i v e b ” but as the opposite of whatever value ” b ” is. Suppose your b is positive; the opposite is negative. What if your original b is already negative? Think: the negative of a negative is a positive; so – b is positive!
Underneath the square-root class, in addition, you have to work with proper care. Possibly b dos was preceded from the a terrible signal, and that means you try squaring all of b, even in the event it is bad. The brand new square regarding a poor try a confident, very b 2 will always be a confident well worth.
The point: work very carefully. Keep track of your signs, work methodically, and skip nothing. Then use a different method to check your work. No matter which method you use, the Quadratic Formula is available to you every time. It will always work. | HuggingFaceTB/finemath | |
# Gibbs fenomen - faktiskt.io • Visa tråd
Debian -- Filförteckning för paketet genius-common/sid/m68k
Let’s do a quick example to verify this. Example 4 Find the Fourier series for f(x) = x2 on − L ≤ x ≤ L. EEL3135: Discrete-Time Signals and Systems Fourier Series Examples - 1 - Fourier Series Examples 1. Introduction In these notes, we derive in detail the Fourier series representation of several continuous-time periodic wave-forms. Recall that we can write almost any periodic, continuous-time signal as an infinite sum of harmoni-cally Fourier Series Sine and cosine waves can make other functions! Here two different sine waves add together to make a new wave: Try "sin (x)+sin (2x)" at the function grapher. The Basics Fourier series Examples Fourier series Let p>0 be a xed number and f(x) be a periodic function with period 2p, de ned on ( p;p).
A series R-C circuit. In Figure 1, there is a source voltage, Vs, in series with a resistor R, and a capacitor C. Se hela listan på mathsisfun.com Solved problems on Fourier series 1.
## Converging Definition - Collection The Ofy
For example, using orthogonality of the roots of a Bessel function of the first kind gives a so-called Fourier-Bessel series. 2018-06-04 · If you go back and take a look at Example 1 in the Fourier sine series section, the same example we used to get the integral out of, you will see that in that example we were finding the Fourier sine series for $$f\left( x \right) = x$$ on $$- L \le x \le L$$. Examples Inspired From Fourier Series. The “Fourier Analysis” is simply the actual process of reverse-engineering, or constructing from scratch (sin & cos) a period function with the setup above — the goal is to solve for coefficients a0, an & bn. The most commonly-seen notation for the Fourier Series looks like the above.
### Datorntverk A lektion 3 MKS B lektion 3 - SlideToDoc.com
Shopping. Tap to unmute. If playback doesn't begin shortly, try restarting your device. Fourier series + differential equations This video shows how to solve differential equations via Fourier series.
Two extensive examples; the Gauss-Bonnet theorem, and the winding number, serves as an in depth review on the result for restricted maximal operators of Vilenkin–Fourier series on martingale Hardy spaces (2019) Damage Detection in Structures – Examples (2019). Important and Fundamental Mathematics Rules with Examples. Fourier series - Wikipedia, the free encyclopedia Fysik Och Matematik, Relativitetsteori, Formatting an essay apa travel essays examples business planning in writing fourier series solved problems examples pdf problem solving >>Web Information retrieval Page Rank Fast Fourier Series Geometrical Transformation These should be given more time with full solved examples during the example, can be found in the book of Pólya and Szegö mentioned above and and by ¯K1 the Fourier transform of K1, then, if the condition ¯K1(u) = 0 holds for Fourier-serie kan dock tillämpas på fyrkantsvågen. than the ones that mathematicians can provide as counter-examples to this presumption. Fall 2004 Figure Sine wave examples Summer 2006 Computer Networks History Introduction to Information Technologies Fall 2004 Example: Fourier series connecting the subject with other areas of mathematics: the Fourier transform treated by contour Numerous examples and applications throughout its four Fourier series animation using phasor addition » Steve on Fourier Series - an understandable introduction.
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e. (x)= � f (x) for 0 ≤ x ≤ L, f (−x) for − L ≤ x<0. Even and odd extensions. • For a function f(x)defined on [0,L], the even extension of f(x)is the function.
This allows us to represent functions that are, for example, entirely above the x−axis. With a sufficient number of harmonics included, our ap- Fourier series: Solved problems °c pHabala 2012 Alternative: It is possible not to memorize the special formula for sine/cosine Fourier, but apply the usual Fourier series to that extended basic shape of f to an odd function (see picture on the left). 📒⏩Comment Below If This Video Helped You 💯Like 👍 & Share With Your Classmates - ALL THE BEST 🔥Do Visit My Second Channel - https://bit.ly/3rMGcSAWhat is A Fourier series on [-L,L] is 2L periodic, and so are all its partial sums. So, what we are really doing when we compute the Fourier series of a function f on the interval [-L,L] is computing the Fourier series of the 2L periodic extension of f.
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### Computational Aspects of Maass waveforms - Yumpu
Fourier Series. Any periodic signal y(t) may be constructed from an infinite sum of sine and cosine terms: with An and Bn being the Fourier coefficients, and T the Nov 30, 2016 Fourier series in 2-D (convergence). Proof of convergence of double Fourier series. 2 Fourier series examples.
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### serie — Translation in English - TechDico
In Figure 1, there is a source voltage, Vs, in series with a resistor R, and a capacitor C. Se hela listan på mathsisfun.com Solved problems on Fourier series 1. Find the Fourier series for (periodic extension of) f(t) = ½ 1, t ∈ [0,2); −1, t ∈ [2,4). Determine the sum of this series. 2. Find the Fourier series for (periodic extension of) f(t) = ½ t−1, t ∈ [0,2); 3−t, t ∈ [2,4). Determine the sum of this series. | HuggingFaceTB/finemath | |
# x+2=16
## Simple and best practice solution for x+2=16 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for x+2=16 equation:
1x + 2 = 16
1x = 16 - 2
1x = 14 (divide both sides by 1 to get x)
1x/1 = 14/1
x = 14
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# Array Problem
Give an array which has n integers,it has both positive and negative integers.Now you need sort this array in a special way.After that,the negative integers should in the front,and the positive integers should in the back.Also the relative position should not be changed.
eg. -1 1 3 -2 2 ans: -1 -2 1 3 2.
O(n)time complexity and O(1) space complexity is required .
What I tried so far …
public static void sortNegPosSwap(int[] arr)
{
int[] neg = new int[arr.length];
int numNeg = 0;
int currNeg = -1;
int numNegSoFar = 0;
for(int i = 0; i < arr.length; i++)
{
if(arr[i] < 0)
{
neg[numNeg++] = arr[i];
}
}
for(int i = arr.length - 1; i >= 0; i--)
{
if(numNegSoFar != 0 && arr[i] >= 0)
{
arr[i + numNegSoFar] = arr[i];
}
if(arr[i] < 0)
{
numNegSoFar++;
}
}
for(int i = 0; i < numNeg; i++)
{
arr[i] = neg[i];
}
}
But the above code is of O(nlogn) complexity . Can someone suggest better ??
What you need is a compare function with std::sort
#include
using namespace std;
bool compare(int i,int j)
{
if(i<0 && j<0)
return false;
else if(i>=0 && j>=0)
return false;
else
return i<j;
}
int main() {
int a[5]={-1,1,3,-2,2};
for(int i=0;i<5;i++)
cout<<a[i]<<" ";
cout<<"\n";
sort(a,a+5,compare);
for(int i=0;i<5;i++)
cout<<a[i]<<" ";
return 0;
}
EDIT
For your complexity constraints , you should point a variable to first positive and other to first negative . Whenever you face a positive and negative in unsorted position and move the pointers to next positive and next negative respectively.
1 Like
You say that relative order must be preserved, right? And just that negative numbers should come first and positive hereafter?
for(i=0;i<n;i++)
{
if(number is negative)
print(arr[i]);
}
for(i=0;i<n;i++)
{
if(number is positive)
print(arr[i]);
}
Did you give this algo a try?
1 Like
I understand this but I need not print the answer directly I have to store it in an array and what I am looking for is an optimized approach of O(n) time complexity . I have reduced the code to O(nlogn) as mentioned above .
Thanks a ton man …I was not able to get this idea … of using modified sort function as I was asked to code it in Java .
Thanks a lot
You are welcome . In my opinion , c++ can greatly help you in your logic development . It is your choice though .
1 Like
You said that
O(1) space complexity is required .
So I went with O(1) approach
1 Like
Yeah that’s true but I want to store it in an array too BTW with storing in an array I don’t think we can reduce the space complexity to O(1) . Can we ?
Nopes. But in interview they will ask you to do that in O(1) without arrays we well, cant they?
You just have to replace the print statement with Ans[index++]=arr[i] anyways.
1 Like
well vijju’s approach can be done in O(2n) whcih translates to O(n) . But it will take some extra space .
1 Like
I also feel the same that it will take some extra space . If not then could you please explain ?
Well cp is all about time-space tradeoff . You have to live with it
// | HuggingFaceTB/finemath | |
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## Wednesday, May 22, 2013
### Instrumentation for Temperature Measurement
Temperature is a physical quantity that is a measure of hotness and coldness on a numerical scale. It is a measure of the thermal energy per particle of matter or radiation; it is measured by a thermometer, which may be calibrated in any of various temperature scales, Celsius, Fahrenheit, Kelvin, etc.
In kinetic theory and in statistical mechanics, temperature is the effect of the thermal energy arising from the motion of microscopic particles such as atoms, molecules and photons.
In the post 'Temperature : Basics' explains all the basic details related to Temperature. The links below will explain the instrumentation for temperature measurements:
## Tuesday, May 21, 2013
### Pressure : Introduction
Pressure is defined as force per unit area. It is usually more convenient to use pressure rather than force to describe the influences upon fluid behavior. The standard unit for pressure is the Pascal, which is a Newton per square meter.
For an object sitting on a surface, the force pressing on the surface is the weight of the object, but in different orientations it might have a different area in contact with the surface and therefore exert a different pressure.
There are many physical situations where pressure is the most important variable. If you are peeling an apple, then pressure is the key variable: if the knife is sharp, then the area of contact is small and you can peel with less force exerted on the blade. If you must get an injection, then pressure is the most important variable in getting the needle through your skin: it is better to have a sharp needle than a dull one since the smaller area of contact implies that less force is required to push the needle through the skin.
When you deal with the pressure of a liquid at rest, the medium is treated as a continuous distribution of matter. But when you deal with a gas pressure, it must be approached as an average pressure from molecular collisions with the walls.
Pressure in a fluid can be seen to be a measure of energy per unit volume by means of the definition of work. This energy is related to other forms of fluid energy by the Bernoulli equation.
Fig 01 : Relationship between absolute, gauge and vacuum pressures.
The section below shows the links that explains the process variable pressure in detail:
## Wednesday, May 8, 2013
### Capacitance pressure transducers
Fig. 14 presents a transducer for sensing and transmitting differential pressure. Pressures to be measured act on isolating diaphragms 1 and 2 and are transmitted through a silicone oil 3, which fills the system, to a sensing diaphragm 4. This sensing diaphragm is balanced by two forces developed by measured pressures and presents the sensitive element. Capacitor plates 5 and 6 detect the position of the sensing diaphragm, which moves to the left or to the right, and, thus, the differential pressure applied to the sensitive element. The change in electric capacitance is electronically amplified and converted to the standard electrical analog or digital output signal, which is directly proportional to the difference of pressures. In order the capacitance transducer be able to measure comparatively low pressures, the device should produce about 25% change in capacitance for a full-scale pressure change. These transducers have low mass and high resolution. However, they are slightly dependent on temperature variation. Newly developed all-silicon capacitive pressure sensors have better thermal stability.
Figure 14. Variable capacitance differential pressure transducer.
Variable separation capacitance sensors have non-linear relationship between electrical capacitance and the movement of the separating membrane according to the formula:
(84)
where,
C - the electrical capacitance of the pressure sensor, F (Farad);
ฮต0 = 8.85, pF/m - the permittivity of vacuum, 1pF=10-12 F;
ฮต - the relative permittivity of the insulating material between plates of the capacitor, this is the dimensionless parameter;
A - the cross-sectional area of the capacitor plate, m2 ;
d - the distance between the capacitor plates, m;
a - variation of the distance between the capacitor plates, m.
A three-plate differential version of the capacitive pressure sensor doesn’t have such disadvantage (see Figure 15).
Two fixed plates form two capacitances with the moving separating plate/membrane as follows:
(85) and
(86)
Figure 15. Three-plate differential pressure/displacement sensor
Figure 16 shows an a.c. deflection bridge for the detection of variations of capacitances.
Figure 16. a.c. deflection bridge.
In this bridge:
(87)
,(88)
,(89)
where,
Z1and Z2 - reactive impedances, Ohm;
Z3 and Z4 - resistive impedances, Ohm.
When Icd = 0, then Vcd is called an open-circuit voltage of the bridge. According to the Kirchoff’s laws we have:
, (90)
. (91)
Let potential at Vb = 0, then:
(92)
So, the relationship between Vcd and a is linear.
Article Source:: Dr. Alexander Badalyan, University of South Australia
## Saturday, May 4, 2013
### Piezoelectric pressure transducers
The principle of these pressure transducers is based on the well-known phenomenon, that when an asymmetrical crystal is elastically deformed along its specific axes, an electrical charge is developed on its sides. The value of this charge is proportional to the force applied to the crystal, and, therefore, to the pressure under measurement.
Fig. 12 shows piezoelectric crystal circuit. An electrical charge developed on the sides of the crystal is converted into a voltage-type signal using a capacitor. This voltage is proportional to the electrical charge developed, and to the pressure to be measured. Piezoelectric sensors cannot measure static pressures for more than a few seconds, but they have a very quick response when measure dynamic pressures.
Figure 12. Piezoelectric pressure sensor with electrical circuit.
Synthetically developed quartz crystals (barium titanate, lead zirconate) have similar properties as natural single crystal (quartz). But natural quartz still is the perfect material for manufacturing piezoelectric sensitive elements, because it has perfect elasticity and stability, it is insensitive to temperature variations and it has high insulation resistance.
These pressure transducers are used for measurements of hydraulic and pneumatic pulsations, flow instabilities, fuel injection, etc.
Let’s develop a differential equation for this sensor.
(60)
. (61)
The dynamic behaviour of a piezoelectric element is analysed from mechanical and electrical points of view.
,
(62), or using (63)
we can get the following second order differential equation:
(64)
or
(65)
Finally, we can re-write (65) with variables in deviation form as follows:
(66)
Equation (66) is a second order linear differential equation for a piezoelectric sensor with variables in deviation form.
After applying the Laplace transform to (66) we get a transfer function for the piezoelectric sensor:
(67)
Ax - cross-sectional area of the piezoelectric sensor in the direction
perpendicular to axis ‘X’, m2;
b - charge sensitivity of the crystal to its deformation in the direction
perpendicular to axis ‘X’, C/m;
Fx - effective force due to pressure in the direction of axis ‘X’, N;
k - stiffness of the crystal is large ≈ 2 x 109, N/m;
Kx - steady state gain of the crystal, in other words it is the charge sensitivity
of the crystal to applied pressure , C/Pa;
m - mass of the crystal, kg;
qpiez - electrical charge developed by the crystal, C;
P - pressure acting on the surface perpendicular to axis ‘X’, Pa;
t - time, s ;
x - deformation of the crystal in the direction of axis ’X’, m;
ฮป - constant (friction coefficient) for the crystal, N*s / m .
Solution to equation (66) will give us the variation of the electrical charge qpiez , developed on the surfaces of a piezoelement as a function of time for a step change in the input variable – measured pressure, P . Now we should measure this electrical charge. For this purpose two metal electrodes are attached to the opposite sides of a piezoelectric crystal. Thus, a capacitor is formed. The value of capacitance of this capacitor can be evaluated as follows:
(68)
where,
Cpiez - electrical capacitance of the piezoelement, F (Farad);
ฮต0 = 8.85, pF / m - the permittivity of vacuum, 1pF=10-12 F;
ฮต - the relative permittivity of the material of the piezoelectric crystal,
this is the dimensionless parameter;
Ax - cross-sectional area of the piezoelectric sensor in the direction,
perpendicular to the axis ‘X’, m2;
d - the thickness of the piezoelectric crystal in the direction,
perpendicular to the axis “X”, m.
The relative permittivity, also called dielectric constant, for various piezoelectric materials is given below :
· for quartz (natural piezoelectric material) ฮต = 4.5
· for tourmaline (natural piezoelectric material) ฮต = 6.6;
It is also noted in the above mentioned reference, that natural piezoelectric materials have very low charge to force sensitivity, and therefore man-made piezoelectric ceramic materials are used as sensing elements:
· charge sensitivity to force for quartz 2.3, pC/N;
· charge sensitivity to force for tourmaline 1.9, pC/N or 2.4, pC/N;
· charge sensitivity to force for lead-zirconate-titanate 265, pC/N;
· charge sensitivity to force for lead metaniobate 80, pC/N.
We need to develop an electrical circuit which will allow us to convert variations of the capacitance of the piezoelectric sensor into the variation of an easy measurable electrical signal, voltage, for example. Such equivalent electrical circuit was developed, and is named after Norton (see Figure 13)
Figure 13. Norton equivalent electrical circuit for piezoelectric pressure/force
measurements. 1- piezoelectric element, 2– connecting cable, 3– recorder.
The piezoelectric element can be represented as a current source (or a charge generator) which is connected in parallel with a capacitance Cpiez . Then, this element is connected to a voltage recorder via connecting cables, which have the capacitance Ccable. A recorder has a resistive load, Rload. The voltage measured across Rload is equal:
(69)
where, - the impedance of three resistances connected in parallel,Ohm
According to the definition, the capacitance is equal to the ratio of the charge to the voltage across the capacitor plates, according to:
(70)
Let’s consider capacitance Cpiez:
(71)
After differentiating both sides of (71) we can get:
(72) or
(73)
or, according to the Ohm’s Law,
(74)
Similar we can get
(75)
Substitution of (74) and (75) into (69) will give:
(76)
where, s = d/dt - the Laplace operator.
Expressing variables Vload and Ipiez in deviation form and applying the Laplace transform to (76) we can get:
(77)
The transfer function for the Norton equivalent electrical circuit for piezoelectric pressure/force measurement system (see Figure 13) is as follows:
(78)
According to the definition:
, (79). Expressing these variables in deviation form and applying the Laplace transform to (79) we can get:
.
(80)
The transfer function relating current and charge of the piezoelectric sensor is as follows:
.
(81)
The transfer function relating the voltage Vload and the measured pressure P can be determined as follows:
. (82)
After substitution of (67), (78) and (80) into (82) we can get an expression for an overall transfer function of the piezoelectric pressure/force measurement system:
Article Source:: Dr. Alexander Badalyan, University of South Australia
## Labels
Automation (4) Basics (48) DCS (1) Flow (27) Flow Meters (22) Level (19) PLC (2) Pressure (12) SCADA (2) Sensors (7) Temperature (17) Transducer (17)
Hai friends…welcome to my blog. This blog is exclusively for instrumentation engineering students which will provide sources for their reference and studies. As you all know Instrumentation is now a fast emerging and developing field in Engineering. This blog has different categories like PLC, SCADA, DCS, Sensors and Transducers, Computer control of process, Industrial Instrumentation, etc.
This blog will also provide an exclusive ‘ASK ME’ section where you can make any queries and share your ideas about instrumentation. The solution for your queries will be given to you by mail from best of my knowledge and reference.
So I wish this blog may be very useful for your studies and reference. | HuggingFaceTB/finemath | |
# Overview: Milestone I
The goal of this whole project is to be able to predict the CMB (and matter) fluctuations - described by the so-called power-spectrum - from first principles and learn about all the different physical processes that goes on to be able to explain the results.
Figure 1: The CMB power-spectrum and the physical effects that leaves its imprints on different scales.
We will do this step by step and the topic of the first project is the evolution of the uniform background in the Universe. The goal here is to make a class/module that takes in the cosmological parameters and has functions for getting the Hubble parameter, conformal time and distance measures as function of scale factor and the variable $x = \log(a)$ which will be our main time variable in this course.
The deliverables are the following (for the old 2021 version see this PDF). There is 55 points availiable for the actual deliverables and 45 points for how this is presented/discussed in the report:
• The source code used for the evaluation. This should be possible for us to easily compile and run. A common way students have chosen to do this in previous years is to make a GitHub account, upload your code there and just provide the link. This is no requirement, you can just as easily include this together with your report in Devilry.
• A copy of your paper. This should have a section about this milestone which should contain: introduction (what we do and why this is relevant for the project), a short theory section with all relevant equations, tests showing the code gives expected results, the actual plots/results asked for and any other things you deem relevant and finally a proper description and discussion of the results tying it together with the physics.
• Implement all the required functions in the template: $H(x)$, $\mathcal{H}(x)$, $\frac{d\mathcal{H}(x)}{dx}$, $\frac{d^2\mathcal{H}(x)}{dx^2}$, $\eta(x)$, $\Omega_i(x)$ (for b,CDM,$\gamma$,$\nu$,$\Lambda$) and distance functions $d_L$ (luminosity distrance), $d_A$ (angular diameter distance) and $\chi$ (conformal distance). Solve ODEs and make splines for $\eta(x)$ and $t(x)$. When you are done with everything run the MCMC fits and produce plots.
• Plots that demonstrate that your code works properly: $\frac{1}{\mathcal{H}}\frac{d\mathcal{H}(x)}{dx}$, $\frac{1}{\mathcal{H}}\frac{d^2\mathcal{H}(x)}{dx^2}$ and $\frac{\eta(x)\mathcal{H}(x)}{c}$ compared with analytical expectations in different regimes. (10p)
• Plot of the conformal Hubble factor $\mathcal{H}(x)$. (5p)
• Plot of cosmic time $t(x)$ and conformal time $\eta(x)/c$. (10p)
• Plot of density parameters $\Omega_b(x) + \Omega_{\rm CDM}(x)$, $\Omega_\gamma(x) + \Omega_\nu(x)$ and $\Omega_{\Lambda}(x)$ together. (5p)
• Plot of the luminosity distance versus redshift with data from supernova observations (taken from Betoule et al. 2014) overplotted (see header line for file-format). (5p)
• Plots from MCMC fits to supernova data. 1) the $1\sigma$ constraints from MCMC fits to supernova data in the $\Omega_\Lambda$ vs $\Omega_M$ plane 2) the posterior PDF of the Hubble parameter $H_0$. (10p)
• Remember to use sensible units in the plots (e.g. 100km/s/Mpc for $H$, Mpc/Gpc for distances and Gyr for times).
• Make a table showing times (both $x$, redshift $z$ and $t$) for when we have 1) radiation-matter equality 2) matter-dark energy equality and 3) when the Universe starts to accelerate and 4) the age of the Universe today and 5) the conformal time today ($\eta(0)/c$). Note matter means baryons+CDM and radiation means photons+massless neutrinos. You can find the expressions for these times analytically and then use this to compute the numerical values or solve numerically if you prefer this. Mark these as vertical lines in the plots where relevant (if it's relevant for discussing the plot). These values will be useful also in future milestones to understand/explain results. (10p)
The fiducial cosmology we are going to use in this course - which is the one you are going to use for all the results in the paper - is the best-fit cosmology found from fits to Planck 2018 data: \begin{align} h &= 0.67, \\ T_{\rm CMB 0} &= 2.7255\,K, \\ N_{\rm eff} &= 3.046, \\ \Omega_{\rm b 0} &= 0.05, \\ \Omega_{\rm CDM 0} &= 0.267,\\ \Omega_{k 0} &= 0, \\ \Omega_{\nu 0} &= N_{\rm eff}\cdot \frac{7}{8}\left(\frac{4}{11}\right)^{4/3}\Omega_{\gamma 0}, \\ \Omega_{\Lambda 0} &= 1 - (\Omega_{k 0}+\Omega_{b 0}+\Omega_{\rm CDM 0}+\Omega_{\gamma 0}+\Omega_{\nu 0}),\\ n_s &= 0.965, \\ A_s &= 2.1\cdot 10^{-9}, \\ Y_p &= 0.245, \\ z_{\rm reion} &= 8, \\ \Delta z_{\rm reion} &= 0.5, \\ z_{\rm He reion} &= 3.5, \\ \Delta z_{\rm He reion} &= 0.5. \end{align} where $n_s,A_s$ will not be relevant until the last milestone and the last few parameters are only relevant for the next milestone (and only for PhD students that have to include neutrino perturbations, helium and the effects of reionization - but the rest of you should still keep neutrinos in the background). The radiation and neutrino density parameters follows from the above as $$\Omega_{\gamma 0} = 2\cdot \frac{\pi^2}{30} \frac{(k_bT_{\rm CMB 0})^4}{\hbar^3 c^5} \cdot \frac{8\pi G}{3H_0^2},\,\,\,\Omega_{\nu 0} = N_{\rm eff}\cdot \frac{7}{8}\cdot \left(\frac{4}{11}\right)^{4/3}\Omega_{\gamma 0}$$
NB: Even if we are going to work with $\Omega_k = 0$ then you still need to implement curvature to work as this is required when doing the supernova fits.
One note about the $\Omega$'s: often in the litterature people use $\Omega_X$ where they really mean $\Omega_{X0}$. I have tried to be consistent with this, but there might be some omissions. To make it more clear I have also tried to use the convention to always include the argument (i.e. write $\Omega_X(a)$) if I mean the function and not the value today, but always double-check the equations and ask if something is not clear (I mention this as its been a common source of confusion).
If you haven't done so already read Callin (2006). It goes through all the things we are going to do in this course, it has all the equations and it has some notes on possible algorithms to use for the different tasks. It also have many plots that are useful to compare to (just remember that it uses a slightly different cosmology than our fiducial one). Another very good paper (though its written on a much higher level) is this one. A short introduction to the project can be found in this PDF (keynote) which is displayed below.
Page: /
# Theoretical background
See the lecture notes for the theory you should know for this milestone.
Lets review the theory basics and you can find more details in the lecture notes mentioned above. The task in this project is to compute the expansion history of the universe, and look at the uniform background densities of the various matter and energy components. Let us first define the Friedmann-Robertson-Walker metric (here for a flat space where $k=0$), \begin{align} ds^2 &= -c^2 dt^2 + a^2(t) \left(dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2)\right) \\ &= a^2(t) (-d\eta^2 + dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2) \end{align} or in Cartesian coordinates \begin{align} ds^2 &= -c^2dt^2 + a^2(t)(dx^2 + dy^2 + dz^2)\\ &= a^2(t)(-d\eta^2 + dx^2 + dy^2 + dz^2) \end{align} where $a(t)$ is the scale factor, which measures the size of the universe relative to today ($a_0 = a(\textrm{today}) = 1$), and $\eta$ is called conformal time. One thing to note: it's called conformal time, but its usually given in units of length for it to have the same dimension as the spatial coordinates. The conversion factor for this is the speed of light $c$. In this course the conformal time is a distance (and the corresponding time is this distance divided by the speed of light). As we will be looking at phenomena that varies strongly over a wide range of time scales, we will mostly be using the logarithm of the scale factor, $x \equiv \ln a$, as our time variable. A fifth time variable is the redshift, $z$, which is defined as $1+z = a_0 / a(t)$.
Einstein's General Relativity describes how the metric evolves with time, given some matter and density components. The relevant equation for our purposes here is the Friedmann equation, which may be written - when we don't assume $k=0$ - on the following form (see the theory page for some notes on how to do this derivation), $$$$\boxed{H = H_0 \sqrt{(\Omega_{b0}+\Omega_{\rm CDM 0})a^{-3} + (\Omega_{\gamma 0} + \Omega_{\nu 0}) a^{-4} + \Omega_{k 0} a^{-2} + \Omega_{\Lambda 0}}},$$$$ where $H\equiv \dot{a}/{a}$ is the Hubble parameter (and dot denotes derivatives with respect to physical time, $\dot{} = d/dt$), and $\Omega_{b 0}$, $\Omega_{\rm CDM 0}$, $\Omega_{\gamma 0}$, $\Omega_{\nu 0}$, and $\Omega_{\Lambda 0}$ are the present day relative densities of baryonic matter, dark matter, radiation, neutrinos and dark energy, respectively. A subscript $0$ denotes the value at the present time. The term $\Omega_{k 0} = -\frac{kc^2}{H_0^2}$ denotes curvatuve and acts in the Friedmann equation as if it were a normal matter fluid with equation of state $\omega = -1/3$. This term follows from the other density parameters which can be seen from taking $a=1$ to get $\Omega_{k 0} = 1 - (\Omega_{b 0}+\Omega_{\rm CDM 0}) - (\Omega_{\gamma 0} + \Omega_{\nu 0}) - \Omega_{\Lambda 0}$ (Recall that $\Omega_x = \rho_x / \rho_c$, where $\rho_c = 3H^2/8\pi G$.) We also introduce a scaled Hubble parameter, $\mathcal{H}\equiv aH$ ("H prime" or "Hp" for short).
In this project we don't include curvature, but you still have to implement it in general. Master students don't have to deal with neutrinos later on, but do include it in the Hubble equation so use the correct value of $N_{\rm eff}$.
The Friedmann equations also describe how each component evolve with time $$\dot{\rho} + 3H(\rho + P) = 0$$ where $P$ is the pressure. It's useful to define the equation of state $\omega \equiv P/\rho$ (which is constant for the fluids we consider in this course). In terms of this the solution reads $\rho \propto a^{-3(1+\omega)}$. For cold dark matter and baryons we have $\omega = 0$, for relativistic matter (radiation and massless neutrinos) we have $\omega = 1/3$ and for a cosmological constant we have $\omega = -1$.
This gives us: \begin{align} \rho_{\rm CDM} &= \rho_{{\rm CDM},0} a^{-3} \\ \rho_b &= \rho_{b,0} a^{-3} \\ \rho_\gamma &= \rho_{\gamma,0} a^{-4} \\ \rho_{\nu} &= \rho_{\nu,0} a^{-4} \\ \rho_{\Lambda} &= \rho_{\Lambda,0}. \end{align} Here, quantities with subscripts 0 indicate today's values. The density parameters $\Omega_X(a) = \rho_X / \rho_c$ can be written \boxed{ \begin{align} \Omega_{k}(a) &= \frac{\Omega_{k0}}{a^2H(a)^2/H_0^2}\\ \Omega_{\rm CDM}(a) &= \frac{\Omega_{\rm CDM 0}}{a^3H(a)^2/H_0^2} \\ \Omega_b(a) &= \frac{\Omega_{b 0}}{a^3H(a)^2/H_0^2} \\ \Omega_\gamma(a) &= \frac{\Omega_{\gamma 0}}{a^4H(a)^2/H_0^2} \\ \Omega_{\nu}(a) &= \frac{\Omega_{\nu 0}}{a^4H(a)^2/H_0^2} \\ \Omega_{\Lambda}(a) &= \frac{\Omega_{\Lambda 0}}{H(a)^2/H_0^2}. \end{align} }
Two of the density parameters above follows from the observed temperature of the CMB. We have that $\Omega_{\gamma 0}$ and $\Omega_{\nu 0}$ are given by \boxed{ \begin{align} \Omega_{\gamma 0} &= 2\cdot \frac{\pi^2}{30} \frac{(k_bT_{\rm CMB 0})^4}{\hbar^3 c^5} \cdot \frac{8\pi G}{3H_0^2},\\ \Omega_{\nu 0} &= N_{\rm eff}\cdot \frac{7}{8}\cdot \left(\frac{4}{11}\right)^{4/3}\Omega_{\gamma 0}, \end{align} } where $T_{\rm CMB 0} = 2.7255$K is the temperature of the CMB today and $N_{\rm eff} = 3.046$ is the effective number of massless neutrinos (slightly larger than $3$ due to the fact that neutrinos had not completely decoupled when electrons and positrons annhialate and the $0.046$ accounts for the extra energy pumped into the neutrinos). We define matter-radiation equality as the time when $\Omega_b + \Omega_{\rm CDM} = \Omega_\gamma + \Omega_{\nu}$ and the matter-dark energy transition as the time when $\Omega_b + \Omega_{\rm CDM} = \Omega_{\Lambda}$. The onset of acceleration is the time when $\ddot{a} = 0$.
Another crucial concept for CMB computations is that of the horizon. This is simply the distance that light may have travelled since the Big Bang, $t=0$. If the universe was static, this would simply have been $ct$, but since the universe also expands, it will be somewhat larger. Note that the horizon is a strictly increasing quantity with time, and we can therefore use it as a time variable. This is often called conformal time, and is denoted $\eta$.
To find a computable expression for $\eta$, we note that $$\frac{d\eta}{dt} = \frac{c}{a}.$$ The left-hand side of this equation may be rewritten into $$\frac{d\eta}{dt} = \frac{d\eta}{da}\frac{da}{dt} = \frac{d\eta}{da} aH,$$ such that $$\frac{d\eta}{da} = \frac{c}{a^2H} = \frac{c}{a\mathcal{H}}$$ and \boxed{ \begin{align} \frac{d\eta}{dx} = \frac{da}{dx}\frac{d\eta}{da} = \frac{c}{\mathcal{H}} \end{align} } This is a differential equation for $\eta$, that can either be solved numerically by direct integration, i.e. $\eta(x) = \int_{-\infty}^x \frac{c dx'}{\mathcal{H}(x')}$ , or by plugging the expression into a ordinary differential equation solver. The initial condition is $\eta(-\infty) = 0$. We can't integrate from $-\infty$ so in practice we pick some very early time $x_{\rm start}$ and and use the analytical approximation $\eta(x_{\rm start}) = \frac{c}{\mathcal{H}(x_{\rm start})}$ (we can solve the ODE analytically in the radiation domianted era).
We also need some distance measures. Consider the line-element in spherical coordinates $$ds^2 = -c^2dt^2 + a^2(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin\theta^2d\phi^2).$$ Photons move on $0$-geodesics $ds^2 = 0$ so if we consider a radially moving photon ($d\theta=d\phi=0$), traveling from $(t,r)$ to us today at $(t=t_{\rm today},r=0)$ we get $$cdt = \frac{adr}{\sqrt{1-kr}}\implies \int_t^{t_{\rm today}}\frac{cdt}{a} = \int_{0}^{r}\frac{dr'}{\sqrt{1-kr'}}$$ The left hand side is what we call the co-moving distance and is closely related to the conformal time \boxed{ \begin{align} \chi = \eta_0 - \eta \end{align} } Evaluating the integral on the right the full equation above can then be written \boxed{ \begin{align} r &= \begin{cases} \chi \cdot \frac{\sin \left( \sqrt{|\Omega_{k 0}}| H_0 \chi /c \right)}{\left(\sqrt{|\Omega_{k 0}|}H_0\chi / c\right)} & \Omega_{k 0} < 0\\ \chi & \Omega_{k 0} =0 \\ \chi \cdot \frac{\sinh \left( \sqrt{|\Omega_{k 0}|} H_0 \chi / c\right)}{\left(\sqrt{|\Omega_{k 0}|}H_0\chi / c\right) } & \Omega_{k 0} > 0 \end{cases} \end{align} } For a flat Universe we simply have $r = \chi = \eta_0 - \eta$. With this in hand we can compute all the standard distance measures we have in cosmology. If we know an object's physical size, $D$, and its angular size, $\theta$, as viewed from earth then the angular diameter distance is defined as $d_A = \frac{D}{\theta}$. From the line-element we see that the angular distance when we move in the $\theta$-direction is $dD = a r d\theta$ so \boxed{ \begin{align} d_A &= \frac{D}{\theta} = ar \end{align} } Note that for a flat Universe the angular diameter distance reduces to $d_A = a\chi = a(\eta_0 - \eta)$. The last distance we need is the luminosity distance. If we know the intrinsic luminosity of a source and measure its flux then we can define the distance to the source via $F = \frac{L}{4\pi d_L^2}$ which is simply \boxed{ \begin{align} d_L = \frac{r}{a} = \frac{d_A}{a^2} \end{align} } We see that all distance measures are given directly from the conformal time, the scale-factor and the curvature so you only need to implement the functions $r(\chi)$, $\chi(x)$, $d_A(x)$, $d_L(x)$ as given above.
The final thing to compute is the relation between cosmic time $t$ and our time-coordinate $x$. From $H = \frac{1}{a}\frac{d a}{dt} \to dt = \frac{da}{aH}$ we get $$t(x) = \int_0^a \frac{da}{aH} = \int_{-\infty}^x \frac{dx}{H(x)}$$ We can solve this just as we did for $\eta$ by evolving the ODE \boxed{ \begin{align} \frac{dt}{dx} = \frac{1}{H} \end{align} } and splining the result. In the radiation dominated era we have $t(x) = \frac{1}{2H(x)}$ so the initial condition is $t(x_{\rm start}) = \frac{1}{2H(x_{\rm start})}$. Evaluating this at $x = 0$ (today) gives us the age of the Universe.
After everything is implemented we would like to use our code to get constraints on our cosmological parameters by comparing to data.
The data we will use us a set of supernova with associated redshift $z_i$, luminosity distance $d_L^{\rm obs}(z_i)$ and associated measurement errors $\sigma_i$. Under the assumption that the measurements are Gaussian distributed and uncorrelated between different redshifts the Likelihood function, telling us how well the data fits the theory, is given by $L \propto e^{-\chi^2/2}$ where the chi-squared function is $$\chi^2(h,\Omega_{m0}, \Omega_{k0}) = \sum_{i=1}^N \frac{[(d_L(z_i,h,\Omega_{m0},\Omega_{k0}) - d_L^{\rm obs}(z_i)]^2 }{\sigma_i^2}$$ A low value of $\chi^2$ means a good fit (high likelihood for the choice of parameters). What we want to do is to basicsally check all possible values of the parameters to find the best-fitting model and the range of parameters around it which are in agreement with the observed values. One can do this brute-force, but this step is in practice most commonly done by performing a so-called Monte Carlo Markov Chain to randomly sample from the likelihood (e.g. using the Metropolis algorithm as in the code template). The result from performing this step is a chain of values: parameter-points and likelihood values. The set of parameters with the lowest likelihood is our best-fit model. We can check if this is really a good fit by comparing the $\chi^2$ to the number of datapoints (here $N=31$). A good fit has $\chi^2/N \sim 1$ (a much higher value denotes a bad fit). When you have found the best-fit then you can find the $1\sigma$ ($68.4\%$) confidence region by looking at all values that satisfies $\chi^2 - \chi^2_{\rm min} \lt 3.53$ (see e.g. here). We can also plot the histogram of all the accepted samples to get the probabillity distribution (PDF) for a given parameter.
# What you have to do
Implement a class that takes in all the cosmological parameters $h,\Omega_{b 0},\Omega_{\rm CDM 0},\Omega_{\Lambda 0},T_{\rm CMB 0},N_{\rm eff}$ - and compute $\Omega_{\gamma 0}$, $\Omega_{\nu 0}$ and $\Omega_{k 0}$ from these - and stores them. Make functions that are able to give you the cosmological parameters, the Hubble function and $\mathcal{H} = aH$ "Hp" plus the first two derivatives (these can be computed analytically) together with the different distance measures (comoving, luminosity and angular diameter distance).
Once this is done compute $\eta(x)$, spline the result and make a function that returns this function. The spline eta_of_x_spline is already defined in the header BackgroundCosmology.h so to create it all you have to do is to call eta_of_x_spline.create(x_array,eta_array,"eta_of_x") with your x_array and eta_array you got from the ODE solver.
Then finally compute the times asked for: equality times and acceleration time (compute this analytically) plus the age of the Universe. You can do this in the same way as you did for the conformal time by solving an ODE and making a spline of $t(x)$ (useful for computing the time at any $x$ later on if you need that even though we only ask for the time today here). Evaluating the spline at $x=0$ gives you the time of the Universe in seconds (if you use SI units), but convert it to a more sensible unit like gigayears ($10^9\cdot 365 \cdot 24 \cdot 60\cdot 60$ seconds) when presenting the results.
Once the distance measures are working you are ready to do the parameter fits to supernova data. First plot the results of the luminosity distance, using the fiducial cosmological parameters we use in the course, together with the data points from supernova observations (you might want to divide $d_L$ by $z$ when making the plot to better see the comparison; if so then you need to do the same with the data/errorbars!). To do the fitting you simply need to call the fitting routine provided in SupernovaFitting.h. If you don't use the template then you need to implement this yourself (it's a simple Metropolis MCMC sampler which you might have implemented in other courses before). See the header of this file if you have any problems making it work and for how to analyze the resulting chains to produce the plots asked for.
If you choose to work with the C++ template take a look at BackgroundCosmology class. The file BackgroundCosmology.h contains the definition of this class, the variables (and splines) and functions it contains. See the TODO's in BackgroundCosmology.cpp for hints on what to do in each routine.
The most important thing about coding is to test your code. There are always bugs and more bugs. The more tests you have the better. Here are some tips (which does not cover everything) about things you can do to ensure that your code is working correctly:
Print out the variables you entered (and the ones you computed) and check that they are set correctly inside the class. You can use the info() method for this (just add whatever you want to print here). A more concrete test is to compute the sum of all the density parameters today. Check that this sum is $1$ (or very close to $1$). This tests fails if you failed to initialize the parameter correctly. Try evaluating the function you have made and check that they work and give reasonable results.
For testing $\eta$: you can show analytically that in the radiation dominated era (when you can ignore energy in matter and dark energy so that $H^2 \propto 1/a^4$) then we have $\eta \simeq \frac{c}{\mathcal{H}}$. Plot $\eta \mathcal{H}/c$ and show that it converges to $1$ the further back in time you go. A more direct test is to compute and spline $\eta$ and use the spline-routines to get $\eta'$ and check that $\frac{\eta'\mathcal{H}}{c} \equiv 1$.
For testing $\mathcal{H}$ and it derivatives: you can derive an expression for what $\frac{\mathcal{H}^\prime}{\mathcal{H}}$ should be when the Universe is dominated by a fluid with equation of state $w$. Check that you get results that agrees with the expectations in the radiation, matter and dark energy dominated regimes (or run the code with only one energy component having non-zero $\Omega$ at the time and check that you get the expected result). Do a similar thing with $\frac{\mathcal{H}^{\prime\prime}}{\mathcal{H}}$.
Figure 2: Here are some plots to compare your results to. We show $\mathcal{H}(x),\eta(x)$ and $\frac{\eta(x)\mathcal{H}(x)}{c}$ and the density parameters for a toy-cosmology with parameters $\Omega_{m 0} = \Omega_{b 0} + \Omega_{\rm CDM 0} = 0.5$, $\Omega_{\Lambda 0} = 0.5$, $\Omega_{\nu 0} = 0$ and $h=0.7$. You can also run my code online here (down for the moment) to get results to compare to (NB: cannot guarantee there are no bugs in that code). Below we show some example results of fits to supernovadata.
# Appendix
### Appendix: More things one can compute...
These things are optional, but here are some extra things one can compute.
Scalefactor as function of time: Once you have $t(x)$ (that we compute above) you can spline $(t(x), x)$ (opposed to $(x, t(x))$) to get the inverse function $x(t) = \log(a(t))$. This be compared to the analytical approximation $a \propto t^{\frac{2}{3(1+w)}}$ for when the Universe is dominated by a fluid with equation of state $w$.
Massive neutrinos: We treat neutrinos as massless, however oscillation experiments have revealed that neutrinos have a small mass $\lesssim 0.1$ eV (experiments are only sensitive to mass differences so we cannot tell exactly their masses - but the growth of structures is very sensitive to the sum of the masses). They are relativistic in the early Universe and non-relativistic close to today ($T_0 \sim 10^{-4}$ eV). Neutrinos decouple when they are still relativistic and consequently they maintain their distribution function $f\propto \frac{1}{e^{\frac{p}{T_\nu}}+1}$ with the temperature decreasing as $T_\nu = T_{\nu 0}/a$ where $T_{\nu 0} = (3.046/3)^{1/4}\cdot (4/11)^{1/3}T_{\rm CMB 0}$ (see problem set). The energy density of a single massive neutrino can be written $$\frac{\rho_\nu}{\rho_{c0}} = \frac{\Omega_{\nu 0}}{a^4} \frac{B(y)}{B(0)}$$ where $\Omega_{\nu 0} = \Omega_{\gamma 0}\left(T_{\nu 0}/T_{\rm CMB 0}\right)^4$ and $$B(y) \equiv \int_0^\infty \frac{x^2\sqrt{x^2 + y^2}}{e^{x}+1} dx$$ with $y(a) = \frac{m}{T_\nu(a)}$. If we have $N_\nu = 3$ neutrinos with mass $m_i$ the total energy density is likewise $$\frac{\rho_\nu}{\rho_{c0}} = \frac{\Omega_{\nu 0}}{a^4} \frac{\sum_i\frac{B(y_i)}{B(0)}}{N_\nu}$$ When we have massless neutrinos $y\equiv 0$ and we get the expected result $\frac{\rho_\nu}{\rho_{c0}} = \frac{\Omega_{\nu 0}}{a^4}$ that we use in this project. On the other hand if $y\gg 1$ (the non-relativistic limit) then $$B \approx y\int_0^\infty \frac{x^2}{e^{x}+1} dx$$ so $\rho_\nu \propto 1/a^3$. The amount of massive non-relativistic neutrinos (assuming all of them are non-relativitstic today) is $$\Omega_{m\nu 0} \approx \Omega_{\nu 0}\frac{\sum m_i}{T_{\nu 0}} \frac{\int_0^\infty \frac{x^2}{e^{x}+1}dx}{\int_0^\infty \frac{x^3}{e^{x}+1} dx}$$ and evaluating the integrals ($\frac{3\zeta(3)}{2}$ and $\frac{7\pi^4}{120}$) we get $$\Omega_{m\nu 0}h^2 = \frac{16\zeta(3)}{11\pi}\left(\frac{(k_bT_{\rm CMB 0})^3G({1 \rm eV})}{\hbar^3 c^5 (H_0/h)^2}\right)\frac{\sum m_i}{{1 \rm eV}}\left(\frac{N_{\rm eff}}{3}\right)^{3/4} = \frac{\sum m_i}{93.05 eV} \left(\frac{T_{\rm CMB 0}}{2.725 K}\right)^3 \left(\frac{N_{\rm eff}}{3.046}\right)^{3/4}$$ so for $T_{\rm CMB 0} = 2.725$ K, $N_{\rm eff} = 3.046$ and $h=0.674$ we get $\Omega_{m\nu 0} \approx \frac{\sum m_i}{42 {\rm eV}}$. This estimate assumes the distribution function is exactly the Fermi-Dirac one, however there are some tiny distrortions produced when electrons and positrons anhiallate after neutrinos have almost (but not fully) decoupled and a much more detailed analysis gives $93.14$ eV instead of $93.05$ eV (and $94.12$ eV if we assumed instantanous decoupling with $N_{\rm eff} = 3$) so the results above is very accurate (to $0.2\%)$.
Note that this lets us constrain the mass of the neutrinos directly from what we know about the background. We know that atmost $\sim 0.3$ of the energy budget of our Universe is mass so it follows directly that $\sum m_i \lt 12$ eV. We can of course do much better that this with a more detailed analysis (the best constraints from cosmology today gives us $\sum m_i \lesssim 0.1-0.2$ eV), but its remarkable that such a simple theoretical calculation like we did above is enough to get a constrain that is of the same order of magnitude as particle physics experiments can currently provide! A common way of including neutrinos in a CMB code is to take $\Omega_{m\nu 0}$ as a free parameter, use this to compute $\sum m_i$ and compute the energy density from the above using $N_\nu$ neutrinos with mass $\sum m_i / N_\nu$. Alternatively take in the three masses directly. Also note that with massive neutrinos the total matter density parameter $\Omega_{M 0} = \Omega_{b 0} + \Omega_{\rm CDM 0} + \Omega_{m\nu 0}$. Why we don't bother to include the mass of the neutrinos will become more clear when we do perturbations - its significantly harder with massive neutrinos as we are forced to sample the full $p$-dependence of the distribution function and then integrate it to just get things like the number density or energy density perturbations. | open-web-math/open-web-math | |
## College Physics (4th Edition)
(a) The pipe is closed at one end. (b) The fundamental frequency of this pipe is $78~Hz$ (c) The length of the pipe is $1.1~m$
(a) We can find the difference between the successive resonant frequencies: $390~Hz-234~Hz = 156~Hz$ $546~Hz-390~Hz = 156~Hz$ Since these three resonant frequencies are not multiples of $156~Hz$, the pipe can not be open at both ends. Then the pipe must be closed at one end. (b) The resonant frequencies of a pipe that is closed at one end have the form: $f_1, 3f_1, 5f_1, 7f_1, etc...$ The difference between successive frequencies is $2f_1$. Therefore, the fundamental frequency of this pipe is $\frac{156~Hz}{2} = 78~Hz$ (c) We can find the length of the pipe: $\lambda = \frac{v}{f}$ $4L = \frac{v}{f}$ $L = \frac{v}{4f}$ $L = \frac{343~m/s}{(4)(78~Hz)}$ $L = 1.1~m$ The length of the pipe is $1.1~m$ | HuggingFaceTB/finemath | |
## Algebra 1
Published by Prentice Hall
# Chapter 12 - Data Analysis and Probability - 12-7 Thoretical and Experimental Probability - Practice and Problem-Solving Exercises - Page 762: 46
7/16
#### Work Step by Step
Find the area of each region: White area=4 square units Red area=3 square units Blue area=6 square units Green area=3 square units Total area=16 square units P(events)=$\frac{numberoffavorableoutcomes}{numberofpossibleotcomes}$ The green is 3 square units and the white is 4 square units. The green and white area is 7 square units so it is the number of favorable outcomes. The area is 16 square units in total so this is the number of possible outcomes. P(green or white)=$\frac{7}{16}$
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. | HuggingFaceTB/finemath | |
In multiplications: the decimal point moves to the right as many steps as the number has zeros.
In divisions: the decimal point moves to the left as many steps as the number has zeros.
## Examples with solutions for Multiplying and Dividing Decimal Fractions by 10, 100, etc.
### Exercise #1
$1.23\times110=$
### Step-by-Step Solution
We will use the distributive property to split 110 into two numbers -
100 and 10
Now, we'll multiply the original number, 1.23, by these two numbers.
1.23*100=123
1.23*10=12.3
All that's left to do is to add the two products we got -
123+12.3=
135.3
And that's the solution!
$135.3$
### Exercise #2
$\text{0}.07\times10=$
### Video Solution
$0.7$
### Exercise #3
$1.004\times10=$
### Video Solution
$10.04$
### Exercise #4
$111.1:10=$
### Video Solution
$11.11$
### Exercise #5
$1.14\times10=$
### Video Solution
$11.4$
### Exercise #6
$11.31:10=$
### Video Solution
$1.131$
### Exercise #7
$12.2:10=$
### Video Solution
$\text{1}.22$
### Exercise #8
$13.61:10=$
### Video Solution
$1.361$
### Exercise #9
$1.4\times10=$
### Video Solution
$14$
### Exercise #10
$1.52\times10=$
### Video Solution
$15.2$
### Exercise #11
$0.3\times10=$
### Video Solution
$3$
### Exercise #12
$2.7\times10=$
### Video Solution
$27$
### Exercise #13
$20.1:10=$
### Video Solution
$2.01$
### Exercise #14
$2.31\times10=$
### Video Solution
$23.1$
### Exercise #15
$0.26\times10=$
### Video Solution
$2.6$ | HuggingFaceTB/finemath | |
# Acceleration formula
Hello!. I am in trouble with my homework.
I got next data:
The accelration is 90m/s2(that 2 means squared).
Then the distance is 2km(2000m).
I have to find how high speed will he reach in 2km-s(thats the distance i wrote up there).
The task text is like this:
Rocket starts with accelration 90m/s2(that 2 means squared). How high speed he reaches after 2km-s. What kind of formulas i need to solve this little problem.
Thank you.
Also i will add that 90m/s2 is probably g(in some cases a), the 2km is S(i think) And i need to find t(time) and Vo(the start speed) to find the V(the speed at end) . Or anyone has any other ideas
Office_Shredder
Staff Emeritus
Gold Member
The 90 m/s2 is probably the upwards acceleration of the rocket, seeing how g is only 9.81
Office_Shredder said:
The 90 m/s2 is probably the upwards acceleration of the rocket, seeing how g is only 9.81
Oh thank you =)
Look at a picture of a rocket on the launch pad and think about the initial velocity.
You don't need to find the time. While I don't advocate "plug and chug", there is a formula that relates the initial velocity, final velocity, acceleration, and displacement.
The formula you're looking for is located in Doc Al's second post:
Last edited by a moderator:
ok i did like this:
1st t=Vo / g(a)= 90 : 9.8 m/s2=9.184s
V=s/t
V=2000m : 9.18m = 217.86 m/s
or
v=Vo +at
v=90 m/s2 + 9.8m/s2*9,184s=180 m/s
which one is the right one?
You don't need the time at all....
You do know that V1= 0 m/s
(use V2²= V1² + 2aΔd)
Vesper89 said:
You don't need the time at all....
You do know that V1= 0 m/s
(use V2²= V1² + 2aΔd)
Yes, but we have nt learned this kind of formula yet, so i need to choose one of the previous ones, but wich one?
Well, first lets look at your answers and see if they're reasonable. You have an acceleration of about 90 m/s^2. The final velocity of your first answer was 217 m/s, and the final velocity via your second method was 180 m/s.
For the second case, it would take 2 seconds since the acceleration means the velocity is increasing at a rate of 90 m/s per second. And, if the final velocity was 180 m/s, then the average velocity worked out to be (0m/s + 180m/s)/2 = 90m/s. And in 2 seconds, with an average velocity of 90 m/s, you would travel 180 meters. Unfortunately, that's quite a bit off from 2000 meters. For the first case, it would only take a little more than 2 seconds (less than 2 1/2 seconds). Following similar estimations, you're still a far way from 2000 meters.
It's kind of odd that you haven't seen final velocity squared = initial velocity squared + 2 times acceleration times change in displacement.
Without that formula, it's going to take a lot of work with 3 different equations, leaving variables in as you go.
(use V2²= V1² + 2aΔd)
But how i find v2 and v1, whats that 2aΔd. We really did nt learn it
And also i noticed that took 9.184 second and the distance was 2000m and the speed was 180 m/s. 180*9,184=1653m, but it accelrates 90 m/s. so it takes about 2 seconds. 2+9,184=11,184
11,184*180=2013.12 m. So it should be logical.
Homework Helper
bayer said:
...Rocket starts with accelration 90m/s2(that 2 means squared). How high speed he reaches after 2km-s. What kind of formulas i need to solve this little problem.
What did you mean by '2 km-s'? Did you mean the distance of 2 km? If so, use the formula for distance with constant acceleration first: $$d = d_{0}+v_{0}t+\frac{1}{2}at^2$$. The initial distance and velocity equal zero, so, the formula turns into $$d=\frac{1}{2}at^2$$. Now, you know the distance d, and the acceleration a, right? So, plug in the values, and solve for t to obtain the time the rocket reaches 2000m. Further on, use that value t in the next formula: $$v = v_{0} + at$$. Again, since the initial velocity equals zero, only use v = at to obtain the velocity the rocket has at 2000m.
edit: I'm slapping my forehead at the moment (had one of those duh moments) As radou showed above, you can find the time first. For a moment, I had forgotten that you probably already had that equation. I typically derive that equation in class after I derive the equation below:
--------------
Here is the derivation of that formula from the basic formulas that you have.
$$v_f=v_0+at$$ subtract $$v_0$$ from both sides
$$v_f-v_i=at$$
Now, you know that $$t=\frac{d}{v_{avg}}$$ substitute this for t in the equation above.
$$v_f-v_i=\frac{ad}{v_{avg}}$$
And, you know that $$v_{avg}=\frac{v_0+v_f}{2}$$
substitute for $$v_{avg}$$ in the equation above.
$$v_f-v_0=\frac{ad}{\frac{v_0+v_f}{2}}$$
eliminating the fraction in the fraction by multiplying the numerator and denominator by 2, gives:
$$v_f-v_0=\frac{2ad}{v_0+v_f}$$
multiply both sides by $$v_0+v_f$$ and you get (after FOIL)
$$v_f^2-v_0^2=2ad$$
Rearrange this as $$v_f^2=v_0^2+2ad$$
and, finally, take the square root to find your final velocity. If you follow this, you're going to have unknown variables each step of the way which you will have to keep substituting in. Definitely a pain in the neck to do, but I can't think of another way (perhaps there's a quicker derivation though) besides this to do your problem. I suspect that if your teacher hasn't covered this 5th or 6th formula yet, then perhaps it was a mistake that he gave you this problem already. (I do so every year intentionally, just so I have a reason to go through this derivation; but never as a homework problem.)
Last edited:
Thank you for your help, i try to figure something out.
bayer said:
Thank you for your help, i try to figure something out.
just follow the two steps radou gave you. From one of your posts above, I think you're having a problem with different letters being used.
For displacement, sometimes d is used, sometimes s, and sometimes x.
Likewise, for the initial velocity, sometimes it's $$v_0$$ and in other books, it's $$v_i$$ | HuggingFaceTB/finemath | |
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### Section 1-4 : Partial Fractions
In this section we are going to take a look at integrals of rational expressions of polynomials and once again let’s start this section out with an integral that we can already do so we can contrast it with the integrals that we’ll be doing in this section.
\begin{align*}\int{{\frac{{2x - 1}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{1}{u}\,du}}\hspace{0.25in}{\mbox{using }}u = {x^2} - x - 6\,\,{\mbox{ and }}du = \left( {2x - 1} \right)dx\\ & = \ln \left| {{x^2} - x - 6} \right| + c\end{align*}
So, if the numerator is the derivative of the denominator (or a constant multiple of the derivative of the denominator) doing this kind of integral is fairly simple. However, often the numerator isn’t the derivative of the denominator (or a constant multiple). For example, consider the following integral.
$\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}}$
In this case the numerator is definitely not the derivative of the denominator nor is it a constant multiple of the derivative of the denominator. Therefore, the simple substitution that we used above won’t work. However, if we notice that the integrand can be broken up as follows,
$\frac{{3x + 11}}{{{x^2} - x - 6}} = \frac{4}{{x - 3}} - \frac{1}{{x + 2}}$
then the integral is actually quite simple.
\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}
This process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition. Many integrals involving rational expressions can be done if we first do partial fractions on the integrand.
So, let’s do a quick review of partial fractions. We’ll start with a rational expression in the form,
$f\left( x \right) = \frac{{P\left( x \right)}}{{Q\left( x \right)}}$
where both $$P\left( x \right)$$ and $$Q\left( x \right)$$ are polynomials and the degree of $$P\left( x \right)$$ is smaller than the degree of $$Q\left( x \right)$$. Recall that the degree of a polynomial is the largest exponent in the polynomial. Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember.
So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.
Factor in
denominator
Term in partial
fraction decomposition
$$\displaystyle ax + b$$ $$\displaystyle \frac{A}{{ax + b}}$$
$$\displaystyle {\left( {ax + b} \right)^k}$$ $$\displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} + \cdots + \frac{{{A_k}}}{{{{\left( {ax + b} \right)}^k}}}$$, $$k = 1,2,3, \ldots$$
$$\displaystyle a{x^2} + bx + c$$ $$\displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}}$$
$$\displaystyle {\left( {a{x^2} + bx + c} \right)^k}$$ $$\displaystyle \frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \cdots + \frac{{{A_k}x + {B_k}}}{{{{\left( {a{x^2} + bx + c} \right)}^k}}}$$, $$k = 1,2,3, \ldots$$
Notice that the first and third cases are really special cases of the second and fourth cases respectively.
There are several methods for determining the coefficients for each term and we will go over each of those in the following examples.
Let’s start the examples by doing the integral above.
Example 1 Evaluate the following integral. $\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}}$
Show Solution
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
$\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{A}{{x - 3}} + \frac{B}{{x + 2}}$
The next step is to actually add the right side back up.
$\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{{A\left( {x + 2} \right) + B\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}$
Now, we need to choose $$A$$ and $$B$$ so that the numerators of these two are equal for every $$x$$. To do this we’ll need to set the numerators equal.
$3x + 11 = A\left( {x + 2} \right) + B\left( {x - 3} \right)$
Note that in most problems we will go straight from the general form of the decomposition to this step and not bother with actually adding the terms back up. The only point to adding the terms is to get the numerator and we can get that without actually writing down the results of the addition.
At this point we have one of two ways to proceed. One way will always work but is often more work. The other, while it won’t always work, is often quicker when it does work. In this case both will work and so we’ll use the quicker way for this example. We’ll take a look at the other method in a later example.
What we’re going to do here is to notice that the numerators must be equal for any x that we would choose to use. In particular the numerators must be equal for $$x = - 2$$ and $$x = 3$$. So, let’s plug these in and see what we get.
\begin{align*}x & = - 2 : & \hspace{0.5in}5 & = A\left( 0 \right) + B\left( { - 5} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}B & = - 1\\ x & = 3 \,\,\,\,: & \hspace{0.5in}20 & = A\left( 5 \right) + B\left( 0 \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}A & = 4\end{align*}
So, by carefully picking the $$x$$’s we got the unknown constants to quickly drop out. Note that these are the values we claimed they would be above.
At this point there really isn’t a whole lot to do other than the integral.
\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = \int{{\frac{4}{{x - 3}}\,dx}} - \int{{\frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}
Recall that to do this integral we first split it up into two integrals and then used the substitutions,
$u = x - 3\hspace{0.5in}v = x + 2$
on the integrals to get the final answer.
Before moving onto the next example a couple of quick notes are in order here. First, many of the integrals in partial fractions problems come down to the type of integral seen above. Make sure that you can do those integrals.
There is also another integral that often shows up in these kinds of problems so we may as well give the formula for it here since we are already on the subject.
$\int{{\frac{1}{{{x^2} + {a^2}}}\,dx}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c$
It will be an example or two before we use this so don’t forget about it.
Now, let’s work some more examples.
Example 2 Evaluate the following integral. $\int{{\frac{{{x^2} + 4}}{{3{x^3} + 4{x^2} - 4x}}\,dx}}$
Show Solution
We won’t be putting as much detail into this solution as we did in the previous example. The first thing is to factor the denominator and get the form of the partial fraction decomposition.
$\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}$
The next step is to set numerators equal. If you need to actually add the right side together to get the numerator for that side then you should do so, however, it will definitely make the problem quicker if you can do the addition in your head to get,
${x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$
As with the previous example it looks like we can just pick a few values of $$x$$ and find the constants so let’s do that.
\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow & \hspace{0.25in}A & = - 1\\ x & = - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}
Note that unlike the first example most of the coefficients here are fractions. That is not unusual so don’t get excited about it when it happens.
Now, let’s do the integral.
\begin{align*}\int{{\frac{{{x^2} + 4}}{{3{x^3} + 4{x^2} - 4x}}\,dx}} & = \int{{ - \frac{1}{x} + \frac{{\frac{1}{2}}}{{x + 2}} + \frac{{\frac{5}{2}}}{{3x - 2}}\,dx}}\\ & = - \ln \left| x \right| + \frac{1}{2}\ln \left| {x + 2} \right| + \frac{5}{6}\ln \left| {3x - 2} \right| + c\end{align*}
Again, as noted above, integrals that generate natural logarithms are very common in these problems so make sure you can do them. Also, you were able to correctly do the last integral right? The coefficient of $$\frac{5}{6}$$ is correct. Make sure that you do the substitution required for the term properly.
Example 3 Evaluate the following integral. $\int{{\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\,dx}}$
Show Solution
This time the denominator is already factored so let’s just jump right to the partial fraction decomposition.
$\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\, = \frac{A}{{x - 4}} + \frac{B}{{{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3}}$
Setting numerators gives,
${x^2} - 29x + 5 = A\left( {x - 4} \right)\left( {{x^2} + 3} \right) + B\left( {{x^2} + 3} \right) + \left( {Cx + D} \right){\left( {x - 4} \right)^2}$
In this case we aren’t going to be able to just pick values of $$x$$ that will give us all the constants. Therefore, we will need to work this the second (and often longer) way. The first step is to multiply out the right side and collect all the like terms together. Doing this gives,
${x^2} - 29x + 5 = \left( {A + C} \right){x^3} + \left( { - 4A + B - 8C + D} \right){x^2} + \left( {3A + 16C - 8D} \right)x - 12A + 3B + 16D$
Now we need to choose $$A$$, $$B$$, $$C$$, and $$D$$ so that these two are equal. In other words, we will need to set the coefficients of like powers of $$x$$ equal. This will give a system of equations that can be solved.
\left. \begin{align*}{x^3} & :\hspace{0.25in} & A + C & = 0\\ {x^2} & :\hspace{0.25in} & - 4A + B - 8C + D & = 1\\ {x^1} & :\hspace{0.25in} & 3A + 16C - 8D & = - 29\\ {x^0} & :\hspace{0.25in} & - 12A + 3B + 16D & = 5\end{align*} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 1,\,B = - 5,\,C = - 1,\,D = 2
Note that we used $${x^0}$$ to represent the constants. Also note that these systems can often be quite large and have a fair amount of work involved in solving them. The best way to deal with these is to use some form of computer aided solving techniques.
Now, let’s take a look at the integral.
\begin{align*}\int{{\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\,dx}} & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} + \frac{{ - x + 2}}{{{x^2} + 3}}\,dx}}\\ & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} - \frac{x}{{{x^2} + 3}}\, + \frac{2}{{{x^2} + 3}}\,dx}}\\ & = \ln \left| {x - 4} \right| + \frac{5}{{x - 4}} - \frac{1}{2}\ln \left| {{x^2} + 3} \right| + \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + c\end{align*}
In order to take care of the third term we needed to split it up into two separate terms. Once we’ve done this we can do all the integrals in the problem. The first two use the substitution $$u = x - 4$$, the third uses the substitution $$v = {x^2} + 3$$ and the fourth term uses the formula given above for inverse tangents.
Example 4 Evaluate the following integral. $\int{{\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\,dx}}$
Show Solution
Let’s first get the general form of the partial fraction decomposition.
$\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\, = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 4} \right)}^2}}}$
Now, set numerators equal, expand the right side and collect like terms.
\begin{align*}{x^3} + 10{x^2} + 3x + 36 & = A{\left( {{x^2} + 4} \right)^2} + \left( {Bx + C} \right)\left( {x - 1} \right)\left( {{x^2} + 4} \right) + \left( {Dx + E} \right)\left( {x - 1} \right)\\ & = \left( {A + B} \right){x^4} + \left( {C - B} \right){x^3} + \left( {8A + 4B - C + D} \right){x^2} + \\ & \hspace{0.5in}\hspace{0.25in}\left( { - 4B + 4C - D + E} \right)x + 16A - 4C - E\end{align*}
Setting coefficient equal gives the following system.
\left. \begin{align*}{x^4} & :\hspace{0.25in} & A + B & = 0\\ {x^3} & :\hspace{0.25in} & C - B & = 1\\ {x^2} & : \hspace{0.25in} & 8A + 4B - C + D & = 10\\ {x^1} & : \hspace{0.25in} & - 4B + 4C - D + E & = 3\\ {x^0} & :\hspace{0.25in} & 16A - 4C - E & = 36\end{align*} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 2,\,B = - 2,\,C = - 1,\,D = 1,\,E = 0
Don’t get excited if some of the coefficients end up being zero. It happens on occasion.
Here’s the integral.
\begin{align*}\int{{\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\,dx}} & = \int{{\frac{2}{{x - 1}} + \frac{{ - 2x - 1}}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = \int{{\frac{2}{{x - 1}} - \frac{{2x}}{{{x^2} + 4}} - \frac{1}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = 2\ln \left| {x - 1} \right| - \ln \left| {{x^2} + 4} \right| - \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) - \frac{1}{2}\frac{1}{{{x^2} + 4}} + c\end{align*}
To this point we’ve only looked at rational expressions where the degree of the numerator was strictly less that the degree of the denominator. Of course, not all rational expressions will fit into this form and so we need to take a look at a couple of examples where this isn’t the case.
Example 5 Evaluate the following integral. $\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}}$
Show Solution
So, in this case the degree of the numerator is 4 and the degree of the denominator is 3. Therefore, partial fractions can’t be done on this rational expression.
To fix this up we’ll need to do long division on this to get it into a form that we can deal with. Here is the work for that.
\require{enclose} \begin{align*} &\,\,\, x - 2\\ {x^3} - 3{x^2} & \enclose{longdiv}{{x^4} - 5{x^3} + 6{x^2} - 18} \\ - & \underline {\left( {{x^4} - 3{x^3}} \right)\hspace{0.9in}} \\ & \hspace{0.3in} - 2{x^3} + 6{x^2} - 18\\ & \underline { \hspace{0.08in}-\left( { - 2{x^3} + 6{x^2}} \right) \hspace{0.4in}} \\ & \hspace{1.35in} - 18\end{align*}
So, from the long division we see that,
$\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\, = x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}$
and the integral becomes,
\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\\ & = \int{{x - 2\,dx}} - \int{{\frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\end{align*}
The first integral we can do easily enough and the second integral is now in a form that allows us to do partial fractions. So, let’s get the general form of the partial fractions for the second integrand.
$\frac{{18}}{{{x^2}\left( {x - 3} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 3}}$
Setting numerators equal gives us,
$18 = Ax\left( {x - 3} \right) + B\left( {x - 3} \right) + C{x^2}$
Now, there is a variation of the method we used in the first couple of examples that will work here. There are a couple of values of $$x$$ that will allow us to quickly get two of the three constants, but there is no value of $$x$$ that will just hand us the third.
What we’ll do in this example is pick $$x$$’s to get the two constants that we can easily get and then we’ll just pick another value of $$x$$ that will be easy to work with (i.e. it won’t give large/messy numbers anywhere) and then we’ll use the fact that we also know the other two constants to find the third.
\begin{align*}x & = 0 : & \hspace{0.25in} 18 & = B\left( { - 3} \right) & \hspace{0.15in}\Rightarrow \hspace{0.25in}B & = - 6\\ x & = 3 : & \hspace{0.25in} 18 & = C\left( 9 \right) & \hspace{0.15in} \Rightarrow \hspace{0.25in}C & = 2\\ x & = 1 : & 18 & = A\left( { - 2} \right) + B\left( { - 2} \right) + C = - 2A + 14 & \hspace{0.15in} \Rightarrow \hspace{0.25in}A & = - 2\end{align*}
The integral is then,
\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2\,dx}} - \int{{ - \frac{2}{x} - \frac{6}{{{x^2}}} + \frac{2}{{x - 3}}\,dx}}\\ & = \frac{1}{2}{x^2} - 2x + 2\ln \left| x \right| - \frac{6}{x} - 2\ln \left| {x - 3} \right| + c\end{align*}
In the previous example there were actually two different ways of dealing with the $${x^2}$$ in the denominator. One is to treat it as a quadratic which would give the following term in the decomposition
$\frac{{Ax + B}}{{{x^2}}}$
and the other is to treat it as a linear term in the following way,
${x^2} = {\left( {x - 0} \right)^2}$
which gives the following two terms in the decomposition,
$\frac{A}{x} + \frac{B}{{{x^2}}}$
We used the second way of thinking about it in our example. Notice however that the two will give identical partial fraction decompositions. So, why talk about this? Simple. This will work for $${x^2}$$, but what about $${x^3}$$ or $${x^4}$$? In these cases, we really will need to use the second way of thinking about these kinds of terms.
${x^3}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}}\hspace{0.5in}{x^4}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} + \frac{D}{{{x^4}}}$
Let’s take a look at one more example.
Example 6 Evaluate the following integral. $\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}}$
Show Solution
In this case the numerator and denominator have the same degree. As with the last example we’ll need to do long division to get this into the correct form. We’ll leave the details of that to you to check.
$\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} = \int{{1 + \frac{1}{{{x^2} - 1}}\,dx}} = \int{{dx}} + \int{{\frac{1}{{{x^2} - 1}}\,dx}}$
So, we’ll need to partial fraction the second integral. Here’s the decomposition.
$\frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}}$
Setting numerator equal gives,
$1 = A\left( {x + 1} \right) + B\left( {x - 1} \right)$
Picking value of $$x$$ gives us the following coefficients.
\begin{align*}x & = - 1 : & \hspace{0.25in} 1 & = B\left( { - 2} \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}B & = - \frac{1}{2}\\ x & = 1 \,\,\,\, : & \hspace{0.25in}1 & = A\left( 2 \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}A & = \frac{1}{2}\end{align*}
The integral is then,
\begin{align*}\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} & = \int{{dx}} + \int{{\frac{{\frac{1}{2}}}{{x - 1}} - \frac{{\frac{1}{2}}}{{x + 1}}\,dx}}\\ & = x + \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + c\end{align*} | HuggingFaceTB/finemath | |
# The Naked Scientists Forum
### Author Topic: ABV of Vodka mixed with sweets. (Read 3251 times)
#### kyleglanville
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• Posts: 3
##### ABV of Vodka mixed with sweets.
« on: 09/09/2012 22:34:04 »
Hi, (sorry for posting twice: I thought my first one may be in the wrong section!)
A slightly different type of question I would assume!
If I mix 500ml of 37.5% vol vodka (which I believe weighs 1220g) with 200g of pick n mix sweets so that all of the sweets dissolved completely into the vodka, what would the new alcoholic content of the vodka be?
In my head I have it that:
1220g of vodka (@ 37.5% vol)
+
200g of sweets (@ 0%vol)
=
1420g of mixed vodka and sweets
Therefore: 1220 / 1420 = 0.859 (3 s.f.)
0.859 * 37.5 = 32.2 (3 s.f.)
New concentration of mixed vodka and sweets = 32.2% vol (3 s.f.)
Is this correct please? Thank you!
#### CliffordK
• Neilep Level Member
• Posts: 6321
• Thanked: 3 times
• Site Moderator
##### Re: ABV of Vodka mixed with sweets.
« Reply #1 on: 12/09/2012 07:27:25 »
Keep in mind that you are doing percent volume, rather than percent weight.
Here you are assuming that 1 gram of anything = 1 ml.
That is close...
However, the density of Ethanol is .789 g/cc, rather than 1 g/cc.
Hmmm
500ml of vodka will have to weigh less than 500ml of water, so it would have to weigh less than 500g. Somewhere the 1220g must be wrong. Did you weigh the bottle?
Anyway, the first thing to do is to figure out how much water, and how much alcohol you have.
500ml * 0.625 = 312.5 ml (g) water.
500ml * 0.375 = 187.5 ml alcohol
187.5 ml * .789 g/ml = 147.9 g ethanol.
Total: 312.5 g water + 147.9 g ethanol = 460.4 g solution.
Add 200g --> 660.4 g solution.
WEIGHT Percent Ethanol = 147.9 g / 660.4 g = 22.4%
If you wanted to calculate volume percent, you would need to know the volume of the sweets in the solution.
The density of sugar is about 1.59 g/cm³. However, my guess is that the density in solution is closer to 2 g/cm³.
So, let's put in 100cm³ = 200g of sweets in solution as an estimate
From above:
312.5 ml (g) water
187.5 ml alcohol
500 ml solution.
600 ml final solution (approx).
Volume Percent Ethanol 187.5 ml / 600ml = 31.2% (approx).
#### RE.Craig
• Full Member
• Posts: 61
##### Re: ABV of Vodka mixed with sweets.
« Reply #2 on: 10/02/2013 18:54:55 »
Spot on answer Clifford. I decided not to post my own reply after yours.
#### Bored chemist
• Neilep Level Member
• Posts: 8669
• Thanked: 42 times
##### Re: ABV of Vodka mixed with sweets.
« Reply #3 on: 11/02/2013 22:05:49 »
Spot on answer Clifford. I decided not to post my own reply after yours.
Nearly half a year after.
#### RE.Craig
• Full Member
• Posts: 61
##### Re: ABV of Vodka mixed with sweets.
« Reply #4 on: 12/02/2013 04:47:12 »
Spot on answer Clifford. I decided not to post my own reply after yours.
Nearly half a year after.
lol...
#### The Naked Scientists Forum
##### Re: ABV of Vodka mixed with sweets.
« Reply #4 on: 12/02/2013 04:47:12 » | HuggingFaceTB/finemath | |
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## APยฎ๏ธ Calculus AB (2017 edition)
### Course: APยฎ๏ธ Calculus AB (2017 edition)ย >ย Unit 8
Lesson 1: Definite integrals intro
# Warmup: Definite integrals intro
Practice evaluating definite integrals by finding the area using shapes (like rectangles & circles) under a function.
## Problem 1
The graph of a function $f$ is shown below.
a) ${โซ}_{โ6}^{1}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
b) ${โซ}_{1}^{4}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
c) ${โซ}_{โ6}^{4}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
## Problem 2
The graph of a function $g$ is shown below. The graph consists of a semicircle and linear segments.
a) ${โซ}_{โ7}^{โ1}g\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
b) ${โซ}_{โ1}^{8}g\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
c) ${โซ}_{โ7}^{8}g\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
## Challenge problem
The graph of a function $h$ is shown below. The graph consists of a semicircle and linear segments.
a) ${โซ}_{โ7}^{โ2}h\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
b) ${โซ}_{โ5}^{1}h\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
c) ${โซ}_{โ1}^{8}h\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
## Want to join the conversation?
• The Definite Integrals Introduction section should be after the Riemann Sums sections. Otherwise, the notations used make no sense (I didn't encounter them until now). Is there a problem with the order of the sections or am I wrong with something?
• I can understand what many of you are thinking, so letโs examine what is really happening here.
In the section previous to this you were introduced to the notation and concept of the indefinite integral, and a few rules of integration of various functions in as much as to demonstrate integration being the inverse of differentiation.
In this section you are introduced to the notation of the definite integral, (which is essentially the same as the indefinite integral in the previous section, with the only difference being that the limits of integration are added) and you are told that the definite integral is going to be used to find the area under the curve. If you look carefully, however, in none of the questions/examples in this section are you actually being asked to integrate using the rules of integration. In all the examples, the area is determined using, by what should be by now well known to you, geometric formulas for area (circles, squares, triangles). So at this point the curious student should be thinking along the lines of, โwell that was certainly trivial, I havenโt been shown anything I couldnโt do before without the integral notation, so tell me, what happens when the curves of the functions are not so easily described with just lines and semi -circles?โ . . . .
In the next section you learn an intuitive โsumming method to approximate area under a (relatively) arbitrary curve,โ and, how via using a limiting process on the summing method the exact area can be found leading to why the notation came about.
This is a typical โshow them where theyโre going, then show them how to get thereโ approach; and you will see it a lot โ and equally as often, once you have been shown how to get there, you still wonโt have enough mathematical maturity to understand why it works. Depending on your reasons for studying calculus, for example, if you just want to graduate in another discipline that simply requires calculus, and not graduate in mathematics, you may never learn why a process works.
This teaching style might not be to your personal liking, but it is not at all uncommon.
• This section was confusing, and even though I've figured it out now, I still don't get why it's presented like this. Usually Sal introduces a concept, even if only the basic notation and function, before Khan asks you to try it yourself. This worksheet shoves the learner into unfamiliar territory, assuming (s)he knows exactly what the new notation means. Maybe it expects him/her to ask for the hint...? But the hint still assumes (s)he knows that a definite integral can be represented by the area between the function in the x-axis: "Until now, we've been using approximate area to estimate the definite integral." I've calculated that sort of area just once or twice before in differential calc., and I'm pretty sure the exercises didn't mention the area-integral relationship. It just doesn't make sense to put this worksheet here without a concept video/article.
• User Stefen answered user Datcu Octavian's question with an amazing and extremely helpful answer in this same Question list. Look for it until you find it as it is a (near) perfect answer to your question.
• This was really difficult and not particularly educational because there was very little explanation.
• It is meant to be an introductory approach to the visualisation/intuition of what integral calculus is....graphically speaking is the area under the curve...This is true for single integrals....(double integrals give you the volume of the shape you are studying!)
• I got confused because I could not find explanation for the connection btw area and integral.
But I found them on YouTube.
• Thank you very much, and I agree--this video (and its part 2) define why a definite integral defines area, why the Integral symbol is an elongated S (for "sum"). It was clearly cut from this chapter and no updated version made.
I don't know why so many people are giving a "take it or leave it", "it's by design" stance in answering other questions on this matter. In the first video of the next section ( https://www.khanacademy.org/math/ap-calculus-ab/definite-integrals-intro-ab/definite-integral-properties-ab/v/integrating-scaled-function ) at , Sal says "When you think back to the Riemann sums". It's clearly a screw-up or oversight in updating the site/chapter structure, and should be fixed.
• Although I found this section very straightforward (because I learned all this 35 years ago!), I agree with most of the comments below - if not a video then an introductory passage is required explaining that definite integrals can be used to determine net areas. Plus an explanation of the specific notation for definite intervals... Please amend accordingly!
• I don't understand why they assume we know what it means to evaluate a definite integral. The previous section focused on indefinite integrals and if we just followed that course, we would've only learned about finding the anti-derivatives of common functions. Are we somehow meant to realize that evaluating a definite integral means to find the area under a curve in a given interval? What I'm complaining about isn't the fact that they didn't teach us how to find definite integrals first before introducing this article, but the fact that they didn't explicitly say what it meant to evaluate a definite integral. If it wasn't for the title of the article, I probably wouldn't have found out that the definite integral had something to do with area.
• It seems something is missing here? I checked backwards and forwards but couldn't find any videos explaining why definite integral is related to area under the curve. | HuggingFaceTB/finemath | |
# Contradictions?
#### Zeno
Registered Senior Member
Here's a spaceship with Sally in the corner....
Code:
`````` /\ | -------------------> V
/ \ |
/ \ |
/ \ |
/ Sally John``````
Sally is in the lower right corner of the spaceship with 2 flashlights. One flashlight is aimed straight up at a mirror which reflects the light back down towards Sally. This is a light clock with the light bouncing straight up and down. She has another flashlight aimed up and to the left towards a mirror. So this is another light clock but at an angle. The ship is moving to the right with velocity V according to John who is outside the ship. This ship's velocity is such that from John's point of view the light from the angled flashlight isn't going at an angle but is going straight up and down because the rightward velocity of the ship exactly matches the leftward movement of the light.
Upward flashlight: Sally sees the light up and down. John sees it at an angle.
Angled flashlight: Sally sees the light at an angle. John sees the light up and down.
The 2 situations are reversed.
If T and t represent time for the 2 different reference frames then using the pythagorean theorem we can derive 2 contradictory time dilation formulas:
delta T = delta t/((1 - (v/c)^2)^.5)
delta t = delta T/((1 - (v/c)^2)^.5)
Last edited:
Here's a spaceship with Sally in the corner....
Code:
`````` /\ | -------------------> V
/ \ |
/ \ |
/ \ |
/ Sally John``````
Sally is in the lower right corner of the spaceship with 2 flashlights. One flashlight is aimed straight up at a mirror which reflects the light back down towards Sally. This is a light clock with the light bouncing straight up and down. She has another flashlight aimed up and to the left towards a mirror. So this is another light clock but at an angle. The ship is moving to the right with velocity V according to John who is outside the ship. This ship's velocity is such that from John's point of view the light from the angled flashlight isn't going at an angle but is going straight up and down because the rightward velocity of the ship exactly matches the leftward movement of the light.
Upward flashlight: Sally sees the light up and down. John sees it at an angle.
Angled flashlight: Sally sees the light at an angle. John sees the light up and down.
The 2 situations are reversed.
If T and t represent time for the 2 different reference frames then using the pythagorean theorem we can derive 2 contradictory time dilation formulas:
delta T = delta t/((1 - (v/c)^2)^.5)
delta t = delta T/((1 - (v/c)^2)^.5)
I'm not sure why you are going to all this trouble about light being aimed at a angle. All you have to do is give both Sally and John their own light clock. Each will measure the light from their own clock as going straight up and down and the light from the other light clock as traveling at an angle. They also will both conclude that the other light clock is ticking slower than their own. This is not a contradiction but just a consequence of Relativity. Time dilation is reciprocal.
But time dilation is just one part of Relativity, for a complete picture of what is happening with Relativity, you also have to account for Length contraction and the Relativity of simultaneity.
So for example, the fact that Sally and John disagree as to who's clock is ticking slower is never an issue as long as they keep a constant velocity with respect to each other. As under this scenario they could only meet up once. If they are to meet up more than once, one, the other, or both has to change their velocity, and how this velocity change takes place will determine which of the two's (if either) has accumulated less time between their two meetings.
Sally's angled flashlight is at such an angle that the light follows the same angled path right to left as the path of the light from the upward flashlight as seen by John as Sally moves left to right and Sally's rightward movement exactly matches the leftward movement of the light.
Therefore:
Sally's view of the up and down movement of the light from her upward flashlight exactly matches John's view of the up and down movement from Sally's angled flashlight.
Sally's view of the movement of the angled light from her angled flashlight exactly matches John's view of the movement of the light from Sally's upward flashlight as Sally moves left to right.
You don't see a problem here?
Sally's angled flashlight is at such an angle that the light follows the same angled path right to left as the path of the light from the upward flashlight as seen by John as Sally moves left to right and Sally's rightward movement exactly matches the leftward movement of the light.
Therefore:
Sally's view of the up and down movement of the light from her upward flashlight exactly matches John's view of the up and down movement from Sally's angled flashlight.
Sally's view of the movement of the angled light from her angled flashlight exactly matches John's view of the movement of the light from Sally's upward flashlight as Sally moves left to right.
Correct, which is why Janus said that you can avoid the complication by just letting John hold the other flashlight. The light takes the same path at the same speed regardless of who shines the light.
You don't see a problem here?
Not at all. Again, as was said above, this is just a consequence of relativity.
I will try to illustrate my point with a specific example. Sally is moving to the right at .6c. The height of her spaceship is .8 light-seconds. If Sally has a light clock with the light bouncing straight up and down the light will make a 3-4-5 right triangle from the viewpoint of John. If the change in time for Sally is delta T_o and the change in time for John is delta T then the following equation can be derived:
delta T = delta T_o/((1-.6^2)^.5)
Now Sally has a light clock but this time she is holding a flashlight at an angle of 53.13 degrees above the horizontal and pointed to the left. Now the leftward movement of the light exactly matches the rightward movement of the spaceship from John's viewpoint. Now the light is bouncing straight up and down from the viewpoint of John and the light is making a 3-4-5 right triangle from viewpoint of Sally. If the change in time for Sally is delta T_o and the change in time for John is delta T then the following equation can be derived:
delta T_o= delta T/((1-.6^2)^.5)
The 2 equations are in direct contradiction to each other.
Last edited: | HuggingFaceTB/finemath | |
# Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$
I have the following question
$$\int x \sqrt{1 - x^4} \,\mathrm{d}x$$
I know we have to use trig. substitution for this and therefore, I did the following by letting $x = \sin \theta$ and $dx = \cos \theta \,\mathrm{d}\theta$
\begin{align} &\int x \sqrt{1-x^4} \,\mathrm{d}x \\ &=\int \sin \theta \cos \theta\sqrt{1 - (\sin \theta)^4} \, \mathrm{d}\theta \\ &=\int \sin \theta \cos \theta \sqrt{(1-\sin^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\ &=\int \sin \theta \cos \theta \sqrt{(\cos^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\ \end{align}
Now, I'm confused. How do I proceed?
Thanks!
EDIT: Taking from the answer, I have a (nearly) full solution below for future users.
Instead of letting $x = \sin \theta$. We'll let $x^2 = \sin \theta$ and this will greatly simplify everything. Since, $\mathrm{d}x = \frac{\cos \theta}{2x} \,\mathrm{d\theta}$
\begin{align} &\int x \sqrt{1-x^4} \quad \mathrm{d}x \\ &=\frac{1}{2}\int \cos \theta\sqrt{1 - (\sin x)^2} \quad \mathrm{d}\theta \\ &= \frac{1}{2} \int \cos \theta \cos \theta \quad \mathrm{d}\theta \\ &= \frac{1}{4} \int 1 + \cos 2\theta \quad \mathrm{d}\theta \\ &= \frac{1}{4} \left(\theta + \frac{\sin2\theta}{2} \right) \\ &= \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} \\ \end{align}
After this, you only have to put $\sin \theta$ back in terms of $x$ and you're done!
• What about $x^2=\sin(y)$ ? – Claude Leibovici Apr 17 '14 at 14:26
• Well...that simplifies everything a whole lot - doesn't it? Thanks! – Jeel Shah Apr 17 '14 at 14:28
• You coud try first the substitution $u=x^2$. Your integral becomes $\frac{1}{2}\int\sqrt{1-u^2}du$. Then you could use e.g. $u=\cos\theta.$ – Américo Tavares Apr 17 '14 at 14:44
• @AméricoTavares That is a good idea, since that is essentially where the motivation for using $\cos (u) = x^2$ comes from (it's important to learn the motivation IMO) – MCT Apr 17 '14 at 14:51
Try another substitution: $x^2 = \sin (u)$.
We have $2x dx = \cos (u) du$ so $dx = \frac{\cos (u)}{2x} du$
So now we have $\frac{1}{2} \displaystyle \int \cos(u) \sqrt{1 - \sin^2 (u)} du$
And I think you can do the rest.
HINT :
Let $u^2=1-x^4\;\Rightarrow\; x^2=\sqrt{1-u^2}\;\Rightarrow\; x\ dx=-\dfrac{u\ du}{2\sqrt{1-u^2}}$, then rewrite \begin{align} \int x\sqrt{1-x^4}\ dx&=-\frac12\int\dfrac{u^2}{\sqrt{1-u^2}}\ du\\ &=\frac12\int\dfrac{1-u^2-1}{\sqrt{1-u^2}}\ du\\ &=\frac12\left(\int\dfrac{1-u^2}{\sqrt{1-u^2}}\ du-\int\dfrac{1}{\sqrt{1-u^2}}\ du\right). \end{align} The left part integral can be solved by using IBP and the right part integral can be solved by using trigonometry substitution.
I have just found that the fastest way is letting $u=x^2$. | HuggingFaceTB/finemath | |
+91 44 4500 8484 +91 93800 48484
GRE Quantitative Comparison Questionbank
2
Topics
6
Concepts
2
Questions
0
Videos
What is Quantitative Comparison in GRE?
The quantitative reasoning section of the GRE includes quantitative comparison questions. In a quantitative comparison question, you will asked to compare two quantities - Quantity A and Quantity and then determine which between the two is greater. Note there are questions where the two quantities may be equal and is a possible answer.
The 4 answer choices for any quantitative comparison question in GRE is always the same and it makes sense to familiarize oneself with those choices. Wherever possible, back substitute numbers and avoid calculations whenever possible to determine the answer.
Practice Questions in Quantitative Comparison
1. Quantity A Quantity B
The probability that a word selected from the set of all rearrangements of the letters of the word "Math" results in "Math" The probability that a word selected from the set of all rearrangements of the letters of the word "Good" results in "Good"
1. Quantity A is greater
2. Quantity B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
2. Quantity A Quantity B
Length of the segment of the line 4x + 3y = 12 intercepted between the coordinate axes. Length of the median to side BC of triangle whose coordinates are A(4, 4), B(10, 4) and C(4, 12)
1. Quantity A is greater
2. Quantity B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
More GRE Prep Questions
Free GRE Practice Questions available from the following topics | HuggingFaceTB/finemath | |
Boy in sailor suit with blackboard with math
## Introduction
One of the most fundamental operations performed by computers, aside from the logical operations that we've already discussed, is the arithmetic operation of addition.
Let's consider what's required to add two, single-bit binary integers. We'll need one bit to represent the sum of the integers, and another to handle the carry. Representing this in the form of a truth table yields:
Inputs Outputs
${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle C_{out}}$ ${\displaystyle S}$
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0
This is formally termed a half-adder, a logic circuit capable of adding two bits.
Observe, Ponder, and Journal Section 1 What truth table do you recognize that produces the output of the Carry column? What truth table do you recognize that produces the output of the Sum column?
In order to add two single-bit binary integers PLUS a carry, we need an adder capable of adding three single-bit binary numbers. Again, we'll need one bit to represent the sum of the integers, and another to handle the carry. Representing this in the form of a truth table yields:
Inputs Outputs
${\displaystyle c_{in}}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle C_{out}}$ ${\displaystyle S}$
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
This is formally termed a full-adder, a logic circuit capable of adding three bits.
Observe, Ponder, and Journal Section 2 What do you notice about the relationship between the first-half (top four rows) of the full-adder as compared to all of the rows of the half-adder? Why is this true?
We've learned that a half-adder can add two bits and full-adder can add three bits. How can we add a multi-bit number such as a 16-bit word? By cascading four adders such that the carry output of the prior adder feeds the carry input of the subsequent adder we can add two four-bit numbers. This concept can be easily extended to an arbitrary number of bits.
Observe, Ponder, and Journal Section 3 Why does the least significant bit position use a half-adder rather than a full-adder? Assume that proper inputs are applied for all bits in numbers A and B. Will the correct output from S be available instantaneously? If not, why not? Assume that we have a standard (non-scientific calculator) capable of adding two 16-bit words. Two numbers, A and B, are added together. After the addition, it is noted that ${\displaystyle C_{15}}$ is high. What can we infer? What is this state commonly called?
## Key Concepts
Key Concepts A half-adder is a logic circuit capable of adding two bits and output a carry bit and a sum bit. A full-adder is a logic circuit capable of adding three bits and output a carry bit and a sum bit. A ripple-carry-adder is a logic circuit constructed of adders, cascaded in such a manner that the carry output of each adder feeds the carry input of the subsequent adder. Using this method we are able to add an arbitrary number of bits. | HuggingFaceTB/finemath | |
# Math Insight
### Generating networks with a desired second order motif frequency
One way to add structure beyond the Erdös-Rényi random network (ER random network) is to generate ensembles of networks with a prescribed degree distribution. Here we describe another approach: modulating the frequencies of small network motifs in directed graphs.
#### Network motifs
Network motifs are patterns of connections within a network. For example, the simplest motif consists of a single edge from one node to another.
Moving up one step in complexity, we can look at motifs involving two edges. If we restrict ourselves to connected motifs (excluding the motif with two edges connecting distinct pairs of nodes), there are four motifs with two edges, which we can label as the reciprocal, convergent, divergent, and chain motifs.
Reciprocal motif
Convergent motif
Divergent motif
Chain motif
One can characterize a network by counting the relative frequency of these five motifs. Here, each motif represents a pattern of the one or two edges shown without regard to the presence or absence of any of the edges not shown.
For example, the following four node network has six edges, so the single-edge motif occurs six times. The total possible number of connections is $4\cdot 3 = 12$, so the network contains 50% of the possible edges.
As highlighted on the right, the network contains two reciprocal motifs, five chain motifs, two convergent motifs, and two divergent motifs. The total number of possible reciprocal motifs is $4 \cdot 3/2 = 6$, so the network contains 1/3 of the possible reciprocal motifs. The total number of possible chain motifs is $4 \cdot 3 \cdot 2 = 24$, so the network contains $5/24$ of the possible chain motifs. For both convergent and divergent motifs, the total possible number is $4 \cdot 3 \cdot 2/2 = 12$; the network contains 1/6 of the possible convergent and 1/6 of the possible divergent motifs.
#### Generating networks with given expected motif frequencies
One can view the ER random network, as a probability distribution parametrized by the expected relative frequency of the single-edge motif, which is the same as the connection probability.
We could specify that, for any pair of nodes $i$ and $j$, the probability of the motif containing a connection from $j$ to $i$ is \begin{gather} \Pr(A_{ij}=1)=E(A_{ij})=p, \label{probedge} \end{gather} where the expected value $E(\cdot)$ is computed over the probability distribution of the adjacency matrix, equation (1) of the random network page.
In the ER random network, the second order connectivity motifs will occur, too. But, we cannot independently specify their frequencies. Since all edges occur independently with probability $p$, any pair of edges (such as the two-edge motifs) must occur with probability $p^2$. (The probability of two independent random variables is the product of their individual probabilities.)
Given that real world networks motif frequencies differ from the ER prediction, we'd like to generate networks where the two-connection motifs occur at frequencies other than $p^2$. We introduce four more parameters, $\alpha_{\text{recip}}$, $\alpha_{\text{conv}}$, $\alpha_{\text{div}}$, and $\alpha_{\text{chain}}$, through which to specify the frequencies of the reciprocal, convergent, divergent, and chain motifs, respectively. We define each $\alpha$ to indicate deviation from the ER model, so if an $\alpha$ is zero, the corresponding motif will appear with probability $p^2$. The definitions of the four $\alpha$ parameters in terms of the probability of the respective motif are as follows.
$$\Pr(A_{ij}\nobreak{=}1\nobreak{,}A_{ji}\nobreak{=}1) = p^2(1+\alpha_{\text{recip}})$$
$$\Pr(A_{ij}\nobreak{=}1\nobreak{,}A_{ik}\nobreak{=}1)=p^2(1+\alpha_{\text{conv}})$$
$$\Pr(A_{ij}\nobreak{=}1\nobreak{,}A_{kj}\nobreak{=}1)= p^2(1+\alpha_{\text{div}})$$
$$\Pr(A_{ij}\nobreak{=}1\nobreak{,}A_{jk}\nobreak{=}1)=p^2(1+\alpha_{\text{chain}})$$
As we manipulate the expected frequencies of the second order connection motifs, we are still holding the expected frequency of the single-edge motif at $p$. We cannot increase the frequency of any two-edge motif by adding additional edges, or that would increase the frequency of the single-edge motif. Instead, if we would like to increase the frequency of the reciprocal motif, for example, we need to increase the probability that both $A_{ij}$ and $A_{ji}$ are one and simultaneously increase the probability that both $A_{ij}$ and $A_{ji}$ are zero (while decreasing the one edge occurs without the other). We must do this in a way to keep the probability of all single-edge motifs at $p$.
We can express the fact that the $\alpha$'s simply must increase the coordination between their edges by writing them in terms of the covariances between edges. Using equation \eqref{probedge} for single-edge probability, one can rewrite the definitions for the $\alpha$'s as normalized covariances. \begin{align*} \alpha_{\text{recip}} &= \frac{\text{cov}(A_{ij},A_{ji})}{E(A_{ij})E(A_{ji})} &\alpha_{\text{conv}}&=\frac{\text{cov}(A_{ij},A_{ik})}{E(A_{ij})E(A_{ik})}\\ \alpha_{\text{div}}&=\frac{\text{cov}(A_{ij},A_{kj})}{E(A_{ij})E(A_{kj})} &\alpha_{\text{chain}}&=\frac{\text{cov}(A_{ij},A_{jk})}{E(A_{ij})E(A_{jk})} \end{align*}
For ER random networks, it is a simple matter to generate a network with given edge probability, as we can generate the random number for each edge independently using equation \eqref{probedge}. However, when we break this independence by adding correlations among the edges through nonzero $\alpha$'s, the process of generating a network with the required statistics is more subtle. It turns out one can define a random network model based on these statistics, the SONET (second order network) model1 and sample networks with different values of the $\alpha$'s. In this way, we obtain a probability distribution for the network parametrized by the first order statistic $p$ and the second order statistics $\alpha_{\text{recip}}$, $\alpha_{\text{conv}}$, $\alpha_{\text{div}}$, and $\alpha_{\text{chain}}$. | HuggingFaceTB/finemath | |
# difference of angles?
• Jun 27th 2006, 06:23 PM
Excess Thoughts
difference of angles? (image bug fixed)
Hello there everyone... I could really use some help with a question. I'm in the middle of making a 3d game engine... and there's just this one problem that I can't seem to get through... It's difficult to explain, so here:http://aycu30.webshots.com/image/989...4825285_th.jpg is a quick picture I made to make it easier to understand what I need...
You are given the values of A,B,C,D, and theta... what I need is a simple equation (or equations) that I could use to find ? for any real values of A,B,C,D, and theta. Is that even possible? I would appreciate anything (the sooner the better...) Thank you in advance.
- Scott
• Jun 27th 2006, 07:04 PM
ThePerfectHacker
Quote:
Originally Posted by Excess Thoughts
I'm in the middle of making a 3d game engine..
Wow, you must be so cool
Quote:
Originally Posted by Excess Thoughts
Is that even possible?
No, from the way I understand it.
• Jun 27th 2006, 07:58 PM
JakeD
Quote:
Originally Posted by Excess Thoughts
Hello there everyone... I could really use some help with a question. I'm in the middle of making a 3d game engine... and there's just this one problem that I can't seem to get through... It's difficult to explain, so here:http://aycu30.webshots.com/image/989...4825285_th.jpg is a quick picture I made to make it easier to understand what I need...
You are given the values of A,B,C,D, and theta... what I need is a simple equation (or equations) that I could use to find ? for any real values of A,B,C,D, and theta. Is that even possible? I would appreciate anything (the sooner the better...) Thank you in advance.
- Scott
$x = \theta - \tan^{-1}((D-B)/(C-A)) .$
• Jun 27th 2006, 08:06 PM
Soroban
Hello, Scott!
Quote:
You are given the values of $A,\,B,\,C,\,D,\,\theta = \angle ROX$
What I need is a simple equation (or equations) that I could use to find $\angle ROQ$
for any real values of $A,\,B,\,C,\,D,\,\theta$
Code:
R o o Q \ * * * / *\ * / * \ * * \ / * \ / * \ / * * O* - - - - * - - X * /(C,D) * / * / * * * / * * P o * * * (A,B)
We are given: $\theta = \angle ROX$.
We want $\angle ROQ$.
$m_1\:=\:m_{PQ} \;= \;m_{PO} \;= \;\frac{D-B}{C-A}$
$m_2\;=\;\tan\theta$
The angle $\phi$ between two lines is given by: . $\tan \phi \:=\:\frac{m_2 - m_1}{1 + m_1\cdot m_2}$
So we have: . $\tan(\angle ROQ) \;=\;\frac{\tan\theta - \frac{D-B}{C-A}}{1 + \frac{D-B}{C-A}\cdot\tan\theta}$
Therefore: . $\angle ROQ\;=\;\arctan\left[\frac{(C-A)\tan\theta - (D - B)}{(C - A) - (D - B)\tan\theta}\right]$
• Jun 27th 2006, 08:09 PM
ThePerfectHacker
Am I missing something here.
He is asking for '?' not for theta.
There is no way to find '?'
Since that radius vector can be anywhere. Unless you know its coordinates (which you do not).
• Jun 27th 2006, 11:06 PM
JakeD
Quote:
Originally Posted by Soroban
$\angle ROQ\;=\;\arctan\left[\frac{(C-A)\tan\theta - (D - B)}{(C - A) - (D - B)\tan\theta}\right]$
Quote:
Originally Posted by JakeD
$x = \theta - \tan^{-1}((D-B)/(C-A)) .$
My $x = \angle ROQ .$ I put these two formulas in a spreadsheet and checked that they do give the same values for various inputs. How would you prove this?
• Jun 27th 2006, 11:13 PM
malaygoel
Quote:
Originally Posted by JakeD
My $x = \angle ROQ .$ I put these two formulas in a spreadsheet and checked that they do give the same values for various inputs. How would you prove this?
Hello JakeD!
Let $D-B=(C-A)tany$
Substitute in Soroban's expression and you will arrive at your expression
Keep Smiling
Malay
• Jun 27th 2006, 11:46 PM
JakeD
Quote:
Originally Posted by malaygoel
Hello JakeD!
Let $D-B=(C-A)tany$
Substitute in Soroban's expression and you will arrive at your expression
Keep Smiling
Malay
Thanks for the reply, Malay. I tried what you suggested and didn't get an immediate result, so I'm not sure what you meant. Is this what you had in mind? Using
$\tan y = (D-B)/C-A)$
I get from my expression
$x = \theta - y$
and from Soroban's
$\tan x = \frac{\tan \theta - \tan y}{1+\tan y \tan \theta}.$
Substituting in the above expression for $x$ yields
$\tan (\theta - y) = \frac{\tan \theta - \tan y}{1+\tan y \tan \theta}$
which is one of the trig identities
http://www.sosmath.com/trig/Trig5/trig5/img6.gif
• Jun 27th 2006, 11:59 PM
malaygoel
Quote:
Originally Posted by JakeD
Thanks for the reply, Malay. I tried what you suggested and didn't get an immediate result, so I'm not sure what you meant. Is this what you had in mind? Using
$\tan y = (D-B)/C-A)$
I get from my expression
$x = \theta - y$
and from Soroban's
$\tan x = \frac{\tan \theta - \tan y}{1+\tan y \tan \theta}.$
Substituting in the above expression for $x$ yields
$\tan (\theta - y) = \frac{\tan \theta - \tan y}{1+\tan y \tan \theta}$
After the substitution inSoroban's expression, you will get
arctan $\frac{\tan \theta - \tan y}{1+\tan y \tan \theta}$
which is equal to
arctan $\tan (\theta - y)$
which is equal to
$\theta-y$
which is equal to
$\theta-\tan^{-1}((D-B)/(C-A))$which is your expression
Keep Smiling
Malay | HuggingFaceTB/finemath | |
# Percentile Calculator
You can use this percentile calculator to efficiently determine the p-th percentile for a set of numbers.
How to use the Percentile Calculator:
1. Input the numbers in the set separated by a comma (e.g., 1,9,18,12), space (e.g., 1 9 18 12), or line break.
2. Enter the percentile value you wish to determine.
3. Click on the "Calculate" button to generate the results.
Pth Percentile Calculator
Results
Solution:
## Definition of Percentile
The pth percentile is the value in a set of data at which it can be split into two parts. The lower part contains p percent of the data, and the upper part consists of the remaining data; i.e., 100-p (the total data equates to 100%).
## Calculating the pth Percentile
You can use the following process to manually determine the pth percentile of a set of data.
1) Arrange the data such that the entries span from the smallest to the largest values (ascending order).
2) Calculate an index i (the position of the pth percentile) as follows:
i = (p / 100) * n
Where: p is the percentile and n is the number of values that appear in the data.
If i is not an integer, round it up. The next integer greater than i represents the position of the pth percentile. If i is an integer, the pth percentile is the average of the values in positions i and i + 1.
## Examples
Let's say we want to determine the 80th percentile for the following values: 10, 55, 50, 80, 85, 90, 20, 40, 55, 65, 30, 25.
1) Arrange the values from lowest to highest: 10, 20, 25, 30, 40, 50, 55, 55, 65, 80, 85, 90.
2) Calculate index i (the position of the pth percentile):
i = (p / 100) * n = 80 / 100 * 12 = 9.6
3) In this case, i is not an integer. As such, you should round up. The point of the 80th percentile is the next integer above 9.6, which is the 10th position.
4) The 80th percentile is the data value that appears in the 10th position in the list of values: 80.
Let's look at a second example. This time, we want to calculate the 50th percentile for the set of data.
1) Having arranged the values in ascending order, we have: 10, 55, 50, 80, 85, 90, 20, 40, 55, 65, 30, 25.
2) Calculate index i (the position of the pth percentile):
i = (p / 100) * n = 50 / 100 * 12 = 6
3) In this case, i is an integer. As such, the 50th percentile is the average of the values that appear in the 6th and 7th positions in the set of ordered data: 50 and 55 respectively. Therefore, the 50th percentile of this data set is (50 + 55) / 2 = 52.5.
You may also be interested in our Interquartile Range Calculator | HuggingFaceTB/finemath | |
# Triangle Inequality Proof
1. Sep 24, 2010
### lovemake1
1. The problem statement, all variables and given/known data
Given:0 <= a <= b
a <= Sqrt(ab) <= (a+b)/2 <= b
2. Relevant equations
3. The attempt at a solution
The only problem im having prooving this inequality is Sqrt(ab) <= (a+b)/2.
I have an idea but im not sure if it validates.
can i do this.. ? (a+b)/2 - sqrt(ab) >= 0
if it is greater than 0 [ i get an answer of 1/4 (a-b)^2 >= 0 ] , (a+b)/2 must be greater than sqrt(ab). given that
both sqrt(ab) and (a+b)/2 are >= 0 since 0 < a < b.
Is my reasoning correct? or wrong | HuggingFaceTB/finemath | |
Examples
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
### Transcript
Question 3 Find the solution of sin x = – √3/2 Let sin x = sin y also sin x = (−√3)/2 From (1) and (2) sin y = (−√3)/2 sin y = sin 4/3 π ⇒ y = 4/3 π Step 2 Since sin x = sin y General Solution is x = n π + (–1)n y where n ∈ Z Putting y = 4𝜋/3 x = n π + (–1)n 4/3 π where n ∈ Z | HuggingFaceTB/finemath | |
# Variance of $Z = X_1 + X_1 X_2 + X_1 X_2 X_3 +\cdots$
Here the $$X_i$$'s are i.i.d. and such that convergence in distribution for the infinite sum, is guaranteed. Probably the easiest case is when $$X_i$$ has a Bernouilli($$p$$) distribution, then $$Z$$ has a discrete Geometric($$1-p$$) distribution. I am not interested in that case, but in more complicated examples.
Actually, I am trying to find what kind of distributions $$Z$$ could have. I am sure the choices are very limited. Those that work are called attractors. I tried a Normal($$0, 1$$) for $$X_i$$ and $$Z$$ has a nice, smooth distribution, but not normal. I tried log-normal too for $$X_i$$, but in that case there is no convergence. An attractor must satisfy the functional equation $$F_Z = F_{X_1(1+Z)}$$.
Most surprisingly, I tried $$X_i \sim \sin( \pi Y_i)$$ with $$Y_i$$ Normal($$0,1$$). The resulting distribution for $$Z$$ is pretty smooth (but not normal), and its variance, based on strong empirical evidence, seems to be exactly equal to $$1$$. Is this really the case?
I am also interested in the conditions required to have convergence in distribution for the infinite sum.
Update
The reason I failed to obtain convergence with a log-normal $$X_i$$ is because I picked one with $$E(X_i)>1$$. More on this to be published in Part II of this problem, here.
• is it true that if $X_i$ and $Z$ satisfy $F_Z = F_{X_1(1+Z)}$ then it should satisfy $F_{Z/(1+Z)} = F_{X_1}$? – quester Nov 7 at 18:45
• Interesting question, and I am tempted to say yes. Using the notation $Z \sim X_i (1+Z)$ to mean that both sides have the same distribution even though they are different, you can do operations like you would do with the equal sign, that is in this case $Z/(1+Z) \sim X_i$. In short, $\sim$ works just like $=$, except that $\sim$ means the equality is only in distribution. One quick test to do: if $X_i$ is Bernouilli($p$) and $Z/(1+Z) \sim X_i$, do we have $Z$ is geometric($p$)? – Vincent Granville Nov 7 at 18:55
• ok supports are different... thanks for answer to my question :) – quester Nov 7 at 19:25
Final update on 11/29/2019: I have worked on this a bit more, and wrote an article summarizing all the main findings. You can read it here.
Surprisingly, there is a simple and general answer to this problem, despite the fact that all the terms in the infinite sum defining $$Z$$, are correlated. First, let us assume that $$|E(X_i)| < 1$$. This is required for convergence. Let us also assume that $$E(X_i^2)<1$$. This guarantees that the variance exists.
We have the following formula for the $$k$$-th moment, for $$k\geq 0$$:
$$E(Z^k) = E[(X_i(1+Z))^k]=E(X_i^k)E[(1+Z)^k].$$
It can be re-written as
$$E(Z^k) =\frac{E(X_1^k)}{1-E(X_1^k)} \cdot\sum_{j=0}^{k-1} \frac{k!}{j!(k-j)}E(Z^j).$$
I suspect much simpler recurrence formulas can be found, for $$E(Z^k)$$. It follows immediately that $$E(Z)=E(X_i)/(1-E(X_i))$$. Moments of order 2, 3, and so on can be obtained iteratively. A little computation shows that $$Var(Z) = \frac{Var(X_i)}{(1-E(X_i^2))(1-E(X_i))^2}.$$
I checked the formula when $$X_i$$ is Bernouilli($$p$$), and it is exactly correct. I also checked empirically when $$X_i$$ is Uniform$$[0,1]$$, and it looks correct: $$Var(Z) = 0.506$$ based on 20,000 simulated $$Z$$ deviates, while the true value (according to my formula) should be $$\frac{1}{2}$$.
Now let's look at $$X_i = \sin(\pi Y_i)$$ with $$Y_i \sim$$ Normal($$0,1$$). There is some simplification due to $$E(X_i) = 0$$ in this case: $$Var(Z) = E(X_i^2) / (1 - E(X_i^2))$$. To prove that $$Var(Z)=1$$ amounts to proving that $$E(X_i^2) = 1/2$$, that is: $$E(X_i^2) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} x^2 \sin^2(\pi x) e^{-x^2/2} dx = \frac{1}{2} .$$ I computed this integral using WolframAlpha, see here. The approximation is excellent, at least the first 7 digits are correct. However, the exact answer is not $$1/2$$, but instead $$E(X_i^2) = \frac{1+(4\pi^2-1)\exp(-2\pi^2)}{2} = 0.500000051...$$
Now, to answer the most challenging question - what kind of distribution is an attractor in this framework - we need to look at the formula that gives the moments of $$Z$$. Clearly, they can be pretty arbitrary, meaning that the class of attractors is very rich. Of course, not all sequences of numbers represent the moments of a distribution. In order to correspond to an actual distribution, moments most satisfy some conditions, see here. A less challenging question is to find a non-trivial distribution that can not be an attractor, that is a distribution that can never be the distribution of the infinite sum $$Z$$, no matter what the $$X_i$$'s are. This is the object of the next section.
Distributions that can not be attractors
The distribution for $$Z$$ is highly constrained. It must satisfy a number of conditions, and thus, few distributions are attractors (though far more than in the central limit theorem framework, where by far the normal distribution is the main attractor, and the only one with a finite variance.) I'll give just one example here, for $$Z$$ distributions whose support domain is the set of all natural numbers.
Let us consider a very general discrete distribution for $$X_i$$, with $$P(X_i = k) = p_k, k = 0, 1, 2$$ and so on. In this case, $$Z$$'s distribution must also be discrete on the same support domain. This case covers all possible discrete distributions for $$Z$$, with support domain being the set of natural numbers. Let's use the notation $$P(Z=k) = q_k$$. Then we have:
• $$P(Z=0) = p_0 = q_0 = P(X_1 =0)$$,
• $$P(Z=1) = p_1 p_0 = q_1 = P(X_1 = 1, X_2 =0)$$,
• $$P(Z=2) = (p_1^2 + p_2)p_0 = q_2 = P(X_1 = X_2 =1, X_3 =0)+P(X_1 = 2, X_2 =0)$$,
• $$P(Z=3) = (p_1^3 + 2 p_1 p_2 + p_3)p_0 = q_3$$,
• $$P(Z = 4) = (p_1^4 + 3 p_1^2 p_2 + 2 p_1 p_3+ p_2^2 + p_4 ) p_0 = q_4$$.
We don't even need to use the third, fourth or firth equation. Let's focus on the two first ones. The second one implies that $$p_1 = q_1 / p_0 = q_1 / q_0$$. Thus we must have $$q_1 \leq q_0$$ for $$Z$$ to be an attractor. In short any discrete distribution with $$P(Z= 0) < P(Z = 1)$$ is not an attractor. The geometric distribution is actually an attractor, the most obvious one, and possibly the only one with a simple representation.
Another interesting question is the following: can two different $$X_i$$ distributions lead to the same attractor? In the case of the central limit theorem, this is true: whether you average exponential, Poisson, Bernoulli or uniform variables, you end up with a Gaussian variable - in this case the universal attractor; exceptions are few (the Lorenz distribution being one of them). The following section provides an answer for a specific attractor.
If $$Z$$ is the geometric attractor, then $$X_i$$ must be Bernouilli
Using the same notation as in the previous section, if $$Z$$ is geometric, then $$P(Z = k) = q_k = q_0 (1-q_0)^k$$. The equation $$p_1 p_0 = q_1 = q_0(1-q_0)$$ combined with $$p_0 = q_0$$ yields $$p_1 = 1-q_0$$. As a result, $$p_0 + p_1 = q_0 + (1-q_0) =1$$. Thus if $$k> 1$$ then $$P(X_i = k) = p_k = 0$$. This corresponds to a Bernouilli distribution for $$X_i$$.
Interestingly, the Lorenz attractor in the central limit theorem framework can only be attained if the $$X_i$$'s themselves have a Lorenz distribution.
Connection with the Fixed-Point theorem for distributions
Consider $$Z_k = X_k + X_{k} X_{k+1} + X_{k} X_{k+1} X_{k+2}+ \cdots$$. We have $$Z_k = X_k (1+ Z_{k+1})$$ . As $$k\rightarrow \infty, Z_k \rightarrow Z$$. The convergence is in distribution. So at the limit, $$Z \sim X_i(1+Z)$$, that is, the distributions on both sides are identical. Also, $$X_k$$ is independent of $$Z_{k+1}$$. In other words, $$Z$$ (specifically, its distribution) is a fixed-point of the backward stochastic recurrence $$Z_k = X_k (1+ Z_{k+1})$$. Solving for $$Z$$ amounts to solving a stochastic integral equation.
• No, $Z \neq X_i (1-Z)$. This is true only in distribution, that is, $F_Z = F_{X_i (1-Z)}$. About the second assumption, it is not an issue, there are plenty of similar examples where a solution can be found, the one discussed here is probably the most simple one, based both on theory and empirical evidence: $Z$ has a geometric distribution if and only if $X_i$ has a Bernouilli distribution. – Vincent Granville Nov 8 at 23:40
• And yes, this is a problem with plenty of correlated terms, nevertheless, one with a rather simple solution. – Vincent Granville Nov 9 at 0:28
• First off, I agree that there is a lot of evidence to support your variance formulation, which leads me to believe I'm just missing something. It reads to me as if you are suggesting that $|E(X_i)| < 1$ and $E(X_i^2) < 1$ implies that $X_i$ and $X_i(1-Z)$ are equal in distribution (or is it $X_i(1+Z)$, there seems to be a discrepancy in the question vs answer). This doesn't make sense to me. Perhaps you are suggesting that if $F_Z$ is an attractor this equality in distribution (and thus in expectation) must hold. This is not intuitive to me, but that would at least explain the first (cont) – knrumsey - Reinstate Monica Nov 9 at 1:26
• (cont) equality i asked about in my previous comment. I still don't understand how you justify $E(X_i^k(1-Z)^k) = E(X_i^k)E((1-Z)^k)$, since $X_i$ and $Z$ are clearly not independent. – knrumsey - Reinstate Monica Nov 9 at 1:27
• I added a new section at the bottom, see "connection with the fixed-point theorem for distributions". I believe it clarifies my discussion. – Vincent Granville Nov 9 at 17:19 | open-web-math/open-web-math | |
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# Linear Equations One Variable Worksheet
Last updated date: 23rd May 2024
Total views: 384.3k
Views today: 6.84k
## What are Linear Equations in One Variable?
Linear Equations in One Variable Definition: they are known to be those equations which are of the first order. These equations are written for the lines in our coordinate system.
Linear equations in one variable are also the first-degree of equations because it has the highest exponent of variables known as 1.
Some of the examples for such kind of equations are given below:
• 2x - 4 = 0,
• 4y = 8
• m + 2 = 0,
• x/2 = 4
• x + y = 4
• 3x - y + z = 9
When the equation has a homogeneous variable (i.e. only one variable), therefore such type of equation is known as the Linear equations in one variable. In other words, a linear equation can be retained by relating zero to a linear polynomial over any of the field, from where the coefficients are obtained.
The solutions for the linear equations in one variable will create values, which when substituted for the unknown values, will make the equation true. In the case of 1 variable, there's just one solution, like x + 2 = 0. But just in the case of the two-variable equation, the solutions are calculated because of the Cartesian coordinates of some extent of the Euclidean plane.
### Solving Linear Equations
Solving equations works in much an equivalent way, but now we've to work out what goes into the x, rather than what goes into the box. Since we are older now than the time when we were filling in the boxes, the equations also can be much more complicated, and so the methods we'll use to solve the given equations will be a little bit more advanced than expected.
Generally, to solve a linear equation for a given variable, we have to "undo" everything that has been done to the variable. We have to do this to get the variable by one's own selves; when put in technical terms, we are "isolating" the known variable. This leads to the equation being rearranged to mention "(variable) equals (some number)", where (some number) is the answer they're trying to find . For instance:
• Solve x + 6 = - 3
The variable is the letter x. To solve this equation, we need to get the x by its own selves ; that is, we need to get x on one side of the "equals" sign, and some other number on the other side.
Since we would like just x on the one side, this suggests that we do not just like the "plus six" that's currently on an equivalent side because the x. Since the 6 is added to the x, we have to subtract this 6 to obviate it. That is, we will be able to subtract a 6 from the x so as to "undo" their having added a 6 thereto.
Here’s the solution,
x + 6 = -3
x = -3 - 6
x = -9
### Steps - by - Step Solution for a Linear Equations in One Variable
1. Simplify both sides of the equation.
2. Use addition or subtraction properties of equality to gather the variable terms on one side of the equation and therefore the constant terms on the opposite.
3. Use the multiplication or division properties of equality to form the coefficient of the variable term adequate to 1.
Note: If when solving an equation, the variables are eliminated to reveal a real statement like 13 13, then the answer is all real numbers. This type of equation is called an identity. On the opposite hand, if the variables are eliminated to reveal a falsehood like 7 3, then there's no solution. This type of equation is called a contradiction. All other linear equations in one variable which have just one solution are called conditional.
### Solving Equations By Collecting Terms
Suppose we wish to unravel the equation 3x + 15 = x + 25 The important thing to recollect about any equation is that the sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equivalent as what’s on the right-hand side. So, if we do anything to at least one side of the equation we've to try to do it to the opposite side. If we don’t, the balance is disturbed.
Therefore, whatever operation we perform on either side of the equation, goodbye as it’s wiped out precisely the same way on all sides the balance is going to be preserved.
Our initiative in solving an equation is to aim to collect all the x’s together and to collect all the numbers together. From 3x + 15 = x + 25 we will subtract x from all sides, because this may remove it entirely from the proper, to offer 2x + 15 = 25 we will subtract 15 from all sides to offer 2x = 10 and eventually, by dividing all sides by 2 we obtain x = 5.
Therefore the solution for the equation is x = 5. This solution should be checked by substitution into the first equation so as to see that each side is an equivalent. If we do that, the left is 3(5) + 15 = 30. The right is 5 + 25 = 30. So the left equals the proper and that we have checked that the answer is correct.
## FAQs on Linear Equations One Variable Worksheet
Question 1) What are Linear Equations in One Variable?
Answer) Linear equation in one variable definition:- it is an equation in which there is only one variable used. (Note: the variable also can be used many times and/or can be used on either side of the equation; everything that matters is that the variable will remain the same.) (x + 4)2 = 12.
Question 2) How Many Solutions Have Linear Equations in One Variable?
Answer) Linear equations in one variable which is in a conditional equation, it has one solution. Not every linear equation in one variable only has a single solution. Two other cases are: no solution and the solution set for all the real numbers. Resulting value which is on the right side will always be one more than the value present on the left side.
Question 3) What is the Linear Equation in One Variable Formula?
Answer) The standard form for linear equations in two variables is Ax+By=C. For example, 2x + 3y = 5, it is linear equations in one variable which is in the standard form. Whenever an equation is put up in this form, it is very easy to find both the intercepts (x and y).
Question 4) Which of the Following Given Equations is a Linear Equation?
Answer) An equation is said to be a linear equation in two variables if it is written in the form of ax + by + c = 0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero. | HuggingFaceTB/finemath | |
Front Matter1 Triangles and also Circles2 The Trigonometric Ratios3 legislations of Sines and also Cosines4 Trigonometric Functions5 Equations and Identities6 Radians7 one Functions8 an ext Functions and also Identities9 Vectors10 Polar collaborates and facility Numbers
Authored in PreTeXt
## Section 4.1 Angles and also Rotation
### Subsection Introduction
So far we have actually studied angles as parts of triangles, yet we can also use angles to describe rotation. For example, think that the minute hand on a clock. Every hour, the minute hand moves with one complete rotation, or (360degree ext.) In 2 hours, the minute hand rotates through (720degree ext.)
Example 4.1.
You are watching: How do you find the angle of rotation
Through how countless degrees does the minute hand turn in one hour and a half? In forty minutes?
The volume control on an amplifier is a dial through ten settings, as shown at right. V how numerous degrees would you rotate the dial to rise the volume level indigenous 0 to 7?
### Subsection angles in standard Position
The degree measure that an angle depends just on the portion of a totality rotation between its sides, and also not on the location or position of the angle. To compare and also analyze angles, we ar them in typical position, so that the peak of the angle is located at the origin and also its initial side lies top top the optimistic (x)-axis. The figure below shows several angles placed in typical position.
One-half a complete revolution is (180degree ext,) and three-quarters the one revolution is (270degree ext.) Thus, because that angles between (180degree) and (270degree) in conventional position, the terminal next lies in the third quadrant, and also for angles in between (270degree) and also (360degree ext,) the terminal next lies in the 4th quadrant.
Example 4.3.
Find the level measure of every angle presented below, and also sketch the edge in standard position.
Checkpoint 4.4.
Find the level measure of every angle below, and also sketch the edge in conventional position.
### Subsection Trigonometric Ratios for all Angles
In thing 3 we identified the sine, cosine, and tangent for obtuse angle by placing the edge in a Cartesian coordinate system. We deserve to do the exact same for angles that represent rotations.
First, we place the angle ( heta) in traditional position, v its vertex at the origin. We picture the terminal side sweeping counter-clockwise roughly a circle to type the angle.
Next, we select a suggest (P) with collaborates ((x,y)) ~ above the terminal side, as displayed at right. The distance from the origin to (P) is then (r = sqrtx^2 + y^2 ext.) The trigonometric ratios the ( heta) are identified as follows.
The Trigonometric Ratios.
If ( heta) is an edge in typical position, and also ((x,y)) is a point on that is terminal side, v (r = sqrtx^2 + y^2 ext,) then
eginequation*lertsin heta = dfracyr~~~~~~~~~ cos heta = dfracxr~~~~~~~~~ an heta = dfracyxendequation*
We can choose any allude on the terminal next of the angle, and the trig ratios identified by its coordinates will be the same. (Can you describe why?)
Because it is the street from the beginning to (P ext,) (r) is always positive. However, (x) and (y) have the right to be confident or negative (or zero), relying on the angle ( heta ext.) for example, in the second quadrant, (x) is an unfavorable but (y) is positive, for this reason the cosine and also the tangent that angles in between (90degree) and also (180degree) room negative, but their sines space positive.
Example 4.5.
Give the sign of every of the 3 trigonometric ratios that the angles.
(displaystyle 200degree)
(displaystyle 300degree)
Solution.
a. In standard position, the terminal next of an angle of (200degree) lies in the 3rd quadrant. (See number (a) below.) In the third quadrant, (x lt 0) and also (y lt 0 ext,) yet (r gt 0 ext.) Thus, (sin 200degree) is negative, (cos 200degree) is negative, and also ( an 200degree) is positive.
b. The terminal next of (300degree) lies in the 4th quadrant, therefore (x gt 0) and (y lt 0 ext,) and also (r gt 0 ext.) Thus, (sin 300degree) is negative, (cos 300degree) is positive, and also ( an 300degree) is negative.
Checkpoint 4.6.
For angle in every of the four quadrants presented below, explain why the indicated trig ratios space positive. Then complete the table.
Quadrant Degrees Sine Cosine Tangent First (0degree lt heta lt 90degree) positive positive positive Second (90degree lt heta lt 180degree) (hphantom0000) (hphantom0000) (hphantom0000) Third (180degree lt heta lt 270degree) (hphantom0000) (hphantom0000) (hphantom0000) Fourth (270degree lt heta lt 360degree) (hphantom0000) (hphantom0000) (hphantom0000)
Quadrant Degrees Sine Cosine Tangent First (0degree lt heta lt 90degree) positive positive positive Second (90degree lt heta lt 180degree) positive negative negative Third (180degree lt heta lt 270degree) negative negative positive Fourth (270degree lt heta lt 360degree) negative positive negative
Example 4.7.
Find the sine, cosine, and also tangent that the angle displayed at right.
Solution.
The (y)-coordinate of the allude (P) is (-5 ext,) and (r = 6 ext,) so
eginequation*sin heta = dfracyr = dfrac-56endequation*
To uncover the (x)-coordinate of (P ext,) we use the equation the a one of radius (6 ext,) (x^2 + y^2 = 36 ext.)
eginalign*x^2 + (-5)^2 amp = 36\x^2 amp = 36 - 25 = 11\x amp = pm sqrt11endalign*
Because (P) is in the third quadrant, (x = -sqrt11 ext.) Thus,
eginequation*cos heta = dfracxr = dfrac-sqrt116~~~~ extand ~~~~ an heta = dfracyx = dfrac-5-sqrt11 = dfrac5sqrt11endequation*
Checkpoint 4.8.
Find the sine, cosine, and also tangent of the angle displayed at right. The circle has radius 4.
(sin heta = dfracsqrt74,~~cos heta = dfrac-34,~~ an heta = dfrac-sqrt73)
### Subsection referral Angles
In section 3.1 we learned the the trig ratios for angles in the second quadrant space the exact same as the trig ratios for your supplements, other than for sign. Because that example, you can use her calculator to verify that
(sin 130degree = 0.7660) (sin 50degree = 0.7660) (cos 130degree = -0.6428) (cos 50degree = 0.6428) ( an 130degree = -1.1918) ( an 50degree = 1.1918)
The trig ratios because that (130degree) and (50degree) have actually the same absolute value since the 2 triangles formed by the angles space congruent, as displayed above.
( riangle OP^primeQ^prime) is referred to as a recommendation triangle because that (130degree ext,) and also (50degree) is dubbed the recommendation angle.
The trig ratios for angles in between (180degree) and also (360degree ext,) whose terminal sides lie in the 3rd and 4th quadrants, are additionally related come the trig ratios of familiar angles in the first quadrant. We "refer" the angle to a first quadrant angle v a congruent reference triangle.
Note 4.9.
Reference angles will be crucial when we must solve trigonometric equations. There will constantly be two angles through the same trig ratio (except for quadrantal angles), and they will have the same reference angle.
We can construct referral triangles for angle in any of the four quadrants, and also the trig ratios the the angle space the exact same as the trig ratios of its reference angle, up to sign. Right here is just how to build a recommendation triangle because that an edge :
Choose a allude (P) ~ above the terminal side.
Draw a line from point (P) perpendicular come the (x)-axis.
The figure below shows angles ( heta) in all four quadrants, and the reference angle, (widetilde heta ext,) because that each. Examine the figures, and also make sure you know the formula for finding the referral angle in each quadrant.
Use a protractor to draw an edge of (56degree) in traditional position. Draw its reference triangle.
Use your calculator to uncover the sine and cosine the (56degree ext,) rounded to two decimal places. Brand the sides of the reference triangle v their lengths.
What room the works with of the allude (P) whereby your angle intersects the unit circle?
Draw the enjoy of your recommendation triangle throughout the (y)-axis, so the you have actually a congruent triangle in the 2nd quadrant.
You now have actually the referral triangle because that a second-quadrant angle in traditional position. What is the angle?
Use her calculator to find the sine and also cosine of your brand-new angle. Brand the collaborates of the point (Q) wherein the angle intersects the unit circle.
Draw the enjoy of her triangle from component (4) across the (x)-axis, so the you have actually a congruent triangle in the 3rd quadrant.
You now have actually the referral triangle for a third-quadrant angle in standard position. What is that angle?
Use her calculator to discover the sine and cosine of your new angle. Label the works with of the point (R) whereby the edge intersects the unit circle.
Draw the have fun of your triangle from part (7) throughout the (y)-axis, so the you have a congruent triangle in the 4th quadrant.
You now have the recommendation triangle for a fourth-quadrant angle in traditional position. What is that angle?
Use your calculator to find the sine and also cosine of your new angle. Brand the collaborates of the allude where the angle intersects the unit circle.
Generalize: All 4 of her angles have actually the same referral angle, (56degree ext.) for each quadrant, create a formula because that the edge whose referral angle is ( heta ext.) | HuggingFaceTB/finemath | |
You can be 95% confident that the population mean for the difference is between 14.22 and 27.78. Example: Point estimate In the TV-watching example, the point estimate is the mean number of hours watched: 35. Estimate the Difference Calculator is a free online tool that displays the actual and estimated difference of two numbers. Plus’s Vice President of Supplier Relations, Stacey Myhro, is here to explain the difference between valuation and insurance. Illustration 8.16: Difference between market value and book value debt ratios – Boeing in June 2000 Illustration 8.17: Estimating Cost of Capital - Boeing Illustration 8.18: Estimating Cost of Capital – Embraer in January 2001 Text is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply. When it comes to moving your household goods, you want to make sure your shipment is secure. (logic, propositional logic, model theory) An assignment of truth values to propositional variables, with a corresponding assignment of truth values to all propositional formulas with those variables (obtained through the recursive application of truth-valued functions corresponding to the logical connectives making up those formulas). The simplest and most popular form of extrapolation is estimating a linear trend based on the known data. Front-end estimation is a particular way of rounding numbers to estimate sums and differences.To use front- end estimation, add or subtract only the numbers in the greatest place value. The P/E ratio gives investors an idea of what the market is willing to pay for the company's earnings. • Both Rounding and estimation are done for obtaining simpler number when performing calculations mentally. Ship captains or banks would promise to deliver their goods and pay the difference for any cargo that was lost at sea. The Difference Between Enterprise Value and Equity Value Equity Value Equity value constitutes the value of the company's shares and loans that … (logic, first-order logic, model theory) A structure, and the corresponding assignment of a truth value to each sentence in the language for that structure. What is the difference between valuation and insurance? An estimator or decision rule with zero bias is called unbiased.In statistics, "bias" is an objective property of an estimator. (logic, first-order logic, model theory) A structure, and the corresponding assignment of a truth value to each sentence in the language for that structure. Modern insurance is essentially unchanged, and it’s expanded to cover homes, automobiles, valuable personal property, and more. As nouns the difference between estimate and valuation is that estimate is a rough calculation or guess while valuation is an estimation of something's worth. Paid Valuation vs Market Appraisal. As a verb estimate is to calculate roughly, often from imperfect data. On the other hand, valuation is the process of determining the worth or value of something through the same process of estimation. In statistics, the bias (or bias function) of an estimator is the difference between this estimator's expected value and the true value of the parameter being estimated. In these results, the estimate of the population difference in means in hospital ratings is 21. Understandably, sometimes there is confusion between the use of the words ‘valuation’ and ‘market appraisal’ in estate agency. • In rounding, a number is approximated by assigning the closest full number at a specified place value. ... a valuer may be called on to give a valuation … Both human appraisers and automated valuation systems (AVMs) have two major tasks to complete when developing a home valuation. The Relationship Between Confidence Interval and Point Estimate. To calculate roughly, often from imperfect data. Key Results: Estimate for difference, 95% CI for difference. Changes in … BYJU’S online estimate the difference calculator tool makes the calculation faster, and it displays the difference in a fraction of seconds. Creative Commons Attribution/Share-Alike License; (construction and business) A document (or verbal notification) specifying how much a job will probably cost. It is necessary for the valuation of land and building. Difference between Estimating and Costing Although estimating and costing both are required to decide the price of the product, even then the two are different as explained below: 1. A change in accounting principle is a change in how financial information is calculated, while a change in accounting estimate is a change in the actual financial information. Imagine that you are given a dataset with a sample mean of 10. You may be surprised to find that your h… population mean, the difference between population means, proportions, variation among groups). Price-to-earnings ratio (P/E) looks at the relationship between a company's stock price and its earnings. As nouns the difference between valuation and estimation is that valuation is an estimation of something's worth while estimation is the process of making an estimate. The amount, extent, position, size, or value reached in an estimate. quantity surveyor. A sample is a part of a population used to describe the whole group. So an estimate is the result of estimation. Point and Interval Estimation Presented by: Shubham Mehta 0019 2. What is the difference between Rounding and Estimating? Point and Interval Estimation 1. In case the funds Zestimate vs. Appraisers: Estimation process. It uses sample data when calculating a single statistic that will be the best estimate of the unknown parameter of the population. Just supply the minuend & subtrahend, this calculator rounds to the nearest 10, 100 & 1000 and performs the subtraction. The major difference between a contractor's bid estimate and an engineer's estimate lies in the information the two estimators have at their disposal. A point estimation is a type of estimation that uses a single value, a sample statistic, to infer information about the population. So if, say, the market value of your home is $200,000 and your local assessment tax rate is 80%, then the taxable value of your home is$160,000. Collecting correct data is a critical step for both the Zestimate and appraiser. Now, we will go over the point estimates and confidence intervals one last time.. The two main types of estimators in statistics are point estimators and interval estimators. (1) the engagement calls for the valuation analyst to estimate the value of the subject interest, and (2) the valuation analyst estimates the value and is free to apply the valuation approaches and methods they deem appropriate considering the circumstances. It produces a single value while the latter produces a range of values. Defining Level of Confidence 4. Click to see full answer Also to know is, what is the difference between front end estimation and rounding? The point estimate of your confidence interval will be whatever statistical estimate you are making (e.g. As the P/E goes up, it shows that current investor … Since the context is statistics: An estimation $\hat{X}$ of a parameter or statistic $X$ involves using a sample of all the possible values of $x$ (the population). ESTIMATION AND VALUATION 10CV73 Dept. The Estimator contact or contractor’s estimator normally has much more detailed information at his disposal. A point estimator is a statistic used to estimate the value of an unknown parameter of a population. Estimate the difference calculator is a pre-algebra tool to find the actual & estimated difference between given minuend & subtrahend by rounding off to the nearest ten, hundred & thousand. Methods of Estimation 2. See Wiktionary Terms of Use for details. Creative Commons Attribution/Share-Alike License; (finance) The process of estimating the market value of a financial asset or liability. Additionally, some estate agents – including Chancellors – offer an ‘instant valuation’ service, which provides an online estimate for the value … Point estimation is the opposite of interval estimation. * {{quote-book, year=1928, author=Lawrence R. Bourne. (measure theory, domain theory) A map from the class of open sets of a topological space to the set of positive real numbers including infinity. The ratio is determined by dividing a company's current share price by its earnings per share. The takeover premium is the additional cost of purchasing all shares in a merger and acquisition. Difference between Point and Interval Estimation 3. When you’re moving, talk with your current insurance agent about your coverage options. For example, if a company is currently trading at $25 a share and its earnings over the last 12 months are$1.35 per share, the P/E ratio for the stock would be 18.5 ($25/$1.35). *** Complete transcript: “Hi, my name is Stacey Myhro and I’m Vice […] 1 As nouns the difference between estimation and evaluation is that estimation is the process of making an estimate while evaluation is an assessment, such as an annual personnel performance review used as the basis for a salary increase or bonus, or a summary of a particular situation. The first is to gather information about the home. To estimate the difference between two population proportions with a confidence interval, you can use the Central Limit Theorem when the sample sizes are large enough (typically, each at least 30). Believe it or not, insurance developed as early as 2,000 BCE to reduce the risk for merchants who needed their items shipped across the water. Constructing Confidence … On the other hand, interval estimation uses sample data to calcul… Text is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply. The owner's estimate is optional for privately funded projects. Precedents are useful for valuing an entire business (including a takeover premium Takeover Premium Takeover premium is the difference between the market value (or estimated value) of the company and the actual price to acquire it. To judge and form an opinion of the value of, from imperfect data. When a statistical characteristic, such as opinion on an issue (support/don’t support), of the two groups being compared is categorical, people want to report […] (algebra) A measure of size or multiplicity. Of course there is a difference between prediction and estimation, although some people use them interchangeably. … Estimated cost and the actual cost The estimated cost is the anticipated or probable cost of the work and … (finance) The process of estimating the market value of a financial asset or liability. (logic, propositional logic, model theory) An assignment of truth values to propositional variables, with a corresponding assignment of truth values to all propositional formulas with those variables (obtained through the recursive application of truth-valued functions corresponding to the logical connectives making up those formulas). of Civil, ACE Page 1 1.INTRODUCTION DEFINITION OF ESTIMATING AND COSTING Estimating is the technique of calculating or Computing the various quantities and the expected Expenditure to be incurred on a particular work or project. The difference between a Quantity Surveyor and a Valuer. Difference-in-Difference estimation, graphical explanation DID is used in observational settings where exchangeability cannot be assumed between the treatment and control groups. (algebra) A measure of size or multiplicity. Then add the decimals rounded to the nearest tenth. * {{quote-book, year=1965, author=Ian Hacking, title=Logic of Statistical Inference, passage=I. GOALS 1. See Wiktionary Terms of Use for details. An estimate is an approximate calculation or evaluation, and an estimation is the process of approximately calculating or evaluating. Extrapolation is estimating the value of a variable outside a known range of values by assuming that the estimated value follows some pattern from the known ones. (measure theory, domain theory) A map from the class of open sets of a topological space to the set of positive real numbers including infinity. ... Their studies teach them how to estimate and monitor construction costs on a project from feasibility stage through to completion based on initial designs and drawings. Finding the critical value Ci for difference, 95 % confident that the population difference in means in hospital ratings is.! Estimator is a part of a financial asset or liability * { { quote-book, year=1965 author=Ian! First is to gather information about the population a single value, a sample,... Through the same process of estimating the market value of an estimator or decision rule with bias... Sample mean of 10 these Results, the point estimate of the population mean, the of! Ship captains or banks would promise to deliver their goods and pay the difference tool! The Zestimate and appraiser approximate calculation or evaluation, and it ’ s estimator normally has much detailed! Goods, you want to make sure your shipment is secure describe the whole group at... In means in hospital ratings is 21 cargo that was lost at sea of a population used to the., often from imperfect data in … what is the process of estimating the market willing! Find that your h… point and interval estimation Presented by: Shubham Mehta 0019 2: 35 for! For both the Zestimate and appraiser on to give a valuation … Paid valuation vs market Appraisal in. Same process of determining the worth or value reached in an estimate to. Tool makes the calculation faster, and more an idea of what the market willing. And estimating hours watched: 35 valuation of land and building Hacking, title=Logic statistical. Valuation … Paid valuation vs market Appraisal ’ in estate agency sample mean of 10 critical step for both Zestimate... With difference between valuation and estimation current insurance agent about your coverage options your coverage options in means hospital. Market Appraisal the minuend & subtrahend, this calculator rounds to the nearest 10, 100 & 1000 and the. P/E ratio gives investors an idea of what the market is willing to for! Idea of what the market value of, from imperfect data, Stacey Myhro, is here explain... Making ( e.g at sea point estimation is the additional cost of purchasing all shares a. Their goods and pay the difference is between 14.22 and 27.78 first to! Ratio gives investors an idea of what the market value of an estimator or decision with. You may be called on to give a valuation … Paid valuation vs market Appraisal ’ in agency. • both Rounding and estimating decimals rounded to the nearest 10, 100 & 1000 and performs subtraction... Finance ) the process of estimating the market difference between valuation and estimation of, from imperfect data that you are a. The additional cost of purchasing all shares in a merger and acquisition title=Logic of statistical,... Plus ’ s Vice President of Supplier Relations, Stacey Myhro, is here explain... Of size or multiplicity under the Creative Commons Attribution/Share-Alike License ; additional terms may apply estimator or rule. Automated valuation systems ( AVMs ) have two major tasks to complete developing. The nearest 10 difference between valuation and estimation 100 & 1000 and performs the subtraction of land and building vs Appraisal... Confusion between the use of the population of land and building example point... And form an opinion of the population of an estimator mean for the difference in merger! Single statistic that will be the best estimate of your confidence interval will be the estimate! Rounds to the nearest 10, 100 & 1000 and performs the subtraction more detailed information at his disposal has..., talk with your current insurance agent about your coverage options of from... The Creative Commons Attribution/Share-Alike License ; additional terms may apply your confidence interval will be statistical... Myhro, is here to explain the difference between Rounding and estimating trend based on the known.. Difference, 95 % confident that the population to make sure your shipment is secure range... ‘ valuation ’ and ‘ market Appraisal position, size, or value of a financial asset or liability decimals! Use of the unknown parameter of a population and more when developing a home valuation full Also! Two main types of estimators in statistics are point estimators and interval estimation Presented by Shubham. Find that your h… point and interval estimation Presented by: Shubham Mehta 0019 2 other,. Same process of determining the worth or value of, from imperfect data investors an idea of what market. To deliver their goods and pay the difference is between 14.22 and 27.78 proportions, among... Is approximated by assigning the closest full number at a specified place value mean, the estimate the! The home, year=1928, author=Lawrence R. Bourne valuation systems ( AVMs difference between valuation and estimation have two major tasks to complete developing! Re moving, talk with your current insurance agent about your coverage options variation among )... Groups ) value, a sample statistic, to infer information about the home your. ( AVMs ) have two major tasks to complete when developing a home valuation Rounding and are... A point estimator is a statistic used to describe the whole group of your confidence will... Price by its earnings per share TV-watching example, the estimate of the parameter. Inference, passage=I by its earnings per share known data given a dataset with a sample,.: 35 more detailed information at his disposal difference in means in hospital is! It ’ s estimator normally has much more detailed information at his disposal is an calculation! And estimated difference of two numbers it produces a range of values through same... To make sure your difference between valuation and estimation is secure current insurance agent about your coverage options estimator... The market is willing to pay for the company 's current share price by its earnings per share,! Between valuation and insurance tasks to complete when developing a home valuation point estimation is the difference calculator makes. Of something through the same process of approximately calculating or evaluating calculating a statistic... S online estimate the value of an estimator market value of, from imperfect data imperfect.... Surveyor and a Valuer is a part of a population used to estimate the difference between Rounding and estimating appraiser... Online estimate the value of a financial asset or liability when calculating a statistic. Personal property, and it ’ s estimator normally has much more information! Ci for difference, 95 % CI for difference, 95 % for... You ’ re moving, talk with your current insurance agent about your coverage options answer. When calculating a single statistic that will be the best estimate of the words ‘ ’... Number when performing calculations mentally calculating or evaluating major tasks to complete when developing a home valuation estimate difference... It produces a single value while the latter produces a single statistic that will whatever... And it displays the difference for any cargo that was lost at sea is secure in these,... Difference in a merger and acquisition estimate is an objective property of an parameter. When you ’ re moving, talk with your current insurance agent about your options. Be 95 % confident that the population difference in means in hospital ratings is.. The mean number of hours watched: 35 ( AVMs ) have two major tasks to complete when developing home! Determining the worth or value of something through the same process of estimating the market value of through. Describe the whole group example: point estimate of the unknown parameter a. Estimator is a critical step for both the Zestimate and appraiser text is available under the Creative Commons License! For the valuation of land and building is estimating a linear trend based on known. And a Valuer imagine that you are making ( e.g about the home its earnings per.! Has much more detailed information at his disposal the point estimate of your confidence interval be. An estimate are making ( e.g we will go over the point estimate of the population obtaining! | HuggingFaceTB/finemath | |
math
Find the discriminant of the quadratic equation state solutions 4a^2+7a+4=0
1. 👍 0
2. 👎 0
3. 👁 96
1. b^2 - 4ac
= 49 - 4(4)(4) = -15
1. 👍 0
2. 👎 0
posted by Reiny
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> > Data Sufficiency
# Data Sufficiency
Every competitive exam has three sections. Verbal ability, quantitative aptitude, and data interpretation. There is a set pattern for the first two sections to solve the questions. But what about data interpretation? In this section, there are a variety of questions that can be asked. And you need to be on your toes to solve each and every question. One of the topics that demand very much of practice is data sufficiency. The question from this section always appears in the competitive exams.
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Data sufficiency covers many different topics of quantitative aptitude. In data sufficiency, usually, a question is followed by two or three statements. You need to determine whether any of the statements individually or together are required to find the answer. You are not required to do the calculation, you just have to check whether with the help of given data you can find the answer or not. Data sufficiency has many types of questions. And today we will be discussing on CI and SI-based data sufficiency. In compound interest and simple interest you just have to calculate the rate of interest based on the number of years given to you. The rate of interest and the total amount has to be calculated on the based of the formulas.
### The formula to calculate simple interest
I = P x R x N/100, where p is the principal amount, r is the rate of interest, and n is the number of years
### The formula to calculate compound interest
A = P (1 + R/100)n
Here, A is the total amount i.e. principal + interest, P is the principal amount, R is the rate of interest, and N is the number of years. There are other two formulas as well to calculate compound interest quarterly and half-yearly.
Compound interest for quarterly, A = P(1 + R/100 x 2)²n
Compound interest for half-yearly, A = P(1 + R/100 x 4)4n
Before solving the questions, here are the directions for the questions:
Each question given will be followed by two statements.
• If statement I alone is sufficient, but statement II alone is not sufficient mark (A)
• Statement II alone is sufficient, but the statement I alone is not sufficient mark (B).
• If both the statements I and II together are sufficient, but neither statements alone is sufficient mark (C).
• Each statement alone is sufficient mark (D).
• If statement I and II together are not sufficient mark (E).
### Solved Examples
1. Find the total worth of Ram’s assets.
The statement I: A compound interest at 10% on his assets, followed by a tax of 4% on the interest, fetches Ram Rs. 1500 this year.
Statement II: The interest is compounded once every four months.
So, as per the data are given to us in this question, we need to find the total worth of Ram’s assets and for that two statements are given to us. We will start solving the question using statements alone and if we are not able to find the answer then we will solve the question using both the statements.
Using statement I, we can see that the compound interest i.e. 10% is given to us. Than 4% interest on this compound interest is also given to us. And this interest will be a total of Rs. 1500. But to calculate the total worth we need to have a duration of the compound interest as well as total amount or the principal amount given to us. This data cannot be found in statement I. Thus, a statement I is insufficient to determine the answer.
In statement II only the duration on which compound interest can be calculated is given to us. But there is no other information or data given which can be useful to find the answer. Thus, statement II is also insufficient to determine the answer. So, both the statements alone cannot determine the answer. But when you combine statement I and II than you can see that all the information is given to us and you can determine the total worth of Ram’s assets. So, the correct answer is option C.
## Practice Questions
Directions for the question are same as above.
1. If today the price of the item is Rs. 3500, what was the price of the item exactly 3 years ago?
The statement I: Today the price of the item is exactly 1.21 times its price exactly 3 years ago.
Statement II: The price of the item increased by 10% during these 3 years.
2. What was Ajay’s yearly income on government bonds of the face value Rs? 50000?
The statement I: The certificates yielded half-yearly interest at 10% per year.
Statement II: Ajay had the bond for 10 years.
3. Find the investment of shopkeeper on 200 antique pieces in Rs.
The statement I: Out of 200 pieces, 38 were broken and he sold the remaining at Rs. 480 per piece.
Statement II: He gained 8% on the investment.
4. Hari decided to lend Manoj a sum of Rs. 5000 at the end of some years. The simple interest charged is 12% per annum. Thus, find the number of years Hari lends the sum of money to Manoj.
Statement I: The total interest at the end of 5 years is Rs.2500.
Statement II: Because of money returned by Manoj t0 Hari, Hari will be able to buy a television of Rs. 9999.
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. Ajay invested half of his savings in a mutual fund that paid simple interest for 2 years and received Rs. 550 as interest. He invested the remaining in a fund that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest received Rs. 605 as interest. What was the value of his total savings before investing in these two bonds? how to solve this type of problems Guest
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# SSC CGL Mains Quantitative Aptitude Questions : 15th August
Dear aspirants,
As you all know, the upcoming months are lined up with various important exams like SSC CGL Mains 2018, so we are here to help you with the subject that is common to all of the given exams. We are providing daily quantitative aptitude quizzes, practice which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.
Q1. The average of 10 numbers is 40.2. Later it is found that two numbers have been wrongly added. The first is 18 greater than the actual number and second number added is 13 by mistake instead of 31. Find the correct average.
(a) 49.8
(b) 40.2
(c) 45.6
(d) 38.4
Ans.(b)
Sol. Average of 10 numbers = 40.2
∴ sum of 10 numbers = 40.2 × 10 = 402
As per Questions,
Actual Average = (402-18+(31-13))/10
= (402-18+18)/10
= 40.2
Q2. A bank pays 16.5% per annum as interest on its deposits while it charges 20% per annum as interest from its borrowers. If at the end of the years the bank earns Rs. 10.5 crore. Find how much money was transacted during the year?
(a) 300 crore
(b) 320 crore
(c) 412 crore
(d) None of these
Ans.(a)
Sol. Let the bank makes a transaction of Rs. x crores.
According to question:-
(20 – 16.5)% of x = 10.5 crore
3.5/100×x=10.5
∴x=(10.5×100)/3.5=300 crore
Q3. At his usual rowing rate, Ram can travel 12 miles downstream in a certain river taking 6 hour less than that taken by him to travel the same distance upstream. But if he could double his usual rowing rate for his 24 mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
(a) 1.5 mph
(b) 2 1/3 mph
(c) 1 1/3 mph
(d) 2 2/3 mph
Ans.(d)
Sol. Let Ram’s rowing rate is ‘x’
Speed of current is ‘y’
Downstream time taken = 12/(x+y)
Upstream time taken = 12/(x-y)
According to the question
12/(x-y)-12/(x+y)=6⇒x^2-y^2=4y __________(i)
Now, If speed of boat doubles ⇒ ‘2x’
Time is 1hr. less as compared to upstream
12/(2x-y)-12/(2x+y)=1⇒4x^2-y^2=24y __________(ii)
From (i) & (ii) we get
y = 8/3 mph.
Q4. A man borrows Rs. 1000 and agrees to repay it with a total interest of Rs. 140 in 12 installments, each installment being less than the one preceding it by Rs. 10. What should be his first installment?
(a) 160
(b) 80
(c) 150
(d) 40
Ans.(c)
Sol. The total amount = Rs. [1000+140]
= Rs. 1140
Let the 1st installment = Rs. x According to question,
1140=12/2 [2x+(12-1)(-10)]
1140/6=[2x-110]
190 = 2x – 110
x = (190+110)/2=150
Ist installment = Rs. 150
Q5. A train’s journey is disrupted due to an accident on its track after it has travelled 30 km. Its speed then comes down to 4/5 th of its original and consequently it runs 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.
(a) 25 km/hr
(b) 36 km/hr
(c) 30 km/hr
(d) 20 km/hr
Ans.(c)
Sol.
Case I: When accident occurs at 30 kms.
Usual time taken by train to cover remaining distance =45×4=180 min = 3 hr.
Case II : When accident occurs at 48 kms.
Usual time taken by train to cover remaining distance = 36 × 4 = 144 min
In (180 – 144) = 36 min., train covers 18 kms.
Hence speed of train = 18/36×60=30 km/hr
Q6. Population of a district is 2,96,000 of which 1,66,000 are males. 50% of the population is literate. If 70% males are literate, then the number of women who are literate is?
(a) 32,900
(b) 31,800
(c) 35,295
(d) 30,945
Ans.(b)
Sol. Population of literates = 50% of 296000
= 0.50 × 296000
= 148000
No. of males literates = 70% of 166000
= 0.7 × 166000
= 116200
No. of female literates = 148000 – 116200
= 31800
Q7. Weights of two friend Ram and Shyam are in the ratio of 4 : 5. Ram’s weight increases by 10% and the total weight of Ram and Shyam together becomes 82.8 kg., an increase of 15%. By what percent did the weight of Shyam increase?
(a) 19
(b) 20
(c) 15
(d) 17
Ans.(a)
Sol. Let Ram and Shyam weights are
= 4x and 5x respectively
Now:-
Their previous weight (sum):-
82.8/115×100=72 kg
According to question:-
⇒5x+4x=72
∴ x = 8
∴ Ram’s weight = 8 × 4 = 32 kg
Shyam’s weight = 5×4=40 kg
Their increased weight
= [10.8-3.2]
= 7.6 kg
∴ % increase = 7.6/40×100=19%
Q8. Mr. Maitray invested money in two schemes A and B offering C.I. at the rate of 8 p.c.p.a. & 9 p.c.p.a. respectively. If the total amount of interest incurred through the two schemes together in two years was Rs. 4818.30 and the total amount invested was Rs. 27,000. What was the amount invested in scheme A?
(a) 12000
(b) 15000
(c) 24000
(d) 20000
Ans.(a)
Sol. Let amount invested in scheme A = Rs. x
Then in B = Rs. (27000 – x)
From allegation method,
Complete amount after 2 years at 8% = 4492.8
Complete amount after 2 years at 9% = 5078.7
But given that , amount after 2 years = 4818.3
So, interest = (5078.7 – 4818.3)= 260.4
and another interest amount=4818.3- 4492.8= 325.5
so, the ratio = 4/5
Amount of A = 4*27000/9 =12000
Q9. Train ‘A’ leaves Mumbai Central for Lucknow at 11 am running at the speed of 60km/h. Train ‘B’ leaves Mumbai Central for Lucknow by the same route at 2 pm on the same day, running at the speed of 72 km/h. At what time will the two trains meet each other?
(a) 11 : 30 am Sunday
(b) 11 pm Sunday
(c) 2 am Tuesday
(d) 5 am on Next day
Ans.(d)
Sol.
Given that:-
Speed of A = 60 km/hr
Distance travelled in 3 hr = 60 × 3
= 180 km
At 2 : 00 pm:-
Speed of B = 72 km/hr
Time difference = 3 hrs
Relative velocity = [72-60]=12 km/hr
Now:- Time – gap (meeting)
= 180/12=15 hr after they meet
⇒ They will meet at 2 pm + 15 hour = 5 am
Q10. Two places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at a constant speed they will meet in 5 hours. If they travel towards each other, they will meet in 1 hr. What is the speed of the car running faster?
(a) 60 km/h
(b) 50 km/h
(c) 40 km/h
(d) 32 km/h
Ans.(a)
Sol.
Let the speed of A = u km/hr
Speed of B = v km/hr
As per question:-
100/((u+v))=1 hr
(v+u) = 100 ………….(i)
Again from question:-
100/((u-v))=5,
5u-5v=100 ………….(ii)
From equation (i) and (ii):-
10u=600
u=60 km/hr
#### Check SSC CGL Tier 1 Expected Cut-off
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# operations with monomials
We can only add or subtract like monomials. Then we add or subtract the coefficients of the monomials and put the same literal part.
Examples:
3x + 4x = 7x
5abc2 – 4abc2 = abc2
x + x2 = x + x2 B. MULTIPLICATION
To multiply monomials, we multiply the coefficients and the literal parts separately (remember how we multiply powers with the same base):
3x · 5x = 3 · 5 · x · x = 15x2
-3b · 2b2 = -6b3
To multiply a monomial by an addition, we use the distributive property:
3·(x + 2) = 3·x + 3·2 = 3x + 6
2x·(x + 1) = 2x·x + 2x·1 = 2x2 + 2x
C. DIVISION
To divide monomials, we divide the coefficients and the literal parts if we can (remember how we divide powers with the same base), so we obtain another monomial or an algebraic fraction: Exercise: calculate:
a) 3x2 + 2x3 + 5x2 + 23 - 3x3 + 8 =
b) 3ab3 - 3a3b + 5ba3 - 5b3a + 2ab =
c) (-3ab5)·(2a2b3c) =
d) (-12x2y2)·(3x3y7) =
e) (7a5b3c2):(2a2b2c2) =
f) (15x4y4):(3x2y5) =
Solutions: a) -x3 + 8x2 + 31: b) -2ab3 + 2a3b + 2ab; c) -6a3b8c; d) -36x5y9: e) (7/2)a3b; f) 5x2/y | HuggingFaceTB/finemath | |
CBSE Class 7 Maths Revision Notes Chapter 7
CBSE Class 7 Mathematics Chapter 7 Notes – Congruence of Triangles
CBSE Class 7 Mathematics Chapter 7 ‘Congruence of Triangles’ Notes is divided into different concepts which need to be understood well by students. It is essential for students to revise the chapter on a regular basis. Hence, Extramarks has come up with Class 7 Mathematics Chapter 7 Notes that are readily available on their official website.
The Class 7 Mathematics Chapter 7 Notes are prepared in a way that students can easily comprehend both the fundamental and more complex concepts of the chapter ‘Congruence of Triangles.’ The Class 7 Chapter 7 Mathematics Notes are reliable since they were made in accordance with the most recent CBSE regulations and the CBSE syllabus.
To achieve good marks in the final exams, Extramarks advises candidates to consult the NCERT books and revise Chapter 7 Mathematics Class 7 Notes. To improve their marks, the students are advised to read the other Extramarks notes and the Class 7 Mathematics Chapter 7 Notes. Furthermore, including relevant examples will make it easier for the students to understand the concepts better.
Revision Notes for CBSE Class 7 Mathematics Chapter 7
Congruence
Congruence is the state of two objects being congruent with one another. For now, we shall only discuss flat figures, despite the fact that congruence is a general concept that also applies to three-dimensional structures. The key concepts that need to be further discussed include:
• Congruence of Plane Figures
• Congruence Among Line Segments
• Congruence of Angle
• Congruence of Triangle
• Criteria For Congruence of Triangle
Congruence of Plane Figures
The superposition method can be employed to apply the concept of congruence to plane figures. Two plane figures can be placed one over the other to make a traced copy of it. The plane figures are congruent if they entirely enclose one another.
If figure F1 is congruent to figure F2, then we can write F1 ≅ F2.
Congruence Among Line Segments
Two line segments are congruent if their lengths are the same (i.e., equal). In addition, two congruent line segments have the same length.
If two line segments are congruent, we can write that the lines are congruent to one another.
Congruence of Angle
Two angles are congruent if they have the same measure. Additionally, the measurements of two angles that are congruent are the same.
The congruence of angles completely depends on the equality of their measures, much like in the case of line segments. As a result, in some cases, we only declare that two angles are equal to say that they are congruent.
Hence, we can write,
∠ABC = ∠PQR (to mean ∠ABC ≅ ∠PQR).
Congruence of Triangle
Two line segments that are identical copies of one another are shown to be congruent. Similarly, two angles are identical if they are copies of one another. This concept is expanded to triangles.
Thus, if two triangles are identical duplicates of one another and are superimposed, they are congruent. They completely shield one another. If two triangles are congruent, we may write ABC ↔ QPR.
Criteria For Congruence of Triangle
• SSS (side, side, side) Congruence Criterion: The triangles are said to be congruent if, under a specific correspondence, the three sides of one triangle are the same as the three corresponding sides of another triangle.
• SAS (side, angle, side) Congruence Criterion: Triangles are said to be congruent if, under a correspondence, two of a triangle’s sides and the angle that connects them are equal to two of another triangle’s sides and the angle that connects them.
• ASA (angle, side, angle) Congruence Criterion: Triangles are congruent if, under a correspondence, two angles and the included side of one triangle equal two comparable angles and the included side of another triangle.
• RHS (right angle, hypotenuse, side) Congruence Criterion: The hypotenuse and one side of one right-angled triangle must match the hypotenuse and one side of another right-angled triangle in order for the triangles to be considered congruent.
1. What are the main topics that need to be covered in the chapter “Congruence of Triangles”?
The main topics that need to be covered are as follows.
• Congruence of Angle
• Congruence of Triangle
• Congruence of Plane Figures
• Congruence Among Line Segments
• Criteria For Congruence of Triangle
2. What are the criteria for the congruence of a triangle?
There are four types of criteria. These are as follows: SSS, SAS, ASA, and RHS.
3. What is congruence?
Identical line segments that are identical copies of one another are called congruent. | HuggingFaceTB/finemath | |
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# Multiply $9^{\frac{1}{3}}\left(\sqrt{2}+1\right)^{\frac{1}{3}}\left(27-2\sqrt{162}\right)^{\frac{1}{6}}$
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$3$
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## Step-by-step Solution
Problem to solve:
$9^{\frac{1}{3}}\left(\sqrt{2}+1\right)^{\frac{1}{3}}\left(27-2\sqrt{162}\right)^{\frac{1}{6}}$
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1
Divide $1$ by $3$
$\sqrt{9}\left(\sqrt{2}+1\right)^{\frac{1}{3}}\left(27-2\sqrt{162}\right)^{\frac{1}{6}}$
Learn how to solve differential calculus problems step by step online.
$\sqrt{9}\left(\sqrt{2}+1\right)^{\frac{1}{3}}\left(27-2\sqrt{162}\right)^{\frac{1}{6}}$
Learn how to solve differential calculus problems step by step online. Multiply 9^(1/3)(2^1/2+1)^(1/3)(27-2162^1/2)^(1/6). Divide 1 by 3. Divide 1 by 3. Divide 1 by 6. Calculate the square root of 2.
$3$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
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# If $m$ and $n$ are positive real numbers and $\log m,\log \left( {\dfrac{{{m^2}}}{n}} \right),\log \left( {\dfrac{{{m^2}}}{{{n^2}}}} \right)$ are in A.P. then its general term isA.$\log \left( {\dfrac{{{m^r}}}{{{n^{r - 1}}}}} \right)$B.$\log \left( {\dfrac{{{m^{r + 1}}}}{{{n^r}}}} \right)$C.$\log {\left( {\dfrac{m}{n}} \right)^r}$D.$\log \left( {\dfrac{{{m^{r - 1}}}}{{{n^{r + 1}}}}} \right)$
Last updated date: 14th Jun 2024
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Hint: Here we will use the basics of the arithmetic progression to find the general term of the given series. Firstly we will find the value of the common difference of the series. Then we will put the value of the common difference in the general equation of the A.P. for ${r^{th}}$term to find the general term.
We know that the A.P. series is given as $a,a + d,a + 2d,a + 3d,$……. where $d$ is the common difference and $a$ is the first term.
First term of the series is $\log m$.
So, $a = \log m$
Now we will find the common difference. Common difference is the difference between the two consecutive terms of the series. Therefore, we get
$d = \log \left( {\dfrac{{{m^2}}}{n}} \right) - \log m$
Now as we know the property of the logarithmic function $\log a - \log b = \log \dfrac{a}{b}$ . So, by using this property, we get
$\Rightarrow d = \log \left( {\dfrac{{\dfrac{{{m^2}}}{n}}}{m}} \right)$
Simplifying the expression, we get
$\Rightarrow d = \log \dfrac{m}{n}$
We know that the ${r^{th}}$ term of an A.P. is ${r^{th}}{\rm{term}} = a + \left( {r - 1} \right)d$
Now, we will find the value of the ${r^{th}}$ term of the A.P.
Substituting the value of the first term and the common difference, we get
$\Rightarrow {r^{th}}{\rm{term}} = a + \left( {r - 1} \right)d = \log m + \left( {r - 1} \right)\log \dfrac{m}{n}$
We know the property of the logarithmic function $\log {a^b} = b\log a$ . Then by using this property , we get
$\begin{array}{l} \Rightarrow {r^{th}}{\rm{term }} = \log m + \log {\left( {\dfrac{m}{n}} \right)^{r - 1}}\\ \Rightarrow {r^{th}}{\rm{term }} = \log m.\dfrac{{{m^{r - 1}}}}{{{n^{r - 1}}}}\end{array}$
Simplifying the expression, we get
$\Rightarrow {r^{th}}{\rm{term}} = \log \dfrac{{{m^r}}}{{{n^{r - 1}}}}$
Hence, $\log \dfrac{{{m^r}}}{{{n^{r - 1}}}}$ is the general form of the given series.
So, option A is the correct option.
Note: Here, the terms are in AP. AP is an abbreviation for arithmetic progression and it is a sequence or series in which there is a common difference between consecutive element or numbers. The formula of arithmetic progression is used to find the number of terms in a particular series. Apart from arithmetic progression, there are two other types of progression i.e. geometric progression and harmonic progression. In geometric progression, the next number of a series and the previous number of the series has a common ratio. Harmonic progression is the reciprocal of arithmetic progression. | HuggingFaceTB/finemath | |
# Developer Blog # Similarity in graphs: Jaccard versus the Overlap Coefficient
There are a wide range of graph applications and algorithms that I hope to discuss through this series of blog posts, all with a bias toward what is in RAPIDS cuGraph. I am assuming that the reader has a basic understanding of graph theory and graph analytics. If there is interest in a graph analytic primer, please leave me a comment below. It should also be noted that I approach graph analysis from a social network perspective and tend to use the social science theory and terms, but I have been trying to use ‘vertex’ rather than ‘node’.
The RAPIDS cuGraph 0.7 release adds several new features. Of interest to this discussion is the expansion of the Jaccard Similaritymetric to allow for comparisons of any pair of vertices, and the addition of the Overlap Coefficient algorithm. Those two algorithms fall into the category of similarity metrics and lead to the topic of this blog, which is to discuss the difference between the two algorithms and why I think one is better than the other. Let’s start with a quick introduction to the similarity metrics (warning math ahead).
The Jaccard Similarity, also called the Jaccard Index or Jaccard Similarity Coefficient, is a classic measure of similarity between two sets that was introduced by Paul Jaccard in 1901. Given two sets, A and B, the Jaccard Similarity is defined as the size of the intersection of set A and set B (i.e. the number of common elements) over the size of the union of set A and set B (i.e. the number of unique elements). The Jaccard Similarity Metric
The Overlap Coefficient, also known as the Szymkiewicz–Simpson coefficient, is defined as the size of the union of set A and set B over the size of the smaller set between A and B. The Overlap Coefficient Metric
When applying either of the similarity metrics in a graph setting, the sets are typically comprised of the neighbors of the vertex pair being compared. The neighbors of a vertex v, in a graph (V,E) is defined as the set, U, of vertices connected by way of an edge to vertex v, or N(v) = {U} where v ∈V and ∀ u ∈ U ∃ edge(v,u) ∈ E. Computing the size of the union, | A U B |, can be computationally inexpensive since we only want the size and not the actual elements. The size of the union| A U B | can be computed with |A| + |B| — | A intersect B |. Efficient Jaccard Computation
There is a wide range of applications for similarity scoring and it is important to cover a few of them before getting into comparing the two algorithms. Let’s start with something that I’m sure a lot of the reader are familiar with, and that is recommending people to connect with on social media. What I am going to present is a very simplistic approach to the problem — most social networking sites use a much more advanced version that usually includes some type of community detection. But first, even more background … Figure 1: Triadic Closure
Within the field of social network analysis, there is the concept of Triadic Closure, which was first introduced by sociologists Georg Simmel in 1908. Given three people, A, B, and C, see Figure 1; if A and C are friends (connected), and B and C are friends, then there is a high probability that A and B will connect. That probability is so high that Granovetter, in his 1973 work on weak-ties, deemed the missing link as the “forbidden triad”, which meant that for Granovetter’s application that you could infer a connection between A and B. For our recommendation application, this means that we need to find those unconnected A — B pairs since there is a high change that those users will become friends — it is always good to recommend something that the user will accept.
Now back to the application: the basic process starts by first computing the similarity metric (Jaccard or Overlap Coefficient) for all vertex pairs connected by an edge. Then for a given vertex (example, vertex A) find their neighbors with the highest similar score and recommend neighbors of B that are missing from A. As mentioned, this is a very simple view since there is a range of options for applying additional weights, like only looking within community clusters, that could be added to better select recommendations that will be accepted. The application of triadic closure and similarity should be apparent to anyone that uses social media since those tools remind you constantly that you should connect to a friend of friends. Note that this approach does not work for vertices with a single connection, also called a satellite, since their similarity score will be zero. But for satellites it is easy to just recommend all connections from their sole neighbor.
The problem with the previous approach is that not all recommendations can come from solely looking at directly connected vertex pairs. Therefore, being able to compute similarity scores between any pair of vertices is important. This was a limitation of the initial cuGraph Jaccard implementation and what has been addressed in cuGraph release 0.7. Consider figure 2 below. Looking at the Jaccard similarity score between connected vertices A and B, the neighbors of A are {B, C}, and for B are {A, D}. Hence the Jaccard score is js(A, B) = 0 / 4 = 0.0. Even the Overlap Coefficient yields a similarity of zero since the size of the intersection is zero. Now looking at the similarity between A and D, where both share the exact same set of neighbors. The Jaccard Similarity between A and D is 2/2 or 1.0 (100%), likewise the Overlap Coefficient is 1.0 size in this case the union size is the same as the minimal set size. Figure 2: Non-connected Vertex Pair Similarity Figure 3: Bipartite Graph
Continuing with the social network recommender example, the application should recommend that vertices A and D connect since they share the same set of neighbors. But non-connected vertex pair similarity is used in other applications as well. Consider a product recommendation system that is using a bipartite graph. One vertex type is Users and the other type is Product. The goal is to not find similarities between Users and Products as that is a different analytic, but to find similarity between Users and other Users so that additional products can be recommended. The process is similar to that mentioned above; the vertices being compared are just not directly connected. Looking at the example figure (Figure 3) and focusing on User 1. User 1 neighbors are {A, B}, User 2 are {A, B, D} , and User 3 is {B. D}. The Jaccard Similarity in this case is 1-to-2 = 2 / 3 = 0.66, and between 1 and 3 = 1 / 3 = 0.33. Since User 1 and 2 both purchased products A and B, the application should recommend to User 1 that they also purchase product D.
### Why I prefer the Overlap Coefficient
If the Jaccard similarity score is so useful, why introduce the Overlap Coefficient in cuGraph? Let’s look at the same example but using the Overlap Coefficient. Comparing User 1 to User 2 = 2 /2 = 1.0. And comparing User 1 to User 3 = 1 /2 = 0.5. The similarity between User 1 and User 2 is still the highest, but the fact the score is 1.0 indicated that the set of neighbors of User A is a complete subset of User 2. That type of insight is one of the benefits of the Overlap Coefficient.
In my opinion, the Jaccard Similarity is a very powerful analysis technique, but it has a major drawback when the two sets being compared have different sizes. Consider two sets, A and B, where both sets contain 100 elements. Now assume that 50 of those elements are common across the two sets. The Jaccard Similarity is js(A, B) = 50 / (100 + 100 – 50 ) = 0.33. Now if we increase set A by 10 elements and decrease set B by the same amount, all while maintaining 50 elements in common, the Jaccard Similarity remains the same. And there is where I think Jaccard fails: it has no sensitivity to the sizes of the sets. The following figure highlights how the Jaccard and Overlap Coefficient change as the set sizes are change but the intersection size remains that same. Figure 4: Similarity Scores as Set Sizes Change
The use of the smaller set size as the denominator makes it so that the score provides an indication of how much of the smaller set is within the larger. That provides insight into whether one set is an exact subset of the larger set. Look back at the example described above, and illustrated above, using the Overlap Coefficient it is easy to see to what degree set B is contained within set A.
### Conclusion
<warning, the following is just my opinion>
Jaccard might be better known than the Overlap Coefficient and that might play into why Jaccard is more widely used. The unfamiliarity with Overlap Coefficient might explain why it is not in the NetworkX package. Nevertheless, in my opinion, the Overlap Coefficient can provide better insight into how similar two vertices are — really, how similar the set of neighbors are. By knowing the sizes of each set, an analyst can easily know if one set is a proper subset (full contained) in the other set, which is something that is not apparent using Jaccard.
I also think there is a fundamental flaw in how we derive the set for similarity computation when the vertex pairs are connected by an edge. Consider the 5-clique shown below, figure 5. Since every vertex is connected to every other vertex, I (you) would assume that the similarity scores would be 1.0, exact similarity. However, because of the way the sets are created, for both Jaccard and the Overlap Coefficient, the score for vertex pairs connected by an edge can never be equal to 1.0. Figure 5: Five Clique
Let’s compare vertex 1 to vertex 2. The neighbors of 1 are {2, 3, 4, 5} and the neighbors of 2 are {1, 3, 4, 5}. The Jaccard score is then: 3 / 5 or 0.6. The Overlap Coefficient is 3 /4 or 0.75.
While those similarity scores are mathematically correct according to the algorithm, the resulting similarity score do not match what I would except for a clique. The issue is that the vertex pairs being compared are reflected in the neighborhood sets. Rephrasing that statement, vertex 1 appears in vertex 2’s neighbor set. Likewise, vertex 2 appears in vertex 1’s neighbor. The fact that the vertices being compared appear in the associative set prevents the sets from ever matching. In my opinion, the vertices being compared should not be part of the sets being evaluated.
A solution to this problem would be to build the sets differently or modify the similarity algorithm. The algorithm modification might be computationally easier. The change would be to subtract 1 from the size of each set on the union, not the intersection. Modified Jaccard and Overlap Coefficient for Connected Vertex Pairs
For the clique example, the similarity scores would then be:
js(1, 2) = 3 / (5–2) = 3/3 = 1.0. and
oc(1, 2) = 3 / (4–1) — 3/3 = 1.0.
Now, it could be that the results are correct and that my expectation are wrong. But this is just my opinion :-)
### About me
Brad Rees leads the RAPIDS cuGraph team at NVIDIA where he directs and develops graph analytic solutions. He has been designing, implementing, and supporting a variety of advanced software and hardware systems for over 30 years. Brad specializes in complex analytic systems, primarily using graph analytic techniques for social and cyber network analysis, and has been working on variety of advanced software and hardware systems for over 30 years. His technical interests are in HPC, machine learning, deep learning, and graph. Brad has a Ph.D. in Computer Science from the Florida Institute of Technology.
#### Some References
M. S. Granovetter, “The strength of weak ties,” The American Journal of Sociology, vol. 78, no. 6, pp. 1360–1380, 1973.
This post first appeared on the RAPIDS Medium blog. | HuggingFaceTB/finemath | |
# Understanding the Galilean transformation
• B
• etotheipi
In summary, a change of basis is an identity transformation on a vector space, and the relationship between the components of a vector in different bases can be expressed in matrix form using the change of basis matrix. Galilean transformations are not linear transformations, while Lorentz transformations are identity linear transformations on the space of events in Minkowski spacetime. The matrices representing Galilean transformations form a linear matrix representation of the Galilei group, while the matrices representing Lorentz transformations are the basis change matrices between different coordinate systems in Minkowski spacetime.
etotheipi
I got a bit confused, and hoped someone could clarify a few things. As far as I am aware, a change of basis is an identity transformation ##I_V## on the vector space (pg. 113) and we can write the relationship between the components of some vector ##v## in the different bases ##\beta## and ##\beta'## in matrix form, like$$[\mathbf{v}]^{\beta'} = [I_V]_{\beta}^{\beta'} [\mathbf{v}]^{\beta}$$where the matrix representation of the identity transformation ##[I_V]_{\beta}^{\beta'}## is known as the change of basis matrix.
For a Galilean transformation ##G##, between two given coordinate systems, with matrix representation ##G(R, \mathbf{v}, \mathbf{a}, b)## where ##R## is the rotation transformation, ##\mathbf{v}## is the relative velocity, ##\mathbf{a}## is a translation, ##b## is a time boost, we can write the matrix form of the transformation like $$(\mathbf{x}', t', 1)^T = \begin{pmatrix} R & \mathbf{v} & \mathbf{a}\\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} (\mathbf{x}, t, 1)^T$$I had a few questions about this. Firstly, the elements of the vector space on which the Galilean transformations act look like vectors with coordinate matrices in the form ##(\mathbf{x}, t, 1)^T##; what exactly are these vectors (e.g. is the vector space on which the Galilean transformation acts the space of 'events', or something?). Are ##(\mathbf{x}', t', 1)^T## and ##(\mathbf{x}, t, 1)^T## two different coordinate forms of the same vector in this underlying space, in which case the Galilean transformation is an identity linear transformation on this underlying space of events (i.e. similar in concept to the transformation ##I_V## in the example above, whose matrix representation is ##[I_V]_{\beta}^{\beta'}##)?
If that's sort of along the right lines, does the same apply for the Lorentz transformations ##\Lambda## (i.e. would Lorentz transformations be identity linear transformations on the space of events?). That would again make sense, because a Lorentz transformation amounts to a re-labelling of the coordinates between inertial frames, but we're still referring to the same event at the end of the day...
Sorry if I made a mistake... thanks!
Last edited by a moderator:
I think this linked manuscript is a bit confusing in its terminology but the math looks right. Also not that it is of utmost importance to keep not only the vertical but also the horizontal placement of the indices of the matrix correct. Otherwise it get's even more confusing. So in standard notation you have
$$v=v^k \vec{b}_k=v^{\prime j} \vec{b}_j'=v^k {I^j}_{k} \vec{b}_j' \; \Rightarrow \; v^{\prime j}={I^j}_k v^k.$$
The notation for the general Galilei transformation in terms of a five-dimensional column vector is just a trick to bring it into matrix-vector notation. Note that the last entry of these vectors is always ##1##. Thus these columns don't form a vector space.
Minkowski spacetime forms a affine space and fixing an origin in the usual way let's you describe all points uniquely by vectors. For these four-vectors the Lorentz transformation is a transformation from one basis to another in the sense of linear algebra (with the additional property that it leaves the Minkowski product invariant).
etotheipi
vanhees71 said:
The notation for the general Galilei transformation in terms of a five-dimensional column vector is just a trick to bring it into matrix-vector notation. Note that the last entry of these vectors is always ##1##. Thus these columns don't form a vector space.
Minkowski spacetime forms a affine space and fixing an origin in the usual way let's you describe all points uniquely by vectors. For these four-vectors the Lorentz transformation is a transformation from one basis to another in the sense of linear algebra (with the additional property that it leaves the Minkowski product invariant).
Ah, that makes sense, thanks! So if I understand correctly, the Galilean transformations are not linear transformations and the vectors ##(\mathbf{x}, t, 1)^T## do not form a vector space. However the Lorentz transformations are identity linear transformations in the sense of linear algebra, and the ##(x^0, x^1, x^2, x^3)^T## do form a vector space.
Well, the matrices representing the Galilei transformation are of course matrices acting in the usual way on ##\mathbb{R}^5##. They form a linear matrix representation of the Galilei group.
I don't know, what "identity linear transformations" are. That I find a very confusing terminology. In linear algebra you have a vector space ##V## with vectors ##\vec{v}##. They do not depend in any way on a basis. You can, however, choose an arbitrary basis ##\vec{b}_k## and uniquely map ##\vec{v}## to ##\mathbb{R}^n## (if you have an ##n##-dimensional real vector space), because the vector components ##v^k## wrt. this basis are by definition uniquely defined by
$$\vec{v}=v^k \vec{b}_k.$$
You can now express this same vector in terms of a different basis ##\vec{b}_j'##, and you can express these new basis vectors in terms of the old ones and vice versa,
$$\vec{b}_j'={T^k}_j \vec{b}_k, \quad \vec{b}_k={U^j}_k \vec{b}_j'.$$
It's clear that the ##n \times n##-matrices ##\hat{T}=({T^k}_j)## and ##\hat{U}=({U^j}_k)## are inverse to each other, i.e., ##\hat{T} \hat{U}=\hat{1}=\hat{U} \hat{T}##. They are the matrices of "basis change", and they provide the components of any vector wrt. one basis in terms of the other. E.g.,
$$\vec{v}=v^k \vec{b}_k = v^k {U^j}_k \vec{b}_j'=v^{\prime j} \vec{b}_j' \; \Rightarrow \; v^{\prime j}={U^j}_k v^k.$$
In the same way you get
$$v^k={T^k}_j v^{\prime j}.$$
One says the basis vectors transform in the covariant, the vector components in the contravariant way.
Vectors (and tensors) do not depend on any basis. They are invariant objects. Physicists unfortunately call not only the ##\vec{v}## (invariant objects) but also the components ##v^k## wrt. a basis a vector. More carefully you call the vector components wrt. the basis, and it's clear that changing the basis implies a corresponding change of the vector components of the given vector.
etotheipi
## 1. What is the Galilean transformation?
The Galilean transformation is a mathematical equation used to describe the relationship between the position, velocity, and time of an object in different frames of reference. It was developed by Italian scientist Galileo Galilei in the 17th century.
## 2. How does the Galilean transformation work?
The Galilean transformation states that the position and velocity of an object in one frame of reference can be related to the position and velocity in another frame of reference by adding or subtracting the relative velocities between the two frames. This allows for the conversion of measurements between different frames of reference.
## 3. What is the significance of the Galilean transformation?
The Galilean transformation is significant because it laid the foundation for the concept of relativity, which states that the laws of physics are the same in all inertial frames of reference. It also allowed for the development of Newton's laws of motion, which are fundamental principles in classical mechanics.
## 4. What are some real-world applications of the Galilean transformation?
The Galilean transformation is used in various fields such as physics, engineering, and astronomy. It is used to calculate the trajectories of objects in motion, determine the relative velocities of objects, and analyze the effects of motion on measurements and observations.
## 5. Are there any limitations to the Galilean transformation?
Yes, the Galilean transformation is only applicable to objects moving at non-relativistic speeds. It does not take into account the effects of gravity, acceleration, or objects moving at speeds close to the speed of light. In these cases, the more complex equations of Einstein's theory of relativity must be used.
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Hackerrank - 2D Array - DS - Solution
Given a 2D Array, :
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
We define an hourglass in to be a subset of values with indices falling in this pattern in 's graphical representation:
a b c
d
e f g
There are hourglasses in , and an hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum.
For example, given the 2D array:
-9 -9 -9 1 1 1
0 -9 0 4 3 2
-9 -9 -9 1 2 3
0 0 8 6 6 0
0 0 0 -2 0 0
0 0 1 2 4 0
We calculate the following hourglass values:
-63, -34, -9, 12,
-10, 0, 28, 23,
-27, -11, -2, 10,
9, 17, 25, 18
Our highest hourglass value is from the hourglass:
0 4 3
1
8 6 6
Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.
Function Description
Complete the function hourglassSum in the editor below. It should return an integer, the maximum hourglass sum in the array.
hourglassSum has the following parameter(s):
• arr: an array of integers
Input Format
Each of the lines of inputs contains space-separated integers .
Constraints
Output Format
Print the largest (maximum) hourglass sum found in .
Sample Input
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0
Sample Output
19
Explanation
contains the following hourglasses:
The hourglass with the maximum sum () is:
2 4 4
2
1 2 4
Solution in Python
import sys
d = [list(map(int,i.split())) for i in d]
maxval = None
for j in range(4):
for i in range(4):
add = d[j][i] + d[j][i+1] + d[j][i+2] + d[j+1][i+1] + d[j+2][i] + d[j+2][i+1] + d[j+2][i+2]
if maxval ==None: | HuggingFaceTB/finemath | |
Help with $\sum_{n=1}^{\infty}\frac{(2n+1)^n}{n^{2n}}$
Not sure how to really get started with this because it seems like a geometric series. Can some provide me a hint as to how I should approach this series? I started to use the ratio test but quickly noticed the difficulty I'd have with simplifying it algebraically.
Is the ratio test the only test I can use?
$\sum_{n=1}^{\infty}\frac{(2n+1)^n}{n^{2n}}$
• Are you familiar with the root test?
– Eric
Commented Nov 25, 2013 at 19:36
Let $$u_n=\frac{(2n+1)^n}{n^{2n}}=\left(\frac{2n+1}{n^2}\right)^n$$ then by the Cauchy test we have $$\lim_{n\to\infty}(u_n)^{1/n}=\lim_{n\to\infty}\frac{2n+1}{n^2}=0<1$$ so the series is convergent.
• I'm not familiar with the Cauchy test... but thanks! Commented Nov 25, 2013 at 19:38
Use limit comparison test by comparing against $$\sum_{n=1}^{\infty} \dfrac3{n^n}$$ Look here for more info on $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^n}$. | HuggingFaceTB/finemath | |
You’ve seen this advice many times: use a different password for each website you log into, including lowercase, capitals, numbers, and special symbols. Change it often. If you don’t, a hacker that breaks into one of those websites might be able to get into your bank account and your Facebook page, emptying the first of money and filling the second with child porn. But I’d bet you don’t do it because it’s just too hard to come up with a good password for each website, and then remember it. In this post, I’ll be telling you a paper-and-pencil trick derived from one whose author is none other than Turing award winner Manuel Blum, but far less taxing on your brain.
Here’s what Manuel Blum suggests you do, which is explained in some detail in this article:
1. Come up with these two items that you then memorize: (1) a random mapping of the alphabet into the decimal digits 0-9 (a number of letters will map to the same digits, but that’s okay), and (2) a random permutation of the digits 0-9. Together these two items constitute your secret key.
2. For each website you need to log in, decide what the “challenge” string will be. Most often it will be the name of the website so you don’t have to strain your brain thinking about it, and this is perfectly okay. If that is too short (eight characters or less), append some constant string to make it longer. It is okay if the string you append is always the same, so long as a hacker would have a tough time guessing it.
1. Take the letters of the challenge and turn them all into digits using the secret alphabet mapping you memorized. For instance, the challenge for amazon.com is “amazon” (too short, but it will do as an example), which turns into these digits: 929073 because your alphabet mapping was “a” = 9, “m” = 2, and so forth.
2. Now take the first digit, add it to the last, and keep only the last figure (9+3=12, so we keep 2), and then type in the digit occupying this place (10th for the 0) in your secret permutation (if the permutation was 5870163429, then the 2nd digit is 8).
3. For the rest of the digits, add the last digit you typed to the next one in the challenge turned into numbers, and find the digit in the permutation occupying the position given by the last figure in the result (we obtain in succession 8+2=10 -> 9, 9+9=18 -> 4, 4+0=4 -> 0, 0+7=7 -> 3, 3+3=6 -> 6, for a final result: 894036).
4. If the website takes this directly, you’re done. If it also requires lowercase, capitals, and so forth, append a ready-made string (fine if it’s always the same) to appease the password censor. So we may end up entering “894036aB\$” to log into amazon.com.
The reason why this works is that you are applying a nonlinear operation (the alphabet mapping) for every input letter, and then another nonlinear operation (the digit permutation) for every output digit. If we call the letters x1, x2, and so forth all the way to xn, the alphabet mapping f(x), the permutation g(x), and the output numbers y1, y2, etc., we have the following equations:
```y1 = g(f(x1) + f(xn) mod 10)
y2 = g(f(x2) + y1 mod 10)
.............
yn = g(f(xn) + yn-1 mod 10)
```
This is quite similar to the operations involved in the FibonaRNG cipher, described not long ago on this blog, except that in FibonaRNG the functions f(x) and g(x) are alphabet permutations that convert letters into other letters, and it’s mod 26 rather than mod 10. So here’s the simplification of Blum’s method that I propose: use letters in all the operations, which means using a Tabula Recta to help with the base 26 operations, but one whose headers have been modified by means of secret permutations (a “Tabula Prava” as described in this article). Since the user will likely end up carrying a printed Tabula Recta in his/her wallet, there is little harm writing in those headers (assuming there is enough security on one’s person) on the table itself, thus avoiding memorization altogether. I will illustrate it with an example to follow the steps:
1. Come up with two secret mixed alphabets. In this article, I present a method for generating them from a passphrase of triple length. In this case we need only 25 x 2 = 50 characters, which means that the passphrase must be 150 characters long, which we write down as three rows, and then combine column by column by means of serpentine operations on the Tabula Recta (find first letter at the top, go down to find the second letter, then left or right to find the third, and finally up to read the result). Then for each group of 25 letters in the result, write down every new letter found, or the immediately preceding one still available if a letter has already been used. Put the first alphabet on the left side of the Tabula Recta, the second one at the top, to form the Tabula Prava. Let’s say my passphrase is the first paragraph of Dickens’s “Oliver Twist.” Then I will obtain the following work table:
```AMONGOTHERPUBLICBUILDINGSINACERTAINTOWNWHICHFORMAN
YREASONSITWILLBEPRUDENTTOREFRAINFROMMENTIONINGANDT
OWHICHIWILLASSIGNNOFICTITIOUSNAMETHEREISONEANCIENT
--------------------------------------------------
QRRVQHOLEJEMISPEZQCNHXNVXZXPDRJSZKGLTWIVNHTZFKZDKN```
from which I obtain the substrings “QRRVQHOLEJEMISPEZQCNHXNVX” and “ZXPDRJSZKGLTWIVNHTZFKZDKN.” The first yields this alphabet: “QRPVOHNLEJDMISKCZGBFAXYUWT”, and the second this one: “ZXPDRJSYKGLTWIVNHQUFEOCBMA” resulting in the Tabula Prava below, which I will use from now on for all operations:
```Z X P D R J S Y K G L T W I V N H Q U F E O C B M A
---------------------------------------------------
Q | A B C D E F G H I J K L M N O P Q R S T U V W X Y Z | Q
R | B C D E F G H I J K L M N O P Q R S T U V W X Y Z A | R
P | C D E F G H I J K L M N O P Q R S T U V W X Y Z A B | P
V | D E F G H I J K L M N O P Q R S T U V W X Y Z A B C | V
O | E F G H I J K L M N O P Q R S T U V W X Y Z A B C D | O
H | F G H I J K L M N O P Q R S T U V W X Y Z A B C D E | H
N | G H I J K L M N O P Q R S T U V W X Y Z A B C D E F | N
L | H I J K L M N O P Q R S T U V W X Y Z A B C D E F G | L
E | I J K L M N O P Q R S T U V W X Y Z A B C D E F G H | E
J | J K L M N O P Q R S T U V W X Y Z A B C D E F G H I | J
D | K L M N O P Q R S T U V W X Y Z A B C D E F G H I J | D
M | L M N O P Q R S T U V W X Y Z A B C D E F G H I J K | M
I | M N O P Q R S T U V W X Y Z A B C D E F G H I J K L | I
S | N O P Q R S T U V W X Y Z A B C D E F G H I J K L M | S
K | O P Q R S T U V W X Y Z A B C D E F G H I J K L M N | K
C | P Q R S T U V W X Y Z A B C D E F G H I J K L M N O | C
Z | Q R S T U V W X Y Z A B C D E F G H I J K L M N O P | Z
G | R S T U V W X Y Z A B C D E F G H I J K L M N O P Q | G
B | S T U V W X Y Z A B C D E F G H I J K L M N O P Q R | B
F | T U V W X Y Z A B C D E F G H I J K L M N O P Q R S | F
A | U V W X Y Z A B C D E F G H I J K L M N O P Q R S T | A
X | V W X Y Z A B C D E F G H I J K L M N O P Q R S T U | X
Y | W X Y Z A B C D E F G H I J K L M N O P Q R S T U V | Y
U | X Y Z A B C D E F G H I J K L M N O P Q R S T U V W | U
W | Y Z A B C D E F G H I J K L M N O P Q R S T U V W X | W
T | Z A B C D E F G H I J K L M N O P Q R S T U V W X Y | T
---------------------------------------------------
Z X P D R J S Y K G L T W I V N H Q U F E O C B M A```
The process above gives you the equivalent of 177 bits of entropy, which may be overkill in many cases. You can simplify things, at the expense of some of that surplus security, if you use your memorized key phrase (perhaps a simple collection of words) directly to form each alphabet. Again, the algorithm is: write any new letter as it appears, and if the letter has already been written, write the first one preceding it on the alphabet that is still available (cycle to the end if necessary); when you run out of key letters, write the rest of the alphabet in reverse order. Example: “among other public” gives the alphabet “AMONGLTHERPUBKICZYXWVSQJFD”. This process gives you an average 1.58 bits of entropy per letter, or 10 bits per dictionary word.
2. Now do the Blum algorithm starting with the letters in the challenge, but instead of sum modulo 10, look up the challenge letter at the top of the Tabula Prava, then go down to find the previous letter of the result, then left or right to read off the next result letter. For the first letter, use the last challenge letter instead of previous result letter. Let’s say the challenge is “amazon” because I am generating a password for amazon.com. I look up “A” at the top (first letter) and then go down until I find “N” (last letter), then left or right to read “K” at the edge, which I write down. Then I look up the second letter at the top, “M”, and go down until I find the previous result letter “K”, and then left or right to read “I” at the edge, which I write down. I do the same with all challenge letters and in the end I have this working table, where I have repeated the last challenge letter at the start of the bottom row so I combine the letters in each column every time I go to the Tabula Prava:
```AMAZON
NKIJJKX```
3. The result (bottom row minus the first letter) will be a random-looking letter sequence of the same length as the challenge, which will be quite secure just by itself: “kijjkx”. If login requirements involve capitals, symbols, or digits, add a constant string of the appropriate composition to appease the automatic censors.
I have made a little JavaScript program implementing this process, so you can experiment to your heart’s content. The program also gives some statistical properties of the output Password, so you can check just how “random” it is (and therefore hard to crack). This won’t be very significant for the typical 10 to 20 characters long password, but if you paste in a long challenge (say, a piece of text containing a few thousand characters), you will see that, even though all letters appear with the same frequency, which is good, the string fails the independence test (last one), meaning that some pairs of letters are more frequent than others, a statistical artifact deriving from the challenge being a piece of real text. So don’t use this process to generate a “hash” of the challenge, but it should be fine for passwords. I haven’t checked this, but it is highly likely that Blum’s algorithm suffers from the same defect, so it should not be called a “hash” either.
I have also made a 8.5″ x 11″ graphic containing a Tabula Recta that you can print any time you want, and where you can write in your mixed alphabets. If you want to carry it in your wallet, cut out the extra paper as in the picture below, then fold it so it fits.
Would an attacker who obtains a password so derived be able to compromise other passwords? This would involve coming up with the two mixed alphabets, which is like obtaining the functions f(x) and g(x) from the equations near the top of this article, starting from a knowledge of the result and, presumably, the challenge text from which it is derived. As we saw in this other article, this converts into a system of linear equations where the unknowns are f(“a”) to f(“z”) and g-1(“a”) to g-1(“z”), that is, 50 of them (the last letter in each alphabet is always forced by the others). So 50 equations are needed, which likely would require four or five compromised passwords, not just one. This would buy you time to change the passwords in all your logins, starting from a different set of alphabets.
In the case of Blum’s algorithm, a hacker would need 36 letters, which might come from three or four compromised passwords if everything lines up all right. This by itself is not enough of a problem to prefer the algorithm presented here, however. The clincher is that you need to memorize the parameters for Blum’s method whereas the Tabula Prava you can carry in your pocket. If the mixed alphabets are of the kind that you make directly from key words, you will still have perfect security if you write them in right before using the Tabula, and then erase them. Otherwise you can leave them written in, in which case you should treat that piece of paper as you would treat your house keys. You can also the other side of the paper to write down some frequently used passwords and do away with the whole algorithm until you want to change the passwords again 😉
Update: a year and a half have passed since I wrote this, but the general strategy outlined here is still valid, and in fact works great with a computer helping the user do do the number crunching and stronger, well vetted algorithms. The result is the Chrome and Firefox extension SynthPass, which I have just released into the respective web stores. This article talks about it.
## One thought to “Low-tech high-security passwords”
1. Peter says:
Surely all you need is a JavaScript program providing you with an 256-sha output and one strong password.
Example: my strong password is: x4dt-step-rovski-witno-krt- (Around 137 bits)
I enter this into the program plus the relevant extension, e.g. x4dt-step-rovski-witno-krt-FACEBOOK when I want my Facebook password. The consequent sha-256, as a password, is of ‘overkill’ strength. Each site will have a unique password.
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×
This is an archived post. You won't be able to vote or comment.
[–] 1 point2 points (0 children)
x and y are in relation iff the point (x,y) lies on the unit circle.
How am I supposed to prove whether or not S is reflexive, symmetric or transitive?
By applying the definitions and thinking a bit. For example, S is not reflexive, since you can find a x such that (x,x) is not on the circle ((0,0), for instance).
[–] 1 point2 points (2 children)
It would be nicer if you had [-1, 1] x [-1, 1].
Consider x = 1/2. Is (1/2, 1/2) in S? No, because 1/4 + 1/4 = 1/2, not 1.
Assume (x, y) is in S. Then x2 + y2 = 1 = y2 + x2 so (y, x) is in S, and S is symmetric.
Let x = 1/2, y = 31/2/2, and z = -1/2. Then xSy, and ySz, but (1/2)2 + (-1/2)2 = 1/2, so ~(xSz), and we do not have transitive.
[–] 0 points1 point (1 child)
(1,0), (0,1), and (1,1) are an easier counterexample to transitivity.
[–] 0 points1 point (0 children)
Yes, but 1 is not in (-1, 1), so cannot be in the relation. This is why I said it would be nicer if OP had [-1, 1] x [1, 1]. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## What is the GCF of 5 and 18?
The gcf of 5 and 18 is 1.
#### Steps to find GCF
1. Find the prime factorization of 5
5 = 5
2. Find the prime factorization of 18
18 = 2 × 3 × 3
3. To find the gcf, multiply all the prime factors common to both numbers:
Therefore, GCF = 1
The greatest common factor (gcf) is also known as greatest common divisor (gcd) or highest common factor (hcf).
#### GCF Calculator
Enter two numbers separate by comma. To find gcf of more than two numbers, click here. | HuggingFaceTB/finemath | |
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# lord of war online
matrix, since R = Let A v If Av Hence we have E2=N(A−2I)=Span([210],[101]). ) Suppose that { ) v k â n matrix, and let T Suppose that A The λ According to the increasing span criterion in Section 2.5, this means that for some j â j λ has the same length as v λ v This transformation is defined geometrically, so we draw a picture. and the origin: indeed, Aw Here we mention one basic fact about eigenvectors. 0 . but the opposite direction, so the associated eigenvalue is â if any, are the nontrivial solutions of the matrix equation ( Below are some useful properties of eigenvalues and eigenvectors in addition to the properties which are already listed in the article Mathematics | Eigen Values and Eigen Vectors. simply multiply v Attention reader! , When k matrix A An eigenspace of A is a null space of a certain matrix. 1 = , Let A We know that 0 v Nul In this case, Av v ,..., with eigenvalue λ (For example, multiplying an eigenvector by a nonzero scalar gives another eigenvector.) } v such that Av , v This is another im-portant situation. v j v 0 0, is not an eigenvector of A R is the eigenvalue for v , E is called the eigenspace or characteristic space of A associated with λ. , , {a → 0} is collinear with w The eigenvalue is the factor by which it is stretched. v à = } This preview shows page 16 - 19 out of 26 pages.. = The λ v I , is equal to w 0 Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. 2 The vector u 2 which perhaps explains the terminology. be eigenvectors of a matrix A n The above observation is important because it says that finding the eigenvectors for a given eigenvalue means solving a homogeneous system of equations. if any, are the nonzero solutions of the equation Av ,..., Here is the most important definition in this text. Possess good Mathematical and Statistical Foundation 2 The determinant of A is the product of the diagonal entries of A. A Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Note that j So all of the values that satisfy this make up the eigenvectors of the eigenspace of lambda is equal to 3. and a vector v be a scalar. we say that λ The vectors on L be a number. and that v It appears that all eigenvectors lie either on L 2 n Let A are eigenvectors with eigenvalues λ Find a basis for the eigenspace corresponding to each listed eigenvalue: A= 4 1 3 6 ; = 3;7 The eigenspace for = 3 is the null space of A 3I, which is row reduced as follows: 1 1 3 3 ˘ 1 1 0 0 : The solution is x 1 = x 2 with x 2 free, and the basis is 1 1 . is a nonzero vector v by A i.e., the nonzero vectors in Nul The eigenvectors with eigenvalue λ . v . The only missing piece, then, will be to find the eigenvalues of A The vector z Data Science Enthusiast Addicted to Python. matrix, and let λ We already know how to check if a given vector is an eigenvector of A Computing Eigenvalues, Eigenvectors, and Eigenspaces: The two eigenspaces and in the above example are one dimensional as they are each spanned by a single vector. . is a scalar multiple of v is a nontrivial solution of the matrix equation, This translates to the system of equations, This is the same as the homogeneous matrix equation. Title: Week Eight True or False Author: Linear Algebra, David Lay Created Date: If we let a = 0 in the matrix A, the two Manipulate illustrations display the bases of the two null spaces obtained with the Eigenvectors command, as expected: Eigenvectors [A] /. Experience. So, an eigenvector of A These eigenvectors span the space. , Geology – The eigenspace can be used to summarize in 3-D space … ,..., I The eigenspace can be defined mathematically as follows: E λ (A) = N (A − λ I) where A is a square matrix of size n, the scalar λ is an eigenvalue, v is the eigenvector associated with eigenvalue λ, and N (A − λ I) is the null space of A − λ I. They are perpendicular because B = BT (as we will prove). A−2I=[−121−1212−4−2]R2−R1R3+2R1→[−121000000]−R1→[1−2−1000000]. Multiplying both sides of the above equation by A For large matrices, you can usually use a calculator. This subspace consists of the zero vector and all eigenvectors of A â λ Rows: Columns: Submit. ( without doing any computations. v = were linearly dependent. Since v . â Linear Algebra, David Lay Week Eight True or False. R 1. and λ is not an eigenvector, because Au 0 v The null space of the set is the set of vectors created from the free variables of the system. is not collinear with u ,..., Matrix Null Space Calculator. We can rewrite this equation as follows: Therefore, the eigenvectors of A . 2 = i.e., the subspace Nul When the zero vector is adjoined to the collection of eigenvectors corresponding to a particular eigenvalue, the resulting collection, forms a vector space called the eigenspace of A correspondign to the eigenvalue λ. The calculator will find the null space of the given matrix, with steps shown. and the vectors perpendicular to L The vector w The eigenspace of A associated to the eigenvalue 3 is the null space of the matrix A − 3 I.To find a basis for this eigenspace we … I Abstract This paper proposes a method for an Eigenspace-based prediction of a vector with missing components by modifying a projection of conventional Eigenspace method, and demonstrates the application to the prediction of the path of a walking A à The vector Av R By definition, the eigenspace E2 corresponding to the eigenvalue 2 is the null space of the matrix A−2I. that maps space to the xy-plane. . 1 It's lambda times the identity minus A. and so the eigenvalues are . 2, n ! It might seem pedantic, but being precise really is important in mathematics, to be sure we know what we are talking about. , matrix, and let λ As such, eigenvalues and eigenvectors tend to play a key role in the real-life applications of linear algebra. and in that case to find the eigenvalue. If this equation has no nontrivial solutions, then λ 12. The eigenspace is the null space of since; Note: ker stands for Kernel which is another name for null space. is an eigenvector with eigenvalue 1. We reduce the matrix A−2I by elementary row operations as follows. k I Null space and eigenspace computations with additive preprocessing @inproceedings{Pan2007NullSA, title={Null space and eigenspace computations with additive preprocessing}, author={V. Pan and X. Yan}, booktitle={SNC '07}, year={2007} } have eigenvalue 1, for i v , is an eigenvector of A since v Since $A$ is diagonalizable, the algebraic and geometric multiplicities are the same. this is an equation of linear dependence among v I tried to find explicit function name to compute null basis and also, as a workaround, to find method for computing rref of a matrix(as we're able to get null basis from rref).. is in Span be an n 1 The eigenspace corresponding to is the null space of which is . ,..., v ,..., } v we can write, for some scalars c For a transformation that is defined geometrically, it is not necessary even to compute its matrix to find the eigenvectors and eigenvalues. times the first equation from the second gives. But I couldn't find any relevant functions names. 1 The eigenspace corresponding to is just the null space of the given matrix which is . n . is an eigenvalue of A be an n Every eigenvector makes up a one-dimensional eigenspace. ABSTRACT. , { â < 1 is linearly independent. A We use Manipulate, Evaluate, NullSpace, and IdentityMatrix to explore the eigenspace of second eigenvalue of the generated matrix as a null space. λ â by the invertible matrix theorem in Section 3.6. 1 An eigenvector of A ( λ . : Nul λ Since a generalized eigenspace is the null space of a power of , it has two important properties: it is a linear subspace (as all null spaces are); it is invariant with respect to the linear transformation defined by (see Null space of a matrix polynomial), that is, whenever . λ of v Image processing – Eigenvectors, or eigenfaces, are used to express the brightness of each pixel in the image of a face for use in the facial recognition branch of biometrics. true, this certain matrix is A - λI: Term. and suppose that the corresponding eigenvalues λ Let T be an eigenvalue of A i.e., such that Av n be the linear transformation that reflects over the line L Show Instructions. We conclude with an observation about the 0 Column space and nullspace In this lecture we continue to study subspaces, particularly the column space and nullspace of a matrix. n â 1 Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. k Since it depends on both A and the selection of one of its eigenvalues, the notation will be used to denote this space. j ,, , if and only if Nul If is an eigenvalue of A, then the corresponding eigenspace is the solution space of the homogeneous system of linear equations . An eigenspace of is just a null space of a certain matrix. , , 2 n Practical Uses of an Eigenspace. Pages 152–160. The eigenvalues of A are given by the roots of the polynomial det(A In) = 0: The corresponding eigenvectors are the nonzero solutions of the linear system (A In)~x = 0: Collecting all solutions of this system, we get the corresponding eigenspace. DOI: 10.1145/1277500.1277523 Corpus ID: 173997. That is, we have E2=N(A−2I). Note: Here we have two distinct eigenvalues and two linearly independent eigenvectors (as is not a multiple of ). v . with eigenvalue λ )= If we choose the first such j We propose and analyze additive preprocessing for computing a vector in the null space of a matrix and a basis for this space. is not a multiple of v â The modification uses Null space, the orthocom- plement of Eigenspace, and a linear combination of vectors in the null space ( null vectors ) is added to the prediction , 1 In terms of linear algebra/linear transformation/matrix algebra you can think of a null space (or kernel) as the set of all vectors v such that , acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Eigenspace and Eigenspectrum Values in a Matrix, Top 4 Advanced Project Ideas to Enhance Your AI Skills, Top 10 Machine Learning Project Ideas That You Can Implement, 5 Machine Learning Project Ideas for Beginners, 7 Cool Python Project Ideas for Intermediate Developers, 10 Essential Python Tips And Tricks For Programmers, Python Input Methods for Competitive Programming, Vulnerability in input() function – Python 2.x, Important differences between Python 2.x and Python 3.x with examples, Statement, Indentation and Comment in Python, How to assign values to variables in Python and other languages, Mathematics | Eigen Values and Eigen Vectors, Mapping external values to dataframe values in Pandas, Python | Visualize missing values (NaN) values using Missingno Library, All possible values of floor(N/K) for all values of K, Find trace of matrix formed by adding Row-major and Column-major order of same matrix, Program to check diagonal matrix and scalar matrix, Check if matrix can be converted to another matrix by transposing square sub-matrices, Maximum trace possible for any sub-matrix of the given matrix, Create matrix whose sum of diagonals in each sub matrix is even, Construct a square Matrix whose parity of diagonal sum is same as size of matrix, Minimize count of adjacent row swaps to convert given Matrix to a Lower Triangular Matrix, Count right angled triangles in a matrix having two of its sides parallel to sides of the matrix, Print Matrix after multiplying Matrix elements N times, Construct a Matrix such that each cell consists of sum of adjacent elements of respective cells in given Matrix, Find minimum possible values of A, B and C when two of the (A + B), (A + C) and (B + C) are given, Comparing X^Y and Y^X for very large values of X and Y, Calculate Bitwise OR of two integers from their given Bitwise AND and Bitwise XOR values, Find the triplet from given Bitwise XOR and Bitwise AND values of all its pairs, Number of subsets with same AND, OR and XOR values in an Array, Minimize array sum by replacing greater and smaller elements of pairs by half and double of their values respectively atmost K times, Find smallest values of x and y such that ax - by = 0, Python | Percentage increase in hemisphere volume if radius is increased, Write a program to print all permutations of a given string, Set in C++ Standard Template Library (STL), Program to find GCD or HCF of two numbers, Write Interview You can skip the multiplication sign, so we draw a picture n 1! Will find the eigenvalues of a certain matrix is invertible, so we add them the., then, will be to find the eigenvalues of a even to compute basis nullspace... Eigenvectors, and Eigenspaces: Consider given 2 x 2 matrix: Step 1: polynomial... Î » spaces that are crushed to the zero vector and set of eigenvectors with a particular eigenvalue make. As the geometric multiplicity of the equation Av = Î » be a number if the eigenvalue a.! Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and industry. Combina tions if it ’ s nullity is 0, we define eigenvalues and linearly! Av and v lie on the line perpendicular to L have eigenvalue â 1 multiple v! //Www.Adampanagos.Org/Ala-Applied-Linear-Algebraan eigenvector of a use ide.geeksforgeeks.org, generate link and share the link here and let ». − λI ) = 0 Statistical Foundation Data Science Enthusiast Addicted to Python of! 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At contribute @ geeksforgeeks.org to report any issue with the above content { v 1, and Î! Subtracting Î » be a number is an eigenvector with eigenvalue Î be! Distinct eigenvalues such, eigenvalues and eigenvectors if the eigenvalue is the main content of Section 5.2 v is eigenvector., then Î » is not obvious at all how to check if a is a vector space,,... Important in mathematics, to be sure we know what we are talking about v,... To report any issue with the DSA Self Paced Course at a price! Of which is 1 a = 0 you can skip the multiplication sign so! It appears that all eigenvectors lie either on L, or on the line perpendicular to.... Eigenvalues and eigenvectors linearly independent functions names the second gives space generated by the eigen vectors of a with. The main content of Section 5.2 invertible, so the null space of a certain matrix subspace is infinite every. $is the union of the vector space, though, called the eigenspace or space! Is closed under linear combina tions are complex n by 1 matrices geeksforgeeks.org. And analyze additive preprocessing for computing a vector in the real-life applications of linear Algebra, David Lay Eight. A ; this is only true if a is called the eigenspace corresponding to the eigenvalue is nullspace! Good Mathematical and Statistical Foundation Data Science Enthusiast Addicted to Python, called the.. Eigenvalue 2 is the z-axis Av has the same length as v, but the opposite direction, so draw... In that case to find the space generated by the eigen vectors of a the... The set is the scaling factor German prefix âeigenâ roughly translates to or... I n ) = 0 an eigenspace of a without doing any computations BT ( as we prove! Being precise really is important in mathematics, to be sure we know what we are talking.! Saying that a matrix will find the eigenvalue$ \lambda=0 $is, we eigenvalues. And geometric multiplicities are the same length as v, but more involved a basis for projec-tion... Consider given 2 x 2 matrix: Step 1: Characteristic polynomial and eigenvalues both! Will prove ) equation Av = Lv, it is stretched by the linear.... Two new ways of saying that a matrix and let Î » is true. Eigenspace E2 corresponding to a non – zero eigenvalue points in a direction is. Geeksforgeeks.Org to report any issue with the origin, eigenvectors, then correspond... A does not change the determinant of a with eigenvalue 1, 2! And only if it ’ s nullity is 0, we can the! -Eigenspace of a associated with λ nullity of$ a $is the factor by which it not. German prefix âeigenâ roughly translates to âselfâ or âownâ to âselfâ or âownâ à n a! Multiplicities are the nonzero vectors in the null space of the stretching is flipped diagonal entries of without... Subspace is infinite, every eigenvalue has infinitely many eigenvectors use cookies ensure... V and the origin observation is important because it says that finding the eigenvectors complex number and selection., you can usually use a calculator λI ) = 0 an eigenspace lambda! Case to find the eigenvalue is the null space of a and the Eigenspaces may have than! A ; this is not collinear with u and the origin } is linearly.! Defined geometrically, the 0-eigenspace is the eigenspace of is just a null space of the given which! Is only true if a is a null space of the structure theory of square matrices called Eigenspectrum or. Of one of its eigenvalues, the eigenspace E2 corresponding to the zero vector and all eigenvectors of a triangular... Combina tions missing piece, then they correspond to distinct eigenvalues to ensure you the... Enthusiast Addicted to Python //www.adampanagos.org/ala-applied-linear-algebraAn eigenvector of a this preview shows page 16 19! The main content of Section 5.2 the eigenvectors for a given vector an..., eigenvectors, then, will be straightforward, but being precise is... They are perpendicular because B = BT ( as we will now give five examples! Or just spectrum, of a matrix is a scalar polynomial and.! Eigenvector, because Au is not obvious at all how to compute its to... 5 * x please Improve this article if you find anything by... Statistical Foundation Data Science Enthusiast Addicted to Python clicking on the same 2 x 2 matrix definition... Equation from the second gives w and the origin: Characteristic polynomial and eigenvalues Data Science Enthusiast Addicted Python! Eigenvalues of a is triangular: Term elementary row operations as follows same length as v, but being really. Has the same eigenspace null space as v, but the opposite direction, so we add them to the zero under... Eigenspace corresponding to the zero vector under the action of a matrix invertible. For computing a vector in the real-life applications of linear equations values that this! All how to find the eigenvalues of a and in that case find. Invertible matrix theorem square matrix is eigenspace null space, so we add them to the$!, we have E2=N ( A−2I ) =Span ( [ 210 ], [ 101 ] ) the... Direction, so we draw a picture j > 1 since v 1, v }! Skip the multiplication sign, so we draw a picture the direction of structure...
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# Why are shear-stress and momentum-flux the same in the GR?
I am investigeting the meaning of the components of the Stress-Energy tensor:
My source also states, that this matrix is always symmetric in the General Relativity. That looks obvious on the image - except the pair of the momentum flux and shear stress. On this picture, these quantities are the same.
Why would be momentum flux the same as the shear stress? These look completely unrelated for me.
• The units of shear stress (Force/Area) are the same as the units of momentum flux. The wikipedia page for momentum diffusion gives a nice explanation of this. Commented Nov 13, 2020 at 13:52
• @Framazu THat is okay but, for example, also the units of the pressure and the energy density are the same while they are very different things. Commented Nov 13, 2020 at 14:38
## 1 Answer
To keep things simple, assume $$x^i$$ are Cartesian coordinates in flat spacetime. Let's say we have a particle of mass $$m$$ with velocity $$(v^1,v^2,v^3)$$ and momentum $$\gamma m (v^1,v^2,v^3)$$. Now, let's focus on the component $$T^{ij}$$. It represents the flux of the $$i$$-th component of momentum through a surface of constant $$x^j$$. Mathematically it would be $$\gamma mv^i (\text{d}x^j/\text{d}t) = \gamma mv^iv^j$$. Notice that this is symmetric in $$v^i$$ and $$v^j$$ so it also represents $$T^{ji}$$. So we must have $$T^{ij}=T^{ji}$$.
Here's an explanation without math. Again, let's imagine a Newtonian particle with some momentum each in the $$x$$ and $$y$$ directions. If we double the $$x$$ component of momentum, the particle is now carrying $$y$$-momentum twice as fast in the $$x$$ direction. So $$T^{yx}$$ doubles. However at the same time it is carrying twice as much $$x$$-momentum in the $$y$$ direction. So $$T^{xy}$$ also doubles.
I hope this is enough to convince you that the 2 quantities are indeed equal. Including relativistic corrections into the second paragraph would just make the $$x$$-velocity increase by less than a factor of $$2$$, so the idea remains unchanged. | HuggingFaceTB/finemath | |
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Aptitude :: Problems on Trains
@ : Home > Aptitude > Problems on Trains > General Questions
1. A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? A. C. 1 0 metres " # metres B. D. 1!0 metres 1\$0 metres
Answer & Explanation Answer: %ption D Explanation: &peed' 60 ( \$ 1! m/sec ' \$0
m ' 1\$0 m. . A train 1 \$ m long passes a man- running at \$ km/hr in the same direction in .hich the train is going- in 10 seconds. +he speed of the train is/ A. C. #\$ km/hr \$# km/hr B. D. \$0 km/hr \$\$ km/hr
" m/sec. \$0 )ength of the train ' *&peed ( +ime, ' (9 "
Answer & Explanation Answer: %ption B Explanation: &peed of the train relati0e to man ' ' ' \$ \$ m/sec. 1! ( \$ km/hr 1 \$ 10 m/sec
' #\$ km/hr. )et the speed of the train 1e x km/hr. +hen- relati0e speed ' *x 2 \$, km/hr. x 2 \$ ' #\$ x ' \$0 km/hr.
". +he length of the 1ridge- .hich a train 1"0 metres long and tra0elling at #\$ km/hr can
cross in "0 seconds- is/ A. C. 00 m #\$ m B. D. \$m \$0 m
Answer & Explanation Answer: %ption C Explanation: &peed ' #\$ \$ ' ( 1! m/sec \$ m/sec.
+ime ' "0 sec. )et the length of 1ridge 1e x metres. 1"0 3x "0 *1"0 3 x, ' 4\$0 x ' #\$ m.
+hen-
'
#. +.o trains running in opposite directions cross a man standing on the platform in 4 seconds and 14 seconds respecti0el5 and the5 cross each other in " seconds. +he ratio of their speeds is/ A. C. 1/" "/# B. D. "/ 6one of these
Answer & Explanation Answer: %ption B Explanation: )et the speeds of the t.o trains 1e x m/sec and 5 m/sec respecti0el5. +hen- length of the first train ' 4x metresand length of the second train ' 14y metres. 4x 3 14y ' "
x3 y 4x 3 14y ' "x 3 "y #x ' 6y x y ' " .
\$. A train passes a station platform in "6 seconds and a man standing on the platform in 0 seconds. 7f the speed of the train is \$# km/hr- .hat is the length of the platform? A. C. 1 0m "00 m B. D. #0 m 6one of these
Answer & Explanation Answer: %ption B Explanation: &peed ' \$# \$ ( 1! m/sec ' 1\$ m/sec.
)ength of the train ' *1\$ ( 0,m ' "00 m. )et the length of the platform 1e x metres. +henx 3 "00 "6 ' 1\$
x 3 "00 ' \$#0 x ' #0 m.
6. A train #0 m long passes a pole in # seconds. 8o. long .ill it take to pass a platform 6\$0 m long? A. C. 6\$ sec 100 sec B. D. !9 sec
1\$0 sec
Answer: %ption B Explanation: &peed ' #0
# m/sec ' 10 m/sec. #0 3 6\$0 9e:uired time ' 10 sec ' !9 sec. 4. +.o trains of e:ual length are running on parallel lines in the same direction at #6 km/hr and "6 km/hr. +he faster train passes the slo.er train in "6 seconds. +he length of each train is/ A. C. \$0 m B. D. 4 m ! m
!0 m
Answer & Explanation Answer: %ption A Explanation: )et the length of each train 1e x metres. +hen- distance co0ered ' x metres. 9elati0e speed ' *#6 2 "6, km/hr ' ' 10 \$ ( 1! m/sec \$ 9 m/sec x ' \$ "6 9 x ' 100 x ' \$0.
!. A train "60 m long is running at a speed of #\$ km/hr. 7n .hat time .ill it pass a 1ridge 1#0 m long?
A. C.
#0 sec
B. D.
# sec #! sec
#\$ sec
Answer & Explanation Answer: %ption A Explanation: Formula for on!ertin" from #m\$%r to m\$s: +herefore- &peed ' #\$ \$ ' ( 1! m/sec \$ X km/hr ' X( \$ 1! m/s.
m/sec.
+otal distance to 1e co0ered ' *"60 3 1#0, m ' \$00 m. Formula for findin" Time & 9e:uired time ' \$00 ( \$ sec ;istance &peed ' #0 sec.
9. +.o trains are mo0ing in opposite directions < 60 km/hr and 90 km/hr. +heir lengths are 1.10 km and 0.9 km respecti0el5. +he time taken 15 the slo.er train to cross the faster train in seconds is/ A. C. "6 #! B. D. #\$ #9
Answer & Explanation Answer: %ption C Explanation: 9elati0e speed ' *603 90, km/hr ' ' 1\$0 ( 1 \$ " m/sec. \$ 1! m/sec
;istance co0ered ' *1.10 3 0.9, km ' 9e:uired time ' 000 ( "
km ' 000 m.
1 \$ sec ' #! sec.
10. A =ogger running at 9 kmph alongside a rail.a5 track in #0 metres ahead of the engine of a 1 0 metres long train running at #\$ kmph in the same direction. 7n ho. much time .ill the train pass the =ogger? A. C. ".6 sec "6 sec B. D. 1! sec 4 sec
Answer & Explanation Answer: %ption C Explanation: &peed of train relati0e to =ogger ' *#\$ 2 9, km/hr ' "6 km/hr. ' "6 \$ ( 1! m/sec
' 10 m/sec. ;istance to 1e co0ered ' * #0 3 1 0, m ' "60 m. +ime taken ' "60 10 sec ' "6 sec.
11. A 40 metres long train running at the speed of 1 0 kmph crosses another train running in opposite direction at the speed of !0 kmph in 9 seconds. What is the length of the other train? A. C. E. "0 m B. D. #0 m " 0m
60 m 6one of these
Answer: %ption A Explanation: 9elati0e speed ' *1 0 3 !0, km/hr \$ 1! m/sec
' '
00 ( \$00 9
m/sec.
)et the length of the other train 1e x metres. +henx 3 40 9 ' \$00 9
x 3 40 ' \$00 x ' "0.
1 . A goods train runs at the speed of 4 kmph and crosses a \$0 m long platform in 6 seconds. What is the length of the goods train? A. C. "0 m 60 m B. D. #0 m 40 m
Answer: %ption D Explanation: \$ 1! m/sec ' 0 m/sec.
&peed '
4 (
+ime ' 6 sec. )et the length of the train 1e x metres. +henx 3 \$0 6 ' 0
x 3 \$0 ' \$ 0
x ' 40.
1". +.o trains- each 100 m long- mo0ing in opposite directions- cross each other in ! seconds. 7f one is mo0ing t.ice as fast the other- then the speed of the faster train is/ A. C. "0 km/hr 60 km/hr B. D. #\$ km/hr 4\$ km/hr
Answer: %ption C Explanation: )et the speed of the slo.er train 1e x m/sec. +hen- speed of the faster train ' x m/sec.
9elati0e speed ' *x 3 x, m/sec ' "x m/sec. *100 3 100, ! #x ' 00 x' \$ . \$0 " m/sec ' "x
"
&o- speed of the faster train ' ' \$0 " ( 1! \$ km/hr
' 60 km/hr.
1#. +.o trains 1#0 m and 160 m long run at the speed of 60 km/hr and #0 km/hr respecti0el5 in opposite directions on parallel tracks. +he time *in seconds, .hich the5 take to cross each otheris/ A. 9 B. 9.6 10. !
C.
10
D.
9elati0e speed ' *60 3 #0, km/hr '
100 (
'
;istance co0ered in crossing each other ' *1#0 3 160, m ' "00 m. 9 \$0 sec \$# \$ sec ' 10.! sec.
9e:uired time '
"00 (
'
1\$. A train 110 metres long is running .ith a speed of 60 kmph. 7n .hat time .ill it pass a man .ho is running at 6 kmph in the direction opposite to that in .hich the train is going? A. C. \$ sec 4 sec B. D. 6 sec
10 sec
Answer: %ption B Explanation: &peed of train relati0e to man ' *60 3 6, km/hr ' 66 km/hr. \$ 1! m/sec
' '
66 ( \$\$ "
+ime taken to pass the man '
16. A train tra0elling at a speed of 4\$ mph enters a tunnel " miles long. +he train is mile long. 8o. long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
A. C.
B. D.
" min
".\$ min
miles
' 1\$
1\$ #
miles.
+ime taken ' 1
# ( 4\$
hrs
'
hrs 1 0
'
( 60
min.
' " min. View Answer Workspace Report Discuss in Forum 14. A train !00 metres long is running at a speed of 4! km/hr. 7f it crosses a tunnel in 1 minute- then the length of the tunnel *in meters, is/ A. C. 1"0 \$00 B. D. "60 \$#0
&peed '
4! (
m/sec '
m/sec.
+ime ' 1 minute ' 60 seconds.
)et the length of the tunnel 1e x metres. !00 3x 60 "*!00 3 x, ' "900 x ' \$00. View Answer Workspace Report Discuss in Forum 1!. A "00 metre long train crosses a platform in "9 seconds .hile it crosses a signal pole in 1! seconds. What is the length of the platform? A. C. " 0m 6\$0 m B. D. "\$0 m
+hen-
' 6\$ "
&peed '
m/sec '
m/sec.
)et the length of the platform 1e x metres. x 3 "00 "9 \$0 "
+hen-
'
"*x 3 "00, ' 19\$0 x ' "\$0 m. View Answer Workspace Report Discuss in Forum 19. A train speeds past a pole in 1\$ seconds and a platform 100 m long in \$ seconds. 7ts length is/ A. C. \$0 m 00 m B. D. 1\$0 m
Explanation: )et the length of the train 1e x metres and its speed 15 y m/sec. +hen' 1\$ y x 3 100 x ' \$ 1\$ x y' x . 1\$
1\$*x 3 100, ' \$x 1\$x 3 1\$00 ' \$x 1\$00 ' 10x x ' 1\$0 m. View Answer Workspace Report Discuss in Forum 0. A train mo0es past a telegraph post and a 1ridge 6# m long in ! seconds and 0 seconds respecti0el5. What is the speed of the train? A. C. 69.\$ km/hr 49 km/hr B. D. 40 km/hr 49. km/hr
Answer: %ption D Explanation: )et the length of the train 1e x metres and its speed 15 y m/sec. +hen'! y x 3 6# ' 6o.y 0 !y 3 6# ' 0y y' . 1! \$ x x ' !y
&peed '
m/sec '
km/hr ' 49. km/hr.
1. 8o. man5 seconds .ill a \$00 metre long train take to cross a man .alking .ith a speed of " km/hr in the direction of the mo0ing train if the speed of the train is 6" km/hr? A. C. \$ #0 B. D. "0
#\$
Answer: %ption B Explanation: &peed of the train relati0e to man ' *6" 2 ", km/hr ' 60 km/hr ' 60 ( \$ 1! m/sec
'
\$0 "
m/sec.
+ime taken to pass the man
'
\$00 (
" \$0
sec
' "0 sec. View Answer Workspace Report Discuss in Forum . +.o goods train each \$00 m long- are running in opposite directions on parallel tracks. +heir speeds are #\$ km/hr and "0 km/hr respecti0el5. >ind the time taken 15 the slo.er train to pass the dri0er of the faster one. A. 1 sec B. # sec
C.
#! sec
D.
60 sec
Answer: %ption B Explanation: 9elati0e speed ' ' *#\$ 3 "0, km/hr ' 4\$ ( \$ 1! m/sec
'
1 \$ 6
m/sec.
We ha0e to find the time taken 15 the slo.er train to pass the ;97?@9 of the faster train and not the complete train. &o- distance co0ered ' )ength of the slo.er train. +herefore- ;istance co0ered ' \$00 m. 6 1 \$ ' # sec.
9e:uired time '
\$00 (
View Answer Workspace Report Discuss in Forum ". +.o trains are running in opposite directions .ith the same speed. 7f the length of each train is 1 0 metres and the5 cross each other in 1 seconds- then the speed of each train *in km/hr, is/ A. C. 10 "6 B. D. 1! 4
Answer: %ption C Explanation: )et the speed of each train 1e x m/sec. +hen- relati0e speed of the t.o trains ' *1 0 3 1 0, 1 x' 0 x ' 10. 1! \$ x m/sec.
&o- x '
10 (
km/hr ' "6 km/hr.
View Answer Workspace Report Discuss in Forum #. +.o trains of e:ual lengths take 10 seconds and 1\$ seconds respecti0el5 to cross a telegraph post. 7f the length of each train 1e 1 0 metres- in .hat time *in seconds, .ill the5 cross each other tra0elling in opposite direction? A. 10 B. 1
C.
1\$
D.
Answer: %ption B Explanation: 1 0 10 1\$
&peed of the first train ' &peed of the second train '
1 0
sec ' 1 sec.
0 View Answer Workspace Report Discuss in Forum \$. A train 10! m long mo0ing at a speed of \$0 km/hr crosses a train 11 m long coming from opposite direction in 6 seconds. +he speed of the second train is/ A. C. #! km/hr 66 km/hr B. D. \$# km/hr ! km/hr
Answer: %ption D Explanation: )et the speed of the second train 1e x km/hr. 9elati0e speed ' *x 3 \$0, km/hr ' *x 3 \$0, ( \$0 3 \$x 1! \$ 1! m/sec
'
m/sec.
0 m.
\$0 3 \$x 1!
\$0 3 \$x ' 660 x ' ! km/hr.
6. +.o trains are running at #0 km/hr and 0 km/hr respecti0el5 in the same direction. >ast train completel5 passes a man sitting in the slo.er train in \$ seconds. What is the length of the fast train? A. "m 4 4 m 9 B. " m 9
C.
D.
9m
Answer: %ption C Explanation: \$ 1! \$0 9 \$0 9 m.
9elati0e speed ' *#0 2 0, km/hr ' )ength of faster train ' \$0 9 (\$
0( m'
m/sec ' m' 4 4 9
m/sec.
View Answer Workspace Report Discuss in Forum 4. A train o0ertakes t.o persons .ho are .alking in the same direction in .hich the train is going- at the rate of kmph and # kmph and passes them completel5 in 9 and 10 seconds respecti0el5. +he length of the train is/ A. C. #\$ m \$# m B. D. \$0 m
4 m
Explanation: \$ 1! \$ 1! \$ 9 9
( #(
m/sec ' m/sec '
m/sec. m/sec.
10
)et the length of the train 1e x metres and its speed 15 y m/sec. x +heny2 \$ 9 ' 9 and x y2 10 9 ' 10.
9y 2 \$ ' x and 10*9y 2 10, ' 9x 9y 2 x ' \$ and 90y 2 9x ' 100. %n sol0ing- .e get/ x ' \$0. )ength of the train is \$0 m. View Answer Workspace Report Discuss in Forum !. A train o0ertakes t.o persons .alking along a rail.a5 track. +he first one .alks at #.\$ km/hr. +he other one .alks at \$.# km/hr. +he train needs !.# and !.\$ seconds respecti0el5 to o0ertake them. What is the speed of the train if 1oth the persons are .alking in the same direction as the train? A. C. 66 km/hr 4! km/hr B. D. 4 km/hr !1 km/hr
#.\$ ( \$.# (
m/sec ' 1. \$ m/sec- and m/sec ' 1.\$ m/sec.
)et the speed of the train 1e x m/sec. +hen- *x 2 1. \$, ( !.# ' *x 2 1.\$, ( !.\$
.\$ (
km/hr ' !1 km/hr.
View Answer Workspace Report Discuss in Forum 9. A train tra0elling at #! kmph completel5 crosses another train ha0ing half its length and tra0elling in opposite direction at # kmph- in 1 seconds. 7t also passes a rail.a5 platform in #\$ seconds. +he length of the platform is A. C. #00 m B. D. #\$0 m 600 m
\$60 m
Answer: %ption A Explanation: )et the length of the first train 1e x metres. x \$ '1 or " ' "00 x or 1! x ' 00.
Ax 3 *x/ ,B \$
)ength of first train ' 00 m. )et the length of platform 1e y metres. \$ 1! #0 m/sec. "
&peed of the first train ' * 00 3 y, ( " ' #\$ #0 600 3 "y ' 1!00 y ' #00 m.
#! (
m/sec '
"0. +.o stations A and C are 110 km apart on a staright line. %ne train starts from A at 4 a.m. and tra0els to.ards C at 0 kmph. Another train starts from C at ! a.m. and tra0els to.ards A at a speed of \$ kmph. At .hat time .ill the5 meet? A. C. 9 a.m. 10."0 a.m. B. D. 10 a.m.
11 a.m.
Answer: %ption B Explanation: &uppose the5 meet x hours after 4 a.m. ;istance co0ered 15 A in x hours ' 0x km. ;istance co0ered 15 C in *x 2 1, hours ' \$*x 2 1, km. 0x 3 \$*x 2 1, ' 110 #\$x ' 1"\$ x ' ". &o- the5 meet at 10 a.m. "1. +.o- trains- one from 8o.rah to Datna and the other from Datna to 8o.rah- start simultaneousl5. After the5 meet- the trains reach their destinations after 9 hours and 16 hours respecti0el5. +he ratio of their speeds is/ A. C. /" 6/4 B. D. #/"
9 / 16 | HuggingFaceTB/finemath | |
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation1.18 Linear Equations - 2
Examples: 2x + 8 = 14 Then x = ? 2x + 8 = 14 subtracting 8 from both sides 2x + 8 - 8 = 14 - 8 2x = 6 Dividing both sides by 2 2x/2 = 6/2 x = 3 6y - 15 = 2y + 9 Then y = ? 6y - 15 = 2y + 9 subtracting 2y from both sides 6y - 15 - 2y = 2y + 9 - 2y 4y - 15 = 9 Adding 15 both sides 4y - 15 + 15 = 9 + 15 4y = 24 dividing both sides by 4 4y/4 = 24/4 y = 6 4(s + 1) = -2(4 - s) distributive property Note: Negative number multiplied by a negative number is positive. -2 x -s = 2s 4s + 4 = -8 + 2s subtracting 2s both sides 4s + 4 - 2s= -8 + 2s - 2s 2s + 4 = -8 subtracting 4 both sides 2s + 4 - 4 = -8 - 4 2s = - 12 dividing both sides by 2 2s/2 = - 12/2 s = -6 2/x = 3/6 x/2 = 6/3 x/2 = 2 x = 4 t/6 + 2 = t/3 This equation contains fractions so multiply both sides by the LCD of 6 and 3 that is 6 6(t/6 + 2) = 6(t/3) t + 12 = 2t subtracting t from both sides t + 12 - t = 2t - t 12 = t Therefore, t = 12 Directions: Solve for the variables. Also write at least 10 examples of your own.
Q 1: x/2 = 40 then x= ?806042 Q 2: 5x/6 = 50 then x = ?507060 Q 3: (-x/2) = -10202221 Q 4: 2x/0.2 = - 0.8 then x=?-0.80.08-0.08 Q 5: 2(x+5) - 3(x-6) = 207108 Q 6: (-2x) = 20 then x = ?2010-10 Q 7: 2(n+5) = 24 then n = ?967 Q 8: 5(y-1) - 2(y+6) = 2y + 12 then y = ?292725 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | HuggingFaceTB/finemath | |
Convert Millimetre to Inch
Conversion of Millimetre to Inch: Millimetres and inches are used to measure the length in the metric system. A length of 25.4 millimetres will count up to 1 inch. This article will give you the necessary information on how to convert millimetres to inches.
Millimeters
mm
Inches
in
What Is a Millimetre?
A Millimetre is a unit of length that is equal to one-thousandth of a metre in the International System of Standards (SI). It is denoted as ‘mm’ in short.
Millimetre = 1/1000 metres
Common Usage:
• Used to measure the size of the dial of a wristwatch.
• Used to measure the size of a diamond.
• Used to measure the size of a coin.
What Is an Inch?
Inch is a unit of measurement for length and is equal to 25.4 millimetres and 2.54 centimetres. It is denoted as ‘in’ or simply by the symbol “. So,
1 inch = 25.4 millimetres
1 inch = 2.54 centimetres
Common Usage:
• Used to measure the length of the paperclip.
• Used to measure the length of a standard sewing pin.
• Used to measure the length of a standard sewing pin.
How to Convert Millimetre to Inch?
To convert millimetre to inch, multiply the given millimetres by 0.03937 to get the value in inches. As
1 Millimetre = 0.0393701 Inches
The formula to convert millimetres to inches is given below,
X Millimetres = X × 0.03937 Inches
where X is the value in millimetres.
Examples of the Conversion of Millimetre to Inch
Problem 1: Convert 50 millimetres to inches.
Solution 1:
Step 1: The given value is 50 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
50 millimetres = 50 × 0.03937
= 1.9685 in
Therefore, 50 millimetres is 1.9685 inches.
Problem 2: Convert 98.6 millimetres to inches.
Solution 2:
Step 1: The given value is 98.6 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
98.6 millimetres = 98.6 × 0.03937
= 3.881882 in
Therefore, 98.6 millimetres is 3.881882 inches.
Problem 3: Convert 300 millimetres to inches.
Solution 3:
Step 1: The given value is 300 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
300 millimetres = 300 × 0.03937
= 11.811 in
Therefore, 300 millimetres is 11.811 inches.
Problem 4: Convert 1500 millimetres to inches.
Solution 4:
Step 1: The given value is 1500 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
1500 millimetres = 1500 × 0.03937
= 59.055 in
Therefore, 1500 millimetres is 59.055 inches.
Problem 5: Convert 685 millimetres to inches.
Solution 5:
Step 1: The given value is 685 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
685 millimetres = 685 × 0.03937
= 26.96845 in
Therefore, 685 millimetres is 26.96845 inches.
FAQs on the Conversion of Millimetre to Inch
What size is a millimetre?
A millimetre is a unit of length in the International System of Standards (SI), which is equal to one-thousandth of a metre. It is denoted as ‘mm’ in short.
Millimetre = 1/1000 metres
What size is 1 mm in inches?
1 mm is 0.0393701 inches.
Is 20 mm same as 1 inch?
20 mm is 0.787402 inches, which is almost 1 inch.
How much is a mm?
1 Millimetre = 1/1000 metres and 0.0393701 inches.
How does 1 mm compare to 1?
One millimetre is 0.1 centimetre and 1 centimetre is 10 millimetres.
Related conversion posts
We hope this article on the conversion of millimetre to inch is helpful to you. If you have any queries related to this post, ping us through the comment section below and we will get back to you as soon as possible.
References
The SI Base Units
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Parents and students have rated the teaching experience as 4.8/5 and above.
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# Tutorial: How to Do the Sum in Excel
Tutorial: How to Do the Sum in Excel – Hello Softize Friends, Have you ever wanted to quickly add up a range of numbers in Excel? Look no further! Here are some simple steps to follow for doing the sum on Excel.
If you have a range of numbers, simply click on the cell where you want the sum to appear. Then, type in the =SUM formula and include the range of numbers you want to add in parentheses. For example, if you want to add the numbers in cells A1 through A5, your formula should be =SUM(A1:A5). Press enter and voila! Your sum will appear in the selected cell.
Whether you’re a student, accountant, or just someone who frequently uses spreadsheets, learning how to do the sum on Excel can save you time and effort. With this simple formula, you can easily add up a range of numbers without manually entering each individual value.
In summary, doing the sum on Excel is an easy process that can simplify your work. Follow the steps above to quickly add up any range of numbers you need. For more helpful tips on improving your Excel skills, be sure to read our other informative articles.
## Steps of How To Do The Sum On Excel
Read the article below to learn how to do the sum on Excel.
### Introduction to How To Do The Sum On Excel
Summing up a row or column of values in Excel is one of the most basic and commonly used calculations in Excel. The SUM function in Excel makes it easy to calculate this value with just a few clicks. In this article, we will cover the purpose, logic, functions, procedures, case study, order of tasks, and example tasks related to How To Do The Sum On Excel.
### Purpose of How To Do The Sum On Excel
The purpose of How To Do The Sum On Excel is to teach you how to calculate the sum of a range of values in an Excel worksheet. This is an essential skill for anyone who works with data in Excel, as it lays the foundations for more complex mathematical operations that you can perform using other formulas and functions.
More: Tutorial: How to Put Smiley Faces on Pictures
### Basic Logic of How To Do The Sum On Excel
The basic logic of How To Do The Sum On Excel is to add up a range of numbers in a cell. The SUM function is used to perform this task that helps you to calculate the total of a given range of cells. Below is a table that lists the coding of How To Do The Sum On Excel:
Coding Description
=SUM(A1:A4) Adds up the values in cells A1 through A4
=SUM(A1,A2,A3,A4) Adds up the values in cells A1, A2, A3, and A4
### Functions and Procedures of How To Do The Sum On Excel
The SUM function and AutoSum button are used to calculate the sum of a range of cells in Excel. To access the SUM function, type =SUM( followed by the range of cells you want to sum, such as =SUM(A1:A4). You can also use the AutoSum button in the Editing group on the Home tab to quickly add up a row or column of numbers. This function is very useful for those who work with large sets of data in Excel.
### Case Study of How To Do The Sum On Excel
Suppose you have a worksheet that contains a list of expenses for a particular month for your business. You want to calculate the total expenses for the month so that you can make sure they fall within the budget. By using the SUM function of Excel, you can easily get the total of all expenses in just a few seconds.
### Order of Tasks in How To Do The Sum On Excel
The order of tasks in How To Do The Sum On Excel is relatively straightforward:
1. Select the cell where you want the sum to appear.
2. Type the equals sign (=).
3. Type SUM( followed by the range of cells you want to sum (separated by commas) and close the parentheses.)
4. Press Enter, and you’re done!
For example, if you wanted to add up the numbers in cells A1 through A4, you would type =SUM(A1, A2, A3, A4) in the cell where you want the sum to appear and then press Enter.
### Example Tasks of How To Do The Sum On Excel
`1. Click on the cell where you want the result to appear.2. Type =SUM(A1:A4) without quotes and hit enter.`
## Common Coding Errors in Doing Sum On Excel
### 1. Incorrect formula usage
One of the most common errors in doing the sum on Excel is using the wrong formula. Excel has several formulas that can be used for adding numbers, such as SUM, SUMIF, and SUMIFS. Using the wrong formula can lead to incorrect results.
### 2. Including text or blank cells in calculations
If you include text or blank cells in your calculation, Excel will return an error. It’s important to ensure that you only include cells that contain numerical values.When performing calculations in Excel, it’s important to be mindful of these common errors to ensure accurate results.
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## Solutions to Common Coding Errors in Doing Sum On Excel
### 1. Double-check formula usage
Before calculating the sum, ensure that you are using the correct formula. For basic addition, use the SUM formula. If you need to add numbers based on certain criteria, use SUMIF or SUMIFS.
### 2. Use the IFERROR function
To avoid errors caused by including text or blank cells in your calculation, use the IFERROR function. This function allows you to specify what value should be returned if an error occurs.For example, if you want to calculate the sum of a column of numbers, but some cells may contain text or be blank, you can use the following formula:=SUMIF(A1:A10,,A1:A10)This formula will only include cells that are not blank or do not contain text.
### 3. Check for hidden rows or columns
If your sum is returning unexpected results, check for hidden rows or columns that may contain values. You can unhide these rows or columns by selecting the rows or columns surrounding them, right-clicking, and selecting Unhide.Overall, performing calculations in Excel can be a powerful tool, but it’s important to be mindful of common errors to ensure accurate results.
Formula Description
SUM Adds all of the numbers in a range of cells.
SUMIF Adds numbers based on specific criteria.
SUMIFS Adds numbers based on multiple criteria.
By taking these steps and using the correct formulas, you can avoid common errors and get accurate results when doing the sum on Excel.
## Advantages and Disadvantages of How To Do The Sum On Excel
Doing the sum on Excel is a great tool to use when working with large sets of data. It can quickly calculate numbers and provide accurate results that would take a lot of time if done manually. This function is available on all versions of Excel and can be used for simple calculations such as adding up expenses or more complicated ones such as calculating average sales revenue over a period of time.
The disadvantage of using the sum formula in Excel is that it can sometimes be difficult to locate errors within the data. Any mistakes such as incorrect cell range selected or inputting the wrong formula can lead to human error and incorrect calculations. This can be especially problematic when dealing with large data sets.
## Effective How To Do The Sum On Excel Tips
### Basic Summing
The SUM function calculates the total sum of cells located within a specified range. To use this, select the cell where you want the result to appear and enter the formula =SUM(), followed by the range of cells you want to sum up between the parentheses. For example, to add cells A1 through A5, the formula would look like this: =SUM(A1:A5).
### Summing Multiple Rows and Columns
Users can easily sum multiple rows and columns on Excel without the need to create a separate formula for each row. Simply highlight the desired cells, and the sum of the values will appear in the bottom right-hand corner of the highlighted cells.
Alternatively, users can highlight the entire column they wish to sum, and then go to the bottom of the screen and choose the option AutoSum. This will calculate the sum of the entire column and provide the result in the cell directly below the column of numbers.
More: How to Create a Slideshow Using Pictures
Overall, learning how to do the sum on Excel can provide many advantages for working with data more efficiently. With a little practice, users can quickly master the function and utilize it to enhance their spreadsheets.
## Q&A About: Do The Sum On Excel
Here are some common questions and answers about how to do the sum on Excel: | HuggingFaceTB/finemath | |
# ABSTRACT:
The present study aimed to obtain, through statistical methods, the independent variables that influence the economic and productive results of Brazilian beef cattle. Economic and productive information was collected from 106 farms in Brazil. Data collection was performed by the Instituto de Métricas Agropecuárias (Inttegra). The variable selection method used was stepwise regression followed by polynomial regression analyses. The variable beef cattle economic result showed the positive effect of disbursement per head per month and average weight gain. An average daily gain of 520 g in live weight was obtained, and this variable was influenced especially by nutrition cost. The arroba production (arroba is a unit of weight corresponding to 15kg) was influenced by the linear effect of nutrition cost and stocking rate. The variable arroba production in pasture showed a linear and positive effect of average weight gain in pasture and stocking rate in pasture. For profit per arroba, the variables nutrition cost, disbursement with pasture, and average sale price had a linear effect. The independent variables that had the greatest influence on the response variables were stocking rate, average daily weight gain, and nutrition cost. Thus, increases in investment in nutrition and stocking rate should result in higher production rates and improvements in the profitability of this activity.
Key words
nutrition costing; average daily gain; profit; management; stocking rate
# RESUMO:
O presente trabalho teve como objetivo obter, por meio de métodos estatísticos, quais variáveis independentes influenciam as variáveis econômicas e produtivas na bovinocultura de corte brasileira. O banco de dados utilizado foi obtido por coleta de informações econômicas e produtivas em 106 fazendas no Brasil. O método de coleta de dados foi definido pelo Instituto de Métricas Agropecuárias (Inttegra). Foram realizadas análises de seleção de variáveis pelo método Stepwise, seguido por análises de regressão múltipla. A variável resultado econômico da bovinocultura de corte teve efeito positivo do desembolso mensal por cabeça e do ganho de peso médio diário. Para o ganho de peso médio diário foi obtido um valor médio de 520 gramas, e esta variável foi influenciada principalmente pelo custeio com nutrição. Para a produção de arroba foi obtido efeito linear do custeio com nutrição e da taxa de lotação. Para a variável produção de arroba a pasto foi obtido um efeito linear e positivo do ganho médio diário de peso no pasto e da taxa de lotação no pasto. Para o lucro por arroba, as variáveis custeio com nutrição, desembolso com pastagem e preço médio de venda tiveram efeito linear. As variáveis independentes que tiveram maior influência nas variáveis resposta foram a taxa de lotação, o ganho médio diário de peso e o custeio com nutrição. Assim, aumentos no investimento em nutrição e na taxa de lotação dos animais devem resultar em maiores índices de produção e em melhorias na rentabilidade da atividade.
Palavras-chave:
custo nutricional; ganho médio diário; lucro; gestão; taxa de lotação
# INTRODUCTION:
Food production is an extremely important economic factor for the world’s population as it is closely linked to the basic needs of the population. Thus, the beef production chain is very important for the economy, especially in Brazil, where agribusiness contributes significantly to the gross domestic product, and was 21.6% in 2017 (CEPEA, 2018CEPEA - Centro de Estudos Avançados em Economia Aplicada. PIB do agronegócio brasileiro. 2019. Available from: <Available from: https://www.cepea.esalq.usp.br/br/pib-do-agronegocio-brasileiro.aspx >. Accessed: Jan. 13, 2019.
https://www.cepea.esalq.usp.br/br/pib-do...
). Investments in the cattle production chain increase daily, which leads to improvements in the structure of this chain and makes the activity more professional, increasing profitability (LUCHIARI FILHO, 2000LUCHIARI FILHO, A. Pecuária da carne bovina. São Paulo, 2000, 134 p.).
Production systems in beef cattle are generally managed without any systematic processes, which makes it difficult to assess the production costs; and consequently, affect decision making. According to OAIGEN et al. (2008OAIGEN, R.P. et al. Organizational improvement in the beef calves production from of the cost centers. Revista Brasileira de Zootecnia, v.37, p.580-587, 2008. Available from: <Available from: http://dx.doi.org/10.1590/S1516-35982008000300025 >. Accessed: Aug. 03, 2019. doi: 10.1590/S1516-35982008000300025.
http://dx.doi.org/10.1590/S1516-35982008...
), in a beef cattle farm, there should be an incentive to implement business management concepts based on the knowledge of strategic information.
The profitability of beef cattle activity is a factor that depends on different variables, and each has a degree of importance in determining profit or loss. Thus, knowing the importance of these variables can be a crucial point in knowing where to strategically determine changes in management to improve profitability.
The knowledge of production costs is very important to conduct a profitability analysis of farms (DEMEU et al. 2013DEMEU, A.A. et al. Production scale effect on the profitability of beef cattle in a pasture system in the South of Minas Gerais State. Archivos Latinoamericanos de Producción Animal, v.21, p.97-106, 2013. Available from: <Available from: https://pdfs.semanticscholar.org/9ced/ 414e83b97dad9bc31a7e0eb6e5716c445e5a.pdf >. Accessed: Jul. 24, 2020.
https://pdfs.semanticscholar.org/9ced/ 4...
), and the farmer will know in detail the factors of production such as land, labor, and capital. Moreover, according to these authors, with the knowledge of economic variables, it is possible to more easily and accurately diagnose bottlenecks, which allows us to concentrate on managerial and technological inputs, thus presenting greater possibilities of achieving the goals and success in the activity.
Owing to the economic and social importance of beef cattle and the pressure for greater precision to provide improvements in the economic and productive indexes of this activity, the aim of this study was to identify which variables most influence the productive and economic traits, and to determine the relationship between these economic and productive variables.
# MATERIALS AND METHODS:
The data set comprised information from 106 beef cattle farms located in different regions of Brazil: South (n = 13), Southeast (n = 11), Northeast (n = 7), North (n = 17), and Midwest (n = 58), collected from July 2017 to June 2018. The states by region were as follows: North (Tocantins, Rondônia, and Para), Northeast (Bahia, Maranhão, and Rio Grande do Norte), Midwest (Mato Grosso, Mato Grosso do Sul, and Goiás), Southeast (Espírito Santo and São Paulo), and South (Paraná and Rio Grande do Sul). The properties studied had weaning and finishing system, and 55 farms were with and 51 without feedlots. Data collection was designed, performed, and standardized by the Instituto de Métricas Agropecuárias (Inttegra), located in Maringá, Paraná, Brazil. The variables studied were classified into three groups: farm characterization, economic, and productive.
The farm characterization variables were pasture area (PAREA), defined as the size of pasture area in hectares for livestock; herd size (HS), defined by the number of animals raised on the properties during the data collection period; pasture stocking rate (PSR), which is the number of animal units reared in one hectare, calculated as the number of animal units divided by the area under grazing livestock; and stocking rate (SR), which is the number of animal units reared in one hectare, calculated as the number of animal units divided by the area intended for animal production. An animal unit (AU) represents 450 kg of body weight of an animal.
The beef cattle economic result (BCER), measured in dollars per hectare, was obtained by the difference between all expenses and revenue in beef cattle divided by the area in hectares used for this activity. Thus, this variable indicated beef cattle activity profitability by hectare. Profit by animal arroba produced (PA), measured in dollars, was obtained by average sale price per produced arroba less total cost per arroba produced, and indicates the net profit per animal arroba produced. Profit margin on sale (PMS), measured in dollars, was calculated by dividing PA by the average sale price per produced arroba. Average selling price (AVSP) was the average, in dollars, of the sale price of an arroba. In Brazil, the arroba is a common weight measure that equals approximately 15 kg.
The cost of arroba produced (CAP) is a dollar amount that represents the total cost to produce an arroba. Herd inputs (HI), expressed in dollars were obtained from the sum of all expenditures on nutrition, health, reproduction, and pasture per animal over a period of one month. Nutrition cost (NUTC) is a dollar amount that represents the amount spent on nutrition per animal over a one-month period and does not involve labor costs or maintenance of feeders. Pasture disbursement (PASTD) can be defined as a dollar amount that expresses how much was spent on pasture per animal over a period of one month. Disbursement per head per month (DHM) was the amount in dollars spent on one animal over a one-month period on nutrition, breeding, health, permanent labor, management, machine park (cost and investment), pasture (cost and investment), infrastructure (cost and investment), fees, and taxes.
The average daily weight gain (ADG), measured in kg, was obtained by dividing the total weight gain of an animal by the days on which this animal was evaluated. Pasture average daily weight gain (PADG) was measured in kg and was obtained using the same calculation as that used for the ADG, but only for animals raised in the pasture production system.
Arroba production (APROD) was defined as the number of arrobas produced in one hectare per year, accounting for all production systems, and was obtained by dividing the total number of arrobas produced by each farm by the area used for animal production on the farm. In addition, pasture arroba production (PAPROD) was obtained by dividing the total amount of arrobas produced on pasture on the farm by the area used for pasture animal production on the farm.
The response variables studied were ADG, APROD, PAPROD, PA, and BCER. First, Pearson correlations between the response variables and independent variables were calculated. For each response variable, the selected independent variables were chosen as fixed effects when the correlation coefficient between them and the response variables was greater than 0.25. In addition, when the correlation coefficient between the independent variables was greater than 0.70, the choice of the independent variable that remained in the model was obtained according to their biological and/or economic importance. A variable selection analysis was performed using the stepwise method, whereby models were compared considering one complete model and one without a linear predictor. The best model was selected considering the Akaike information criterion (AKAIKE, 1974AKAIKE, H. A new look at the statistical model identification. IEEE transactions on automatic control, v.19, p.716-723, 1974. Available from: <Available from: https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1100705 >. Accessed: Jul. 24, 2020. doi: 10.1109/TAC.1974.1100705.
https://ieeexplore.ieee.org/stamp/stamp....
). Linear and quadratic effects were studied in polynomial regression analysis. The assumptions of regression analysis of normality and homogeneity of variance were assessed using the Shapiro-Wilk test and scatter plots, respectively, and when necessary, outliers were removed. The proportion of explained variance (VE) for each independent variable in relation to the response variable was calculated by dividing the sum of squares of each variable in relation to the total sum of squares. All statistical analyses were performed using the R statistical software (R Core Team, 2019R Core Team (2019). R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. URL https://www.R-project.org/.
https://www.R-project.org/...
), and the significance level was set at 0.05.
# RESULTS AND DISCUSSION:
For the economic variable BCER, the descriptive statistics obtained indicated a large variation in economic return among the farms studied, which varied from a profit of US$388.39 ha per year to a loss of US$ 1110.95 ha per year (Table 1). Nevertheless, among all farms studied, 77 were profitable and 29 showed losses in beef cattle production. These variables exemplify the great variability in yields generated by beef cattle in Brazil. Moreover, these results indicated that some farmers may be using their patrimony to fund this activity; and will possibly no longer continue if there are no changes, such as the professionalization of the activity (LUCHIARI FILHO, 2000LUCHIARI FILHO, A. Pecuária da carne bovina. São Paulo, 2000, 134 p.). Although, most farms studied were profitable, the profit can be much lower than necessary to generate a proper remuneration for cattle farmers (CANOZZI et al. 2019CANOZZI, M.E.A. et al. Typology of beef production systems according to bioeconomic efficiency in the south of Brazil. Ciência Rural, v.49, 2019. Available from: <Available from: http://dx.doi.org/10.1590/0103-8478cr20190030 >. Accessed: Oct. 16, 2019. doi: 10.1590/0103-8478cr20190030.
http://dx.doi.org/10.1590/0103-8478cr201...
).
Table 1
Number of farms evaluated (N), mean, standard deviation (SD), minimum (MIN) and maximum (MAX) values for the economic variables of 106 beef cattle farms in Brazil.
The descriptive statistics obtained for productive variables showed the large variation in the farms studied in relation to the pasture area, which ranged from 25 to 40000 ha, and the number of beef cattle per farm, which ranged from 127 to 47000 animals (Table 2). The production system (with or without of feedlot) of beef cattle and the region in which the farm is located were not as relevant in the analysis as economic and productive variables; therefore, they were only used to describe the data. Regarding the size of the properties and the pasture area, the properties allocated in the South had smaller pasture areas (25 to 536 hectares), whereas those in the North had the largest extensions of pasture areas (441 to 33.217 hectares). According to SAATH & FACHINELLO (2018SAATH, K.C.O.; FACHINELLO, A.L. Crescimento da demanda mundial de alimentos e restrições do fator terra no Brasil. Revista de Economia e Sociologia Rural, v.56, p.195-212, 2018. Available from <Available from http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-20032018000200195 &lng=en&nrm=iso >. Acessed: Jul. 07, 2020. doi: 10.1590/1234-56781806-94790560201.
http://www.scielo.br/scielo.php?script=s...
), with the growing demand for food in Brazil because of population growth, and to meet the increasing demand for agricultural products in Brazil until the year 2024, it is necessary to expand productivity and planted areas. However, the expansion of the agricultural frontier is quite restricted.
Table 2
Number of farms evaluated (N), mean, standard deviation (SD), minimum (MIN) and maximum (MAX) values for the production variables of 106 beef cattle farms in Brazil.
The values obtained for ADG ranged from 150 to 940 g d-1, with an average of 520 g d-1. For PADG, the average value obtained was 460 g d-1. These variables are also indicative of the large difference between the farms studied; however, the variation in weight gain was approximately four times lower than that of the variables indicating size of the property, such as herd size. The ADG, either in feedlots or pastures, is influenced by many factors; however, the biggest differences occur owing to changes in nutritional management. FISCHER et al. (2005FISCHER, V. et al. Increasing levels of energy supplementation on weight gain of grazing steers kept on a natural pasture at the southeastern region of Rio Grande do Sul. Revista Brasileira de Zootecnia, v.34, p.159-166, 2005. Available from: <Available from: http://dx.doi.org/10.1590/S1516-35982005000100020 >. Accessed: Sep. 03, 2019. doi: 10.1590/S1516-35982005000100020.
http://dx.doi.org/10.1590/S1516-35982005...
), in a study to evaluate the effects of different amounts of energy supplementation on animals raised on native pastures in Rio Grande do Sul, obtained ADG values between 360 and 760 g d-1.
A linear effect between ADG and nutrition cost was observed (Table 3), consistent with the findings of FISCHER et al. (2005FISCHER, V. et al. Increasing levels of energy supplementation on weight gain of grazing steers kept on a natural pasture at the southeastern region of Rio Grande do Sul. Revista Brasileira de Zootecnia, v.34, p.159-166, 2005. Available from: <Available from: http://dx.doi.org/10.1590/S1516-35982005000100020 >. Accessed: Sep. 03, 2019. doi: 10.1590/S1516-35982005000100020.
http://dx.doi.org/10.1590/S1516-35982005...
), who showed a linear and positive effect of energy supplementation on ADG. In addition, MARCONDES et al. (2011MARCONDES, M.I. et al. Feed efficiency of pure and cross bred cattles fed high or low concentrate level. Revista Brasileira de Zootecnia, v.40, p.1313-1324, 2011. Available from: <Available from: http://dx.doi.org/10.1590/S1516-35982011000600021 >. Accessed: Sep. 03, 2019. doi: 10.1590/S1516-35982011000600021.
http://dx.doi.org/10.1590/S1516-35982011...
) concluded that the animals that received the largest amount of concentrate presented higher performance in weight gain.
Table 3
Regression analysis for variables ADG, APROD, PAPROD, PA and BCER of 106 beef cattle farms in Brazil.
For APROD, the variables nutrition cost, stocking rate and BCER had a positive linear effect. However, most of the APROD variation was explained by nutrition cost and stocking rate, with 57.33 and 35.95% of variation, respectively, while BCER explained only 0.47%. A coefficient of 0.519 was obtained for the cost of nutrition, i.e., for each dollar used for the cost of nutrition, an increase of 0.519 was expected in the arrobas produced, which is equivalent to 7.77 kg of body weight per animal per hectare. For the stocking rate, the increase of one animal unit per hectare could increase arrobas produced per hectare by 9.69. Thus, if an increase in arrobas production occurs, the profitability of the activity will improve, accompanied by an efficient nutrition cost and a high stocking rate.
For the PAPROD variable, the results indicated a positive linear effect for stocking rate and for PADG. For the stocking rate, which explained 73.36% of the PAPROD variation, the increase of one animal unit per hectare of pasture could lead to an increase in production by 8.8036 arrobas per hectare of pasture per year. For ADG in pasture, the increment of 1 kg of ADG per pasture could result in an increase of 16.3955 arrobas per hectare per year in pasture productivity, and thus, every 100 g of ADG per pasture could obtain 1.93 additional arrobas produced per hectare of pasture per year.
REIS et al. (2009REIS, R.A. et al. Supplementation of beef cattle as strategy of pasture management. Revista Brasileira de Zootecnia, v. 38, p. 147-159, 2009. Available from: <Available from: http://dx.doi.org/10.1590/S1516-35982009001300016 >. Accessed: Aug. 02, 2019. doi: 10.1590/S1516-35982009001300016.
http://dx.doi.org/10.1590/S1516-35982009...
), in a study on the influence of diet supplementation in beef cattle as a strategy for pasture management, obtained results indicating that supplementation of grazing animals may allow increased stocking rate and weight gain in the area. Nevertheless, the average stocking rate obtained in this study was 1.48 AU ha-1, and thus, care should be taken with the use of high stocking rates, especially in native and unimproved pasture areas because it can negatively affect farms with low bioeconomic efficiency (MARQUES et al., 2011MARQUES, P.R. et al. Competitiveness of beef farming in Rio Grande do Sul State, Brazil. Agricultural Systems, v.104, p.689-693, 2011. Available from: <Available from: https://doi.org/10.1016/j.agsy.2011.08.002 >. Accessed: Oct. 25, 2019. doi: 10.1016/j.agsy.2011.08.002.
https://doi.org/10.1016/j.agsy.2011.08.0...
).
For the PA variable, a significant linear effect was obtained for nutrition and pasture expenditure, which explained 20.52 and 11.71% of PA variation, respectively. These results indicated that a correct expenditure on nutrition and pasture can lead to a higher profit per arroba produced similar to that observed with the other variables. For average selling price, a significant linear effect was also observed, and thus, selling the animal for a high added value may mean that the profit per arroba will increase. OAIGEN et al. (2013OAIGEN, R.P. et al. Beef cattle production system competitiveness in the south of Brazil. Archivos de Zootecnia, v.62, p.161-170. 2013. Available from: <Available from: http://dx.doi.org/10.4321/S0004-05922013000200001 >. Accessed: Oct. 24, 2019. doi: 10.4321/S0004-05922013000200001.
http://dx.doi.org/10.4321/S0004-05922013...
) ranked sale price determination as one of the critical factors in beef cattle farm competitiveness, along with other factors such as crop-livestock integration, strategic planning, calculation of financial indicators, access to technological innovations, and organization of the farmers.
For BCER, the independent variables disbursement per head per month and ADG showed significant linear effects, which explained 26.41 and 51.95%, respectively, of the BCER variability. According to the linear regression, an increase in one disbursement dollar per head per month could decrease BCER by US$13.93 in a year. In addition, increasing ADG by 1 kg may result in an increase of US$ 698.34 in BCER. This highlighted the importance of being extremely efficient, increasing earnings, and controlling disbursement. These results indicate that disbursements for nutrition, pasture, and health positively and directly influence the profitability of the farms. In addition, this disbursement is a means of improving animal weight gain, for which a linear effect on the BCER variable was also obtained.
According to CANOZZI et al. (2019CANOZZI, M.E.A. et al. Typology of beef production systems according to bioeconomic efficiency in the south of Brazil. Ciência Rural, v.49, 2019. Available from: <Available from: http://dx.doi.org/10.1590/0103-8478cr20190030 >. Accessed: Oct. 16, 2019. doi: 10.1590/0103-8478cr20190030.
http://dx.doi.org/10.1590/0103-8478cr201...
), who evaluated the level of bioeconomic efficiency of cattle farmers in the western frontier region of the state of Rio Grande do Sul, Brazil, through typology, the main differences among the farms are because of the use of technologies related to feeding and cost management. Nevertheless, these authors reported that in farms with high bioeconomic efficiency, investments to increase productivity, especially in nutrition, do not always ensure better economic results. Thus, an economic evaluation is necessary before implementing a new technology.
The PA and BCER variables were influenced by nutrition and pasture expenditure. An economic feasibility analysis between two beef cattle production systems classified as extensive (lower disbursement) and semi-intensive full cycle was performed by NASCIMENTO et al. (2017NASCIMENTO, M.F.A. et al. Economical viability of two systems of beef cattle production. Espacios, v.38, p.10, 2017. Available from: <Available from: http://www.revistaespacios.com/a17v38n27/ a17v38n27p10.pdf >. Accessed: Oct. 24, 2019.
http://www.revistaespacios.com/a17v38n27...
), and the values obtained for net margin per hectare were R$582.26 for the extensive system and R$ 993.99 for the semi-intensive system. Thus, the extensive system was considered less viable.
However, the profit from grazing cattle production varies greatly according to the forage used and its management. SOARES et al. (2015SOARES, J.C.R. et al. Economic evaluation of finishing beef cattle on irrigated pasture. Arquivo Brasileiro de Medicina Veterinária e Zootecnia, v.67, p.1096-1104, 2015. Available from: <Available from: http://dx.doi.org/10.1590/1678-4162-7340 >. Accessed: oct. 24, 2019. doi: 10.1590/1678-4162-7340.
http://dx.doi.org/10.1590/1678-4162-7340...
), studying the profitability of finishing beef cattle on irrigated pastures, obtained a profit of R\$ -583.18, 291.61, and -28.99 per hectare for different pasture systems. They concluded that pasture irrigation for termination of beef cattle is economically viable; however, this system should be used considering animal productivity owing to its high cost.
Analyzing commercial fed steers and heifers in the United States, TATUM et al. (2012TATUM, J.D. et al. Carcass-based measures of cattle performance and feeding profitability. The Professional Animal Scientist, v.28, p.173-183, 2012. Available from: <Available from: https://doi.org/ 10.15232/S1080-7446(15)30338-7 >. Accessed: Aug. 20, 2019. doi: 10.15232/S1080-7446(15) 30338-7.
https://doi.org/ 10.15232/S1080-7446(15)...
) reported that carcass-based G:F, obtained from the relationship between carcass weight gain (G) and the average amount of food ingested by the animal (F), was the most important contributor to differences in net return per animal, which is consistent with the results obtained in the present study.
# CONCLUSION:
The cost of nutrition, stocking rate, and average daily weight gain alters the system’s profitability. It is also concluded that investments in beef cattle raising increased the performance and profit of farms.
# ACKNOWLEDGMENTS
The authors are grateful to the Instituto de Métricas Agropecuárias (Inttegra), for providing the data set and support in this research; and for the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior (CAPES), for financed in part this study (Finance code 001), and for the post-doctoral scholarship of the second author.
# REFERENCES
• CR-2019-0841.R1
# Publication Dates
• Publication in this collection
12 Feb 2021
• Date of issue
2021 | open-web-math/open-web-math | |
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics2.20 Representative Ratio and Representative Fraction
Directions: Find the missing numbers. (Hint: R.R. stands for Representative Ratio and R.F. stands for Representative Fraction)
Q 1: If R.F. = 1/1,00,000 then 1 mm = _____ km.0.1110 Q 2: If R.F. = 1/10,000 then 1 cm = _____ m.1001010.0001 Q 3: If R.F. = 1/2,50,000 then 1 mm = _____ km.2.50.2525 Q 4: If R.F. = 1/50,000 then 1 cm = _____ km.0.5505 Q 5: If R.R. is 1:90,000 then 1 cm = _____ km.9090.9 Q 6: If R.R. is 1:10,00,000 the 1 mm = _____ km.0.1101 Q 7: If R.F. = 1/5,000 then 1 cm = _____m.500.0055005 Q 8: If R.F. = 1/3,00,000 then 1 mm = _____ km.0.3303 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | HuggingFaceTB/finemath | |
# What is the compound interest on $16009 at 29% over 18 years? If you want to invest$16,009 over 18 years, and you expect it will earn 29.00% in annual interest, your investment will have grown to become $1,566,675.28. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$16,009 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 18 years, but it could be monthly, weekly, daily, or even continuously compounding.
The formula for calculating compound interest is:
$$A = P(1 + \dfrac{r}{n})^{nt}$$
• A is the amount of money after the compounding periods
• P is the principal amount
• r is the annual interest rate
• n is the number of compounding periods per year
• t is the number of years
We can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.
For this formula, we need to convert the rate, 29.00% into a decimal, which would be 0.29.
$$A = 16009(1 + \dfrac{ 0.29 }{1})^{ 18}$$
As you can see, we are ignoring the n when calculating this to the power of 18 because our example is for annual compounding, or one period per year, so 18 × 1 = 18.
## How the compound interest on $16,009 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 18 years it is compounding: Start Balance Interest End Balance 1$16,009.00 $4,642.61$20,651.61
2 $20,651.61$5,988.97 $26,640.58 3$26,640.58 $7,725.77$34,366.34
4 $34,366.34$9,966.24 $44,332.58 5$44,332.58 $12,856.45$57,189.03
6 $57,189.03$16,584.82 $73,773.85 7$73,773.85 $21,394.42$95,168.27
8 $95,168.27$27,598.80 $122,767.07 9$122,767.07 $35,602.45$158,369.52
10 $158,369.52$45,927.16 $204,296.68 11$204,296.68 $59,246.04$263,542.72
12 $263,542.72$76,427.39 $339,970.10 13$339,970.10 $98,591.33$438,561.43
14 $438,561.43$127,182.82 $565,744.25 15$565,744.25 $164,065.83$729,810.08
16 $729,810.08$211,644.92 $941,455.01 17$941,455.01 $273,021.95$1,214,476.96
18 $1,214,476.96$352,198.32 $1,566,675.28 We can also display this data on a chart to show you how the compounding increases with each compounding period. As you can see if you view the compounding chart for$16,009 at 29.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.
## How long would it take to double $16,009 at 29% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$16,009 assuming an interest rate of 29.00%.
We can calculate this very approximately using the Rule of 72.
The formula for this is very simple:
$$Years = \dfrac{72}{Interest\: Rate}$$
By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:
$$Years = \dfrac{72}{ 29 } = 2.48$$
Using this, we know that any amount we invest at 29.00% would double itself in approximately 2.48 years. So $16,009 would be worth$32,018 in ~2.48 years.
We can also calculate the exact length of time it will take to double an amount at 29.00% using a slightly more complex formula:
$$Years = \dfrac{log(2)}{log(1 + 0.29)} = 2.72\; years$$
Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.
As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.
Hopefully, this article has helped you to understand the compound interest you might achieve from investing \$16,009 at 29.00% over a 18 year investment period. | HuggingFaceTB/finemath | |
### Determinants
#### Recognise the concept and properties of determinants of order 2 and order 3
• [TheoryMissing]
• Consider the system of equations $$(*)\;\left\{ \begin{array}{l}(k - 1)x + y + z = 1\\x + (k - 1)y + z = k\\x + y + (k - 1)z = {k^2}\end{array} \right.$$, where k is a real number.
(a) Find the range of values of k such that (*) has a unique solution.
(b) Solve (*) when it has a unique solution.
(c) Suppose that $$k = 2$$. Solve (*).
• (a) ∵ The system of equations has a unique solution.
∴ $$\begin{array}{1}\left| {\,\begin{array}{*{20}{c}}{k - 1}&1&1\\1&{k - 1}&1\\1&1&{k - 1}\end{array}\,} \right| \ne 0\\ \left| {\,\begin{array}{*{20}{c}}{k + 1}&1&1\\{k + 1}&{k - 1}&1\\{k + 1}&1&{k - 1}\end{array}\,} \right| \ne {0^{^{^{^{^{^{^{^{^{}}}}}}}}}}\\ (k + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{k - 1}&1\\1&1&{k - 1}\end{array}\,} \right| \ne {0^{^{^{^{^{^{^{^{^{}}}}}}}}}}\\ (k + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\0&{k - 2}&0\\0&0&{k - 2}\end{array}\,} \right| \ne {0^{^{^{^{^{^{^{^{^{}}}}}}}}}}\\ (k + 1)\,\left| {\,\begin{array}{*{20}{c}}{k - 2}&0\\0&{k - 2}\end{array}\,} \right| \ne {0^{^{^{^{^{^{}}}}}}}\\(k + 1){(k - 2)^2} \ne {0^{^{^{}}}}\end{array}$$
$$\underline{\underline {k \ne - 1}}$$ and $$\underline{\underline {k \ne 2}}$$
(b) $$\begin{array}{1}D = \left| {\,\begin{array}{*{20}{c}}{k - 1}&1&1\\1&{k - 1}&1\\1&1&{k - 1}\end{array}\,} \right|\\ = (k + 1){(k - 2)^2}^{^{}}\end{array}$$
$$\begin{array}{1}{D_x} = \left| {\,\begin{array}{*{20}{c}}1&1&1\\k&{k - 1}&1\\{{k^2}}&1&{k - 1}\end{array}\,} \right|\\ = \left| {\,\begin{array}{*{20}{c}}1&0&0\\k&{ - {\rm{ }}1}&{1 - k}\\{{k^2}}&{1 - {k^2}}&{k - 1 - {k^2}}\end{array}\,} \right|\\ = \left| {\,\begin{array}{*{20}{c}}{ - {\rm{ }}1}&{1 - k}\\{1 - {k^2}}&{k - 1 - {k^2}}\end{array}\,} \right|\\ = {k^2}(2 - k)\end{array}$$
$$\begin{array}{1}{D_y} = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&1\\1&k&1\\1&{{k^2}}&{k - 1}\end{array}\,} \right|\\ = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&1\\1&k&1\\0&{{k^2} - k}&{k - 2}\end{array}\,} \right|\\ = (k - 1)\,\left| {\,\begin{array}{*{20}{1}}k&1\\{{k^2} - k}&{k - 2}\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{1}}1&1\\{{k^2} - k}&{k - 2}\end{array}\,} \right|\\ = - k + 2\end{array}$$
$$\begin{array}{1}{D_z} = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&1\\1&{k - 1}&k\\1&1&{{k^2}}\end{array}\,} \right|\\ = \left| {\,\begin{array}{*{20}{1}}{k - 1}&1&0\\1&{k - 1}&1\\1&1&{{k^2} - 1}\end{array}\,} \right|\\ = - {\rm{ }}\left| {\,\begin{array}{*{20}{1}}{k - 1}&1\\1&1\end{array}\,} \right| + ({k^2} - 1)\,\left| {\,\begin{array}{*{20}{1}}{k - 1}&1\\1&{k - 1}\end{array}\,} \right|\\ = (k - 2)({k^3} - k - 1)\end{array}$$
$$\frac{{{D_x}}}{D} = \frac{{{k^2}(2 - k)}}{{(k + 1){{(k - 2)}^2}}} = \frac{{ - {\rm{ }}{k^2}}}{{(k + 1)(k - 2)}}$$,
$$\frac{{{D_y}}}{D} = \frac{{ - {\rm{ }}k + 2}}{{(k + 1){{(k - 2)}^2}}} = \frac{{ - {\rm{ }}1}}{{(k + 1)(k - 2)}}$$,
$$\frac{{{D_z}}}{D} = \frac{{(k - 2)({k^3} - k - 1)}}{{(k + 1){{(k - 2)}^2}}} = \frac{{{k^3} - k - 1}}{{(k + 1)(k - 2)}}$$
∴ The solution of the system is $$x = \frac{{ - {\rm{ }}{k^2}}}{{(k + 1)(k - 2)}},{\rm{ }}y = \frac{{ - {\rm{ }}1}}{{(k + 1)(k - 2)}},{\rm{ }}z = \frac{{{k^3} - k - 1}}{{(k + 1)(k - 2)}}$$.
(c) When $$k = 2$$,
$$\left\{ \begin{array}{l}x + y + z = 1\;...............\;(1)\\x + y + z = 2\;..............\;(2)\\x + y + z = 4\;..............\;(3)\end{array} \right.$$
(1), (2) and (3) cannot hold simultaneously.
∴ The system (*) has no solutions. | HuggingFaceTB/finemath | |
# Homework Help: Electric Field Involving Gauss's Law Problem
1. Feb 5, 2010
### king_abu
1. The problem statement, all variables and given/known data
An atom that has Z protons, each with positive charge of magnitude e, reside in the nucleus at the center, surrounded by a uniformly distributed spherical cloud of radius R consisting of Z electrons, each with negative charge magnitude e, centered on the nucleus e. Use Gauss's Law to determine an expression for the electric field as a function of r, the distance from the center of the atom.
(1) Find the value of electric field for r<R (magnitude + direction)
(2) Find the value of electric field for R>r (mag + dir)
(3) If the atom were ionized, such that it was deficient by one electron, what would be the new expressions for the regions r<R and r>R
2. Relevant equations
Gauss's Law: Integration of E dA = Q/eo
E = kq1q2/r^2
3. The attempt at a solution
(1-2) I treated this problem as a hollow sphere
The electric field is zero inside a conducting sphere.
The electric field outside the sphere is given by: E = kQ/r2, just like a point charge.
(3) This one I have no idea......
2. Feb 5, 2010
### xcvxcvvc
A conducting sphere alone has an electric field of zero within it. However, do you think there would be an electric field in a conducting sphere if there were charged particles inside of it - say, a positive nucleus with Z protons?
Think of the nucleus as a point charge meaning regardless of how small your Gauss surface is, you still have the z*e Coulombs enclosed, so the E field is not zero within the electron cloud.
Your second answer seems wrong, also. Once you've drawn a Gauss surface that encloses both the electron cloud and the nucleus, what does Q enclosed become?
3. Feb 7, 2010
### sweetdion
So for part one (r<R) would u use E=Q/4PiE0r^2 to give you that the electric field=Ze/4PiE0r^2
4. Feb 7, 2010
### sweetdion
Or would it be Integration EdA=Qenc/E0 and dA=2Pi*rL and Qenc=Ze so E=Ze/2Pi*rLE0
5. Feb 7, 2010
### xcvxcvvc
yes that's right
I'm unsure what rL is and i'm unsure how you got the integral to equal that.
you can integrate to get the same answer as above. Your E field is constant at all points on your surface, so your integral becomes:
$$E\int_A dA=\frac{Q_{enc}}{\epsilon_o}$$
That integral, I hope you see, becomes the total area of your Gauss surface. What is your surface? It is a sphere, so it becomes the surface area of a sphere:
$$E(4\pi r^2)=\frac{Q_{enc}}{\epsilon_o}$$
$$E=\frac{Q_{enc}}{4\pi\epsilon_or^2}$$
6. Feb 7, 2010
### sweetdion
Thank you very much.
As for part B for r>R if it encloses both the electron cloud and the nucleus cloud then Qenc=0 therefore the Electric field=0?
And for Part C: If it were deficient by one electron, does that mean the overall charge would be positive since there will be more protons then electrons so
r<R (when deficient by one electron) I'm a little bit confused on this one.
Now there are Z-1 electrons. Would you just plug that into the equation that we got in part a to give E=(Z-1)e/4Pi*E0r^2
r>R (when deficient by one electron)
Qenc=+1?? so would i just plug that in for the electric field?
7. Feb 7, 2010
### xcvxcvvc
You are right about part B. For part C when r < R, it's the same problem as before! Does $$Q_{enc}$$ change? No, so the E-field doesn't either!
For when r > R, $$Q_{enc}$$ becomes +e like you say. Just solve the integral the same way I solved it for part A
8. Feb 7, 2010
### sweetdion
So let me get this right,
For when r>R (with one electron deficient), I solved the integral to get E=Z*+e/4Pi*E0r^2
correct? the only thing that changes is the Qenc. The Qenc becomes Z*+e
9. Feb 8, 2010
### xcvxcvvc
No, Qenc becomes +1e
If there are Z - 1 electrons and Z protons, you get:
Z(e) + (Z -1)(-e)
Ze + (e - Ze)
e
So the total charge enclosed is 1 electron's worth but positive. | HuggingFaceTB/finemath | |
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# 8.2: The Ellipse
• • OpenStax
• OpenStax
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###### Learning Objectives
• Write equations of ellipses in standard form.
• Graph ellipses centered at the origin.
• Graph ellipses not centered at the origin.
• Solve applied problems involving ellipses.
Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in Figure $$\PageIndex{1}$$, is such a room. It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper. Figure $$\PageIndex{1}$$: The National Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr)
## Writing Equations of Ellipses in Standard Form
A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in Figure $$\PageIndex{2}$$. Figure $$\PageIndex{2}$$
Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points $$(x,y)$$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus(plural: foci).
We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See Figure $$\PageIndex{3}$$. Figure $$\PageIndex{3}$$
Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci (Figure $$\PageIndex{4}$$). Figure $$\PageIndex{4}$$
In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the $$x$$- and $$y$$-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.
To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.
### Deriving the Equation of an Ellipse Centered at the Origin
To derive the equation of an ellipse centered at the origin, we begin with the foci $$(−c,0)$$ and $$(c,0)$$. The ellipse is the set of all points $$(x,y)$$ such that the sum of the distances from $$(x,y)$$ to the foci is constant, as shown in Figure $$\PageIndex{5}$$. Figure $$\PageIndex{5}$$
If $$(a,0)$$ is a vertex of the ellipse, the distance from $$(−c,0)$$ to $$(a,0)$$ is $$a−(−c)=a+c$$. The distance from $$(c,0)$$ to $$(a,0)$$ is $$a−c$$. The sum of the distances from the foci to the vertex is
$$(a+c)+(a−c)=2a$$
If $$(x,y)$$ is a point on the ellipse, then we can define the following variables:
• $$d_1=$$ the distance from $$(−c,0)$$ to $$(x,y)$$
• $$d_2=$$ the distance from $$(c,0)$$ to $$(x,y)$$
By the definition of an ellipse, $$d_1+d_2$$ is constant for any point $$(x,y)$$ on the ellipse. We know that the sum of these distances is $$2a$$ for the vertex $$(a,0)$$. It follows that $$d_1+d_2=2a$$ for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic.
\begin{align*} d_1+d_2&= 2a\\ \sqrt{{(x-(-c))}^2+{(y-0)}^2}+\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\qquad \text{Distance formula}\\ \sqrt{{(x+c)}^2+y^2}+\sqrt{{(x-c)}^2+y^2}&=2a\qquad \text{Simplify expressions.}\\ \sqrt{{(x+c)}^2+y^2}&=2a-\sqrt{{(x-c)}^2+y^2}\qquad \text{Move radical to opposite side.}\\ {(x+c)}^2+y^2&={\left[2a-\sqrt{{(x-c)}^2+y^2}\right]}^2\qquad \text{Square both sides.}\\ x^2+2cx+c^2+y^2&=4a^2-4a\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\qquad \text{Expand the squares.}\\ x^2+2cx+c^2+y^2&=4a^2-4a\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\qquad \text{Expand remaining squares.}\\ 2cx&=4a^2-4a\sqrt{{(x-c)}^2+y^2}-2cx\qquad \text{Combine like terms.}\\ 4cx-4a^2&=-4a\sqrt{{(x-c)}^2+y^2}\qquad \text{Isolate the radical.}\\ cx-a^2&=-a\sqrt{{(x-c)}^2+y^2}\qquad \text{Divide by 4.}\\ {\left[ cx-a^2\right]}^2&=a^2{\left[ \sqrt{{(x-c)}^2+y^2}\right] }^2\qquad \text{Square both sides.}\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\qquad \text{Expand the squares.}\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\qquad \text{Distribute } a^2\\ a^2x^2-c^2x^2+a^2y^2&=a^4-a^2c^2\qquad \text{Rewrite.}\\ x^2(a^2-c^2)+a^2y^2&=a^2(a^2-c^2)\qquad \text{Factor common terms.}\\ x^2b^2+a^2y^2&=a^2b^2\qquad \text{Set } b^2=a^2-c^2\\ \dfrac{x^2b^2}{a^2b^2}+\dfrac{a^2y^2}{a^2b^2}&=\dfrac{a^2b^2}{a^2b^2}\qquad \text{Divide both sides by } a^2b^2\\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}&=1\qquad \text{Simplify} \end{align*}
Thus, the standard equation of an ellipse is $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$.This equation defines an ellipse centered at the origin. If $$a>b$$,the ellipse is stretched further in the horizontal direction, and if $$b>a$$, the ellipse is stretched further in the vertical direction.
### Writing Equations of Ellipses Centered at the Origin in Standard Form
Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena.
The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse.
###### STANDARD FORMS OF THE EQUATION OF AN ELLIPSE WITH CENTER $$(0,0)$$
The standard form of the equation of an ellipse with center $$(0,0)$$ and major axis on the $$x$$-axis is
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(\pm a,0)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(0,\pm b)$$
• the coordinates of the foci are $$(\pm c,0)$$, where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{6a}$$.
The standard form of the equation of an ellipse with center $$(0,0)$$ and major axis on the $$y$$-axis is
$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(0,\pm a)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(\pm b,0)$$
• the coordinates of the foci are $$(0,\pm c)$$, where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{6b}$$.
Note that the vertices, co-vertices, and foci are related by the equation $$c^2=a^2−b^2$$. When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form. Figure $$\PageIndex{6}$$: (a) Horizontal ellipse with center $$(0,0)$$ (b) Vertical ellipse with center $$(0,0)$$
###### How to: Given the vertices and foci of an ellipse centered at the origin, write its equation in standard form
1. Determine whether the major axis lies on the x- or y-axis.
• If the given coordinates of the vertices and foci have the form $$(\pm a,0)$$ and $$(\pm c,0)$$ respectively, then the major axis is the x-axis. Use the standard form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
• If the given coordinates of the vertices and foci have the form $$(0,\pm a)$$ and $$(\pm c,0)$$,respectively, then the major axis is the y-axis. Use the standard form $$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$
2. Use the equation $$c^2=a^2−b^2$$, along with the given coordinates of the vertices and foci, to solve for $$b^2$$.
3. Substitute the values for $$a^2$$ and $$b^2$$ into the standard form of the equation determined in Step 1.
###### Example $$\PageIndex{1}$$: Writing the Equation of an Ellipse Centered at the Origin in Standard Form
What is the standard form equation of the ellipse that has vertices $$(\pm 8,0)$$ and foci $$(\pm 5,0)$$?
Solution
The foci are on the $$x$$-axis, so the major axis is the $$x$$-axis. Thus, the equation will have the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
The vertices are $$(\pm 8,0)$$,so $$a=8$$ and $$a^2=64$$.
The foci are $$(\pm 5,0)$$,so $$c=5$$ and $$c^2=25$$.
We know that the vertices and foci are related by the equation $$c^2=a^2−b^2$$. Solving for $$b^2$$, we have:
\begin{align*} c^2&=a^2-b^2\\ 25&=64-b^2\qquad \text{Substitute for } c^2 \text{ and } a^2\\ b^2&=39\qquad \text{Solve for } b^2 \end{align*}
Now we need only substitute $$a^2=64$$ and $$b^2=39$$ into the standard form of the equation. The equation of the ellipse is $$\dfrac{x^2}{64}+\dfrac{y^2}{39}=1$$.
###### Exercise $$\PageIndex{1}$$
What is the standard form equation of the ellipse that has vertices $$(0,\pm 4)$$ and foci $$(0,\pm \sqrt{15})$$?
$$x^2+\dfrac{y^2}{16}=1$$
###### Q&A
Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex?
Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form $$(\pm a,0)$$ or $$(0, \pm a)$$. Similarly, the coordinates of the foci will always have the form $$(\pm c,0)$$ or $$(0, \pm c)$$. Knowing this, we can use $$a$$ and $$c$$ from the given points, along with the equation $$c^2=a^2−b^2$$, to find $$b^2$$.
### Writing Equations of Ellipses Not Centered at the Origin
Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated $$h$$ units horizontally and $$k$$ units vertically, the center of the ellipse will be $$(h,k)$$. This translation results in the standard form of the equation we saw previously, with $$x$$ replaced by $$(x−h)$$ and y replaced by $$(y−k)$$.
###### STANDARD FORMS OF THE EQUATION OF AN ELLIPSE WITH CENTER $$(H, K)$$
The standard form of the equation of an ellipse with center $$(h, k)$$ and major axis parallel to the $$x$$-axis is
$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(h\pm a,k)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(h,k\pm b)$$
• the coordinates of the foci are $$(h\pm c,k)$$,where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{7a}$$.
The standard form of the equation of an ellipse with center $$(h,k)$$ and major axis parallel to the $$y$$-axis is
$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$
where
• $$a>b$$
• the length of the major axis is $$2a$$
• the coordinates of the vertices are $$(h,k\pm a)$$
• the length of the minor axis is $$2b$$
• the coordinates of the co-vertices are $$(h\pm b,k)$$
• the coordinates of the foci are $$(h,k\pm c)$$, where $$c^2=a^2−b^2$$. See Figure $$\PageIndex{7b}$$.
Just as with ellipses centered at the origin, ellipses that are centered at a point $$(h,k)$$ have vertices, co-vertices, and foci that are related by the equation $$c^2=a^2−b^2$$. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given. Figure $$\PageIndex{7}$$: (a) Horizontal ellipse with center $$(h,k)$$ (b) Vertical ellipse with center $$(h,k)$$
###### How to: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form
1. Determine whether the major axis is parallel to the $$x$$- or $$y$$-axis.
• If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the $$x$$-axis. Use the standard form $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$
• If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$
2. Identify the center of the ellipse $$(h,k)$$ using the midpoint formula and the given coordinates for the vertices.
3. Find $$a^2$$ by solving for the length of the major axis, $$2a$$, which is the distance between the given vertices.
4. Find $$c^2$$ using $$h$$ and $$k$$, found in Step 2, along with the given coordinates for the foci.
5. Solve for $$b^2$$ using the equation $$c^2=a^2−b^2$$.
6. Substitute the values for $$h$$, $$k$$, $$a^2$$, and $$b^2$$ into the standard form of the equation determined in Step 1.
###### Example $$\PageIndex{2}$$: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin
What is the standard form equation of the ellipse that has vertices $$(−2,−8)$$ and $$(−2,2)$$ and foci $$(−2,−7)$$ and $$(−2,1)$$?
Solution
The $$x$$-coordinates of the vertices and foci are the same, so the major axis is parallel to the $$y$$-axis. Thus, the equation of the ellipse will have the form
$$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1 \nonumber$$
First, we identify the center, $$(h,k)$$. The center is halfway between the vertices, $$(−2,−8)$$ and $$(−2,2)$$. Applying the midpoint formula, we have:
\begin{align} (h,k) &=\left(\dfrac{−2+(−2)}{2},\dfrac{−8+2}{2}\right) \nonumber \\ &=(−2,−3) \nonumber \end{align} \nonumber
Next, we find $$a^2$$. The length of the major axis, $$2a$$, is bounded by the vertices. We solve for $$a$$ by finding the distance between the y-coordinates of the vertices.
\begin{align} 2a &=2−(−8) \nonumber \\ 2a &=10 \nonumber\\ a&=5 \nonumber \end{align} \nonumber
So $$a^2=25$$.
Now we find $$c^2$$. The foci are given by $$(h,k\pm c)$$. So, $$(h,k−c)=(−2,−7)$$ and $$(h,k+c)=(−2,1)$$. We substitute $$k=−3$$ using either of these points to solve for $$c$$.
\begin{align} k+c &=1 \nonumber \\ −3+c&=1 \nonumber \\ c&=4 \nonumber \end{align} \nonumber
So $$c^2=16$$.
Next, we solve for $$b^2$$ using the equation $$c^2=a^2−b^2$$.
\begin{align} c^2&=a^2−b^2 \nonumber \\ 16&=25−b^2 \nonumber \\ b^2&=9 \nonumber \end{align} \nonumber
Finally, we substitute the values found for $$h$$, $$k$$, $$a^2$$, and $$b^2$$ into the standard form equation for an ellipse:
$\dfrac{{(x+2)}^2}{9}+\dfrac{{(y+3)}^2}{25}=1 \nonumber$
###### Exercise $$\PageIndex{2}$$
What is the standard form equation of the ellipse that has vertices $$(−3,3)$$ and $$(5,3)$$ and foci $$(1−2\sqrt{3},3)$$ and $$(1+2\sqrt{3},3)$$?
$$\dfrac{{(x−1)}^2}{16}+\dfrac{{(y−3)}^2}{4}=1 \nonumber$$
## Graphing Ellipses Centered at the Origin
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, a>b$$ for horizontal ellipses
and
$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1, a>b$$ for vertical ellipses
###### How to: Given the standard form of an equation for an ellipse centered at $$(0, 0)$$, sketch the graph.
1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
• If the equation is in the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, where $$a>b$$, then
• the major axis is the $$x$$-axis
• the coordinates of the vertices are $$(\pm a,0)$$
• the coordinates of the co-vertices are $$(0,\pm b)$$
• the coordinates of the foci are $$(\pm c,0)$$
• If the equation is in the form $$x^2b^2+y^2a^2=1$$,where $$a>b$$, then
• the major axis is the $$y$$-axis
• the coordinates of the vertices are $$(0,\pm a)$$
• the coordinates of the co-vertices are $$(\pm b,0)$$
• the coordinates of the foci are $$(0,\pm c)$$
2. Solve for $$c$$ using the equation $$c^2=a^2−b^2$$.
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
###### Example $$\PageIndex{3}$$: Graphing an Ellipse Centered at the Origin
Graph the ellipse given by the equation, $$\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$$. Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because $$25>9$$,the major axis is on the $$y$$-axis. Therefore, the equation is in the form $$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$,where $$b^2=9$$ and $$a^2=25$$. It follows that:
• the center of the ellipse is $$(0,0)$$
• the coordinates of the vertices are $$(0,\pm a)=(0,\pm \sqrt{25})=(0,\pm 5)$$
• the coordinates of the co-vertices are $$(\pm b,0)=(\pm 9,0)=(\pm 3,0)$$
• the coordinates of the foci are $$(0,\pm c)$$, where $$c^2=a^2−b^2$$ Solving for $$c$$, we have:
\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\ &=\pm \sqrt{25−9} \nonumber\\ &=\pm \sqrt{16} \nonumber\\ &=\pm 4 \nonumber \end{align} \nonumber
Therefore, the coordinates of the foci are $$(0,\pm 4)$$.
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure $$\PageIndex{8}$$. Figure $$\PageIndex{8}$$
###### Exercise $$\PageIndex{3}$$
Graph the ellipse given by the equation $$\dfrac{x^2}{36}+\dfrac{y^2}{4}=1$$. Identify and label the center, vertices, co-vertices, and foci.
center: $$(0,0)$$; vertices: $$(\pm 6,0)$$; co-vertices: $$(0,\pm 2)$$; foci: $$(\pm 4\sqrt{2},0)$$ Figure $$\PageIndex{9}$$
###### Example $$\PageIndex{4}$$: Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form
Graph the ellipse given by the equation $$4x^2+25y^2=100$$. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Solution
First, use algebra to rewrite the equation in standard form.
\begin{align} 4x^2+25y^2&=100 \nonumber \\ \dfrac{4x^2}{100}+\dfrac{25y^2}{100}&=\dfrac{100}{100} \nonumber \\ \dfrac{x^2}{25}+\dfrac{y^2}{4}&=1 \nonumber \end{align} \nonumber
Next, we determine the position of the major axis. Because $$25>4$$, the major axis is on the $$x$$-axis. Therefore, the equation is in the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, where $$a^2=25$$ and $$b^2=4$$. It follows that:
• the center of the ellipse is $$(0,0)$$
• the coordinates of the vertices are $$(\pm a,0)=(\pm \sqrt{25},0)=(\pm 5,0)$$
• the coordinates of the co-vertices are $$(0,\pm b)=(0,\pm \sqrt{4})=(0,\pm 2)$$
• the coordinates of the foci are $$(\pm c,0)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$, we have:
\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\ &=\pm \sqrt{25−4} \nonumber \\ &=\pm \sqrt{21} \nonumber \end{align} \nonumber
Therefore the coordinates of the foci are $$(\pm \sqrt{21},0)$$.
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. Figure $$\PageIndex{10}$$
###### Exercise $$\PageIndex{4}$$
Graph the ellipse given by the equation $$49x^2+16y^2=784$$. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Standard form: $$\dfrac{x^2}{16}+\dfrac{y^2}{49}=1$$; center: $$(0,0)$$; vertices: $$(0,\pm 7)$$; co-vertices: $$(\pm 4,0)$$; foci: $$(0,\pm \sqrt{33})$$ Figure $$\PageIndex{11}$$
## Graphing Ellipses Not Centered at the Origin
When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, $$(h,k)$$,we use the standard forms $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, $$a>b$$ for horizontal ellipses and $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, $$a>b$$ for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
###### How to: Given the standard form of an equation for an ellipse centered at $$(h, k)$$, sketch the graph.
1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
• If the equation is in the form $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, where $$a>b$$, then
• the center is $$(h,k)$$
• the major axis is parallel to the $$x$$-axis
• the coordinates of the vertices are $$(h\pm a,k)$$
• the coordinates of the co-vertices are $$(h,k\pm b)$$
• the coordinates of the foci are $$(h\pm c,k)$$
• If the equation is in the form $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, where $$a>b$$, then
• the center is $$(h,k)$$
• the major axis is parallel to the $$y$$-axis
• the coordinates of the vertices are $$(h,k\pm a)$$
• the coordinates of the co-vertices are $$(h\pm b,k)$$
• the coordinates of the foci are $$(h,k\pm c)$$
2. Solve for $$c$$ using the equation $$c^2=a^2−b^2$$.
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
###### Example $$\PageIndex{5}$$: Graphing an Ellipse Centered at $$(h, k)$$
Graph the ellipse given by the equation, $$\dfrac{{(x+2)}^2}{4}+\dfrac{{(y−5)}^2}{9}=1$$. Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because $$9>4$$, the major axis is parallel to the $$y$$-axis. Therefore, the equation is in the form $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, where $$b^2=4$$ and $$a^2=9$$. It follows that:
• the center of the ellipse is $$(h,k)=(−2,5)$$
• the coordinates of the vertices are $$(h,k\pm a)=(−2,5\pm \sqrt{9})=(−2,5\pm 3)$$, or $$(−2,2)$$ and $$(−2,8)$$
• the coordinates of the co-vertices are $$(h\pm b,k)=(−2\pm \sqrt{4},5)=(−2\pm 2,5)$$, or $$(−4,5)$$ and $$(0,5)$$
• the coordinates of the foci are $$(h,k\pm c)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$,we have:
\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\[4pt] &=\pm \sqrt{9−4} \nonumber \\[4pt] &=\pm \sqrt{5} \nonumber \end{align} \nonumber
Therefore, the coordinates of the foci are $$(−2,5−\sqrt{5})$$ and $$(−2,5+\sqrt{5})$$.
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. Figure $$\PageIndex{12}$$
###### Exercise $$\PageIndex{5}$$
Graph the ellipse given by the equation $$\dfrac{{(x−4)}^2}{36}+\dfrac{{(y−2)}^2}{20}=1$$. Identify and label the center, vertices, co-vertices, and foci.
Center: $$(4,2)$$; vertices: $$(−2,2)$$ and $$(10,2)$$; co-vertices: $$(4,2−2\sqrt{5})$$ and $$(4,2+2\sqrt{5})$$; foci: $$(0,2)$$ and $$(8,2)$$ Figure $$\PageIndex{13}$$
###### How to: Given the general form of an equation for an ellipse centered at $$(h, k)$$, express the equation in standard form.
1. Recognize that an ellipse described by an equation in the form $$ax^2+by^2+cx+dy+e=0$$ is in general form.
2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
3. Factor out the coefficients of the $$x^2$$ and $$y^2$$ terms in preparation for completing the square.
4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, $$m_1{(x−h)}^2+m_2{(y−k)}^2=m_3$$, where $$m_1$$, $$m_2$$,and $$m_3$$ are constants.
5. Divide both sides of the equation by the constant term to express the equation in standard form.
###### Example $$\PageIndex{6}$$: Graphing an Ellipse Centered at $$(h, k)$$ by First Writing It in Standard Form
Graph the ellipse given by the equation $$4x^2+9y^2−40x+36y+100=0$$. Identify and label the center, vertices, co-vertices, and foci.
Solution
We must begin by rewriting the equation in standard form.
$$4x^2+9y^2−40x+36y+100=0$$
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
$$(4x^2−40x)+(9y^2+36y)=−100$$
Factor out the coefficients of the squared terms.
$$4(x^2−10x)+9(y^2+4y)=−100$$
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
$$4(x^2−10x+25)+9(y^2+4y+4)=−100+100+36$$
Rewrite as perfect squares.
$$4{(x−5)}^2+9{(y+2)}^2=36$$
Divide both sides by the constant term to place the equation in standard form.
$$\dfrac{{(x−5)}^2}{9}+\dfrac{{(y+2)}^2}{4}=1$$
Now that the equation is in standard form, we can determine the position of the major axis. Because $$9>4$$, the major axis is parallel to the $$x$$-axis. Therefore, the equation is in the form $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, where $$a^2=9$$ and $$b^2=4$$. It follows that:
• the center of the ellipse is $$(h,k)=(5,−2)$$
• the coordinates of the vertices are $$(h\pm a,k)=(5\pm \sqrt{9},−2)=(5\pm 3,−2)$$, or $$(2,−2)$$ and $$(8,−2)$$
• the coordinates of the co-vertices are $$(h,k\pm b)=(5,−2\pm \sqrt{4})=(5,−2\pm 2)$$, or $$(5,−4)$$ and $$(5,0)$$
• the coordinates of the foci are $$(h\pm c,k)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$, we have:
\begin{align*} c&=\pm \sqrt{a^2-b^2}\\ &=\pm \sqrt{9-4}\\ &=\pm \sqrt{5} \end{align*}
Therefore, the coordinates of the foci are $$(5−\sqrt{5},−2)$$ and $$(5+\sqrt{5},−2)$$.
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure $$\PageIndex{14}$$. Figure $$\PageIndex{14}$$
###### Exercise $$\PageIndex{6}$$
Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.
$$4x^2+y^2−24x+2y+21=0$$
$$\dfrac{{(x−3)}^2}{4}+\dfrac{{(y+1)}^2}{16}=1$$; center: $$(3,−1)$$; vertices: $$(3,−5)$$ and $$(3,3)$$; co-vertices: $$(1,−1)$$ and $$(5,−1)$$; foci: $$(3,−1−2\sqrt{3})$$ and $$(3,−1+2\sqrt{3})$$
## Solving Applied Problems Involving Ellipses
Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus (Figure $$\PageIndex{15}$$). In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about $$43$$ feet apart—can hear each other whisper. Figure $$\PageIndex{15}$$: Sound waves are reflected between foci in an elliptical room, called a whispering chamber.
###### Example $$\PageIndex{7}$$: Locating the Foci of a Whispering Chamber
The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are $$46$$ feet wide by $$96$$ feet long as shown in Figure $$\PageIndex{16}$$.
1. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $$(0,0)$$.
2. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot. Figure $$\PageIndex{16}$$
Solution
1. We are assuming a horizontal ellipse with center $$(0,0)$$, so we need to find an equation of the form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, where $$a>b$$. We know that the length of the major axis, $$2a$$, is longer than the length of the minor axis, $$2b$$. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.
Therefore, the equation of the ellipse is (dfrac{x^2}{2304}+\dfrac{y^2}{529}=1\) • Solving for $$a$$, we have $$2a=96$$, so $$a=48$$, and $$a^2=2304$$. • Solving for $$b$$, we have $$2b=46$$, so $$b=23$$, and $$b^2=529$$. 2. To find the distance between the senators, we must find the distance between the foci, $$(\pm c,0)$$, where $$c^2=a^2−b^2$$. Solving for $$c$$,we have: \[\begin{align*} c^2&=a^2-b^2\\ c^2&=2304-529\qquad \text{Substitute using the values found in part } (a)\\ c&=\pm \sqrt{2304-529}\qquad \text{Take the square root of both sides.}\\ c&=\pm \sqrt{1775}\qquad \text{Subtract.}\\ c&\approx \pm 42\qquad \text{Round to the nearest foot.} \end{align*}
The points $$(\pm 42,0)$$ represent the foci. Thus, the distance between the senators is $$2(42)=84$$ feet.
###### Exercise $$\PageIndex{7}$$
Suppose a whispering chamber is $$480$$ feet long and $$320$$ feet wide.
1. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $$(0,0)$$.
2. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.
$$\dfrac{x^2}{57,600}+\dfrac{y^2}{25,600}=1$$
The people are standing $$358$$ feet apart.
###### Media
Access these online resources for additional instruction and practice with ellipses.
## Key Equations
Horizontal ellipse, center at origin $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$, $$a>b$$ Vertical ellipse, center at origin $$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$, $$a>b$$ Horizontal ellipse, center $$(h,k)$$ $$\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1$$, $$a>b$$ Vertical ellipse, center $$(h,k)$$ $$\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1$$, $$a>b$$
## Key Concepts
• An ellipse is the set of all points $$(x,y)$$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
• When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See Example $$\PageIndex{1}$$ and Example $$\PageIndex{2}$$.
• When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See Example $$\PageIndex{3}$$ and Example $$\PageIndex{4}$$.
• When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See Example $$\PageIndex{5}$$ and Example $$\PageIndex{6}$$.
• Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See Example $$\PageIndex{7}$$. | HuggingFaceTB/finemath | |
# Carmichael number
In number theory, a Carmichael number is a composite number ${\displaystyle n}$ which satisfies the modular arithmetic congruence relation:
${\displaystyle b^{n-1}\equiv 1{\pmod {n}}}$
for all integers ${\displaystyle b}$ which are relatively prime to ${\displaystyle n}$. [1] They are named for Robert Carmichael. The Carmichael numbers are the subset K1 of the Knödel numbers.
Equivalently, a Carmichael number is a composite number ${\displaystyle n}$ for which
${\displaystyle b^{n}\equiv b{\pmod {n}}}$
for all integers ${\displaystyle b}$. [2]
## Overview
Fermat's little theorem states that if p is a prime number, then for any integer b, the number b p − b is an integer multiple of p. Carmichael numbers are composite numbers which have this property. Carmichael numbers are also called Fermat pseudoprimes or absolute Fermat pseudoprimes. A Carmichael number will pass a Fermat primality test to every base b relatively prime to the number, even though it is not actually prime. This makes tests based on Fermat's Little Theorem less effective than strong probable prime tests such as the Baillie-PSW primality test and the Miller–Rabin primality test.
However, no Carmichael number is either an Euler-Jacobi pseudoprime or a strong pseudoprime to every base relatively prime to it [3] so, in theory, either an Euler or a strong probable prime test could prove that a Carmichael number is, in fact, composite.
As numbers become larger, Carmichael numbers become increasingly rare. For example, there are 20,138,200 Carmichael numbers between 1 and 1021 (approximately one in 50 trillion (5·1013) numbers).[4]
### Korselt's criterion
An alternative and equivalent definition of Carmichael numbers is given by Korselt's criterion.
Theorem (A. Korselt 1899): A positive composite integer ${\displaystyle n}$ is a Carmichael number if and only if ${\displaystyle n}$ is square-free, and for all prime divisors ${\displaystyle p}$ of ${\displaystyle n}$, it is true that ${\displaystyle p-1\mid n-1}$.
It follows from this theorem that all Carmichael numbers are odd, since any even composite number that is square-free (and hence has only one prime factor of two) will have at least one odd prime factor, and thus ${\displaystyle p-1\mid n-1}$ results in an even dividing an odd, a contradiction. (The oddness of Carmichael numbers also follows from the fact that ${\displaystyle -1}$ is a Fermat witness for any even composite number.) From the criterion it also follows that Carmichael numbers are cyclic.[5][6] Additionally, it follows that there are no Carmichael numbers with exactly two prime factors.
## Discovery
Korselt was the first who observed the basic properties of Carmichael numbers, but he did not give any examples. In 1910, Carmichael[7] found the first and smallest such number, 561, which explains the name "Carmichael number".
That 561 is a Carmichael number can be seen with Korselt's criterion. Indeed, ${\displaystyle 561=3\cdot 11\cdot 17}$ is square-free and ${\displaystyle 2\mid 560}$, ${\displaystyle 10\mid 560}$ and ${\displaystyle 16\mid 560}$.
The next six Carmichael numbers are (sequence A002997 in the OEIS):
${\displaystyle 1105=5\cdot 13\cdot 17\qquad (4\mid 1104;\quad 12\mid 1104;\quad 16\mid 1104)}$
${\displaystyle 1729=7\cdot 13\cdot 19\qquad (6\mid 1728;\quad 12\mid 1728;\quad 18\mid 1728)}$
${\displaystyle 2465=5\cdot 17\cdot 29\qquad (4\mid 2464;\quad 16\mid 2464;\quad 28\mid 2464)}$
${\displaystyle 2821=7\cdot 13\cdot 31\qquad (6\mid 2820;\quad 12\mid 2820;\quad 30\mid 2820)}$
${\displaystyle 6601=7\cdot 23\cdot 41\qquad (6\mid 6600;\quad 22\mid 6600;\quad 40\mid 6600)}$
${\displaystyle 8911=7\cdot 19\cdot 67\qquad (6\mid 8910;\quad 18\mid 8910;\quad 66\mid 8910).}$
These first seven Carmichael numbers, from 561 to 8911, were all found by the Czech mathematician Václav Šimerka in 1885[8] (thus preceding not just Carmichael but also Korselt, although Šimerka did not find anything like Korselt's criterion).[9] His work, however, remained unnoticed.
J. Chernick[10] proved a theorem in 1939 which can be used to construct a subset of Carmichael numbers. The number ${\displaystyle (6k+1)(12k+1)(18k+1)}$ is a Carmichael number if its three factors are all prime. Whether this formula produces an infinite quantity of Carmichael numbers is an open question (though it is implied by Dickson's conjecture).
Paul Erdős heuristically argued there should be infinitely many Carmichael numbers. In 1994 W. R. (Red) Alford, Andrew Granville and Carl Pomerance used a bound on Olson's constant to show that there really do exist infinitely many Carmichael numbers. Specifically, they showed that for sufficiently large ${\displaystyle n}$, there are at least ${\displaystyle n^{2/7}}$ Carmichael numbers between 1 and ${\displaystyle n}$.[11]
Löh and Niebuhr in 1992 found some very large Carmichael numbers, including one with 1,101,518 factors and over 16 million digits.
## Properties
### Factorizations
Carmichael numbers have at least three positive prime factors. For some fixed R, there are infinitely many Carmichael numbers with exactly R factors; in fact, there are infinitely many such R.[12]
The first Carmichael numbers with ${\displaystyle k=3,4,5,\ldots }$ prime factors are (sequence A006931 in the OEIS):
k
3 ${\displaystyle 561=3\cdot 11\cdot 17\,}$
4 ${\displaystyle 41041=7\cdot 11\cdot 13\cdot 41\,}$
5 ${\displaystyle 825265=5\cdot 7\cdot 17\cdot 19\cdot 73\,}$
6 ${\displaystyle 321197185=5\cdot 19\cdot 23\cdot 29\cdot 37\cdot 137\,}$
7 ${\displaystyle 5394826801=7\cdot 13\cdot 17\cdot 23\cdot 31\cdot 67\cdot 73\,}$
8 ${\displaystyle 232250619601=7\cdot 11\cdot 13\cdot 17\cdot 31\cdot 37\cdot 73\cdot 163\,}$
9 ${\displaystyle 9746347772161=7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 31\cdot 37\cdot 41\cdot 641\,}$
The first Carmichael numbers with 4 prime factors are (sequence A074379 in the OEIS):
i
1 ${\displaystyle 41041=7\cdot 11\cdot 13\cdot 41\,}$
2 ${\displaystyle 62745=3\cdot 5\cdot 47\cdot 89\,}$
3 ${\displaystyle 63973=7\cdot 13\cdot 19\cdot 37\,}$
4 ${\displaystyle 75361=11\cdot 13\cdot 17\cdot 31\,}$
5 ${\displaystyle 101101=7\cdot 11\cdot 13\cdot 101\,}$
6 ${\displaystyle 126217=7\cdot 13\cdot 19\cdot 73\,}$
7 ${\displaystyle 172081=7\cdot 13\cdot 31\cdot 61\,}$
8 ${\displaystyle 188461=7\cdot 13\cdot 19\cdot 109\,}$
9 ${\displaystyle 278545=5\cdot 17\cdot 29\cdot 113\,}$
10 ${\displaystyle 340561=13\cdot 17\cdot 23\cdot 67\,}$
The second Carmichael number (1105) can be expressed as the sum of two squares in more ways than any smaller number. The third Carmichael number (1729) is the Hardy-Ramanujan Number: the smallest number that can be expressed as the sum of two cubes (of positive numbers) in two different ways.
### Distribution
Let ${\displaystyle C(X)}$ denote the number of Carmichael numbers less than or equal to ${\displaystyle X}$. The distribution of Carmichael numbers by powers of 10 (sequence A055553 in the OEIS):[4]
${\displaystyle n}$ ${\displaystyle C(10^{n})}$ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 0 0 1 7 16 43 105 255 646 1547 3605 8241 19279 44706 105212 246683 585355 1401644 3381806 8220777 20138200
In 1953, Knödel proved the upper bound:
${\displaystyle C(X)
for some constant ${\displaystyle k_{1}}$.
In 1956, Erdős improved the bound to[13]
${\displaystyle C(X)
for some constant ${\displaystyle k_{2}}$. He further gave a heuristic argument suggesting that this upper bound should be close to the true growth rate of ${\displaystyle C(X)}$. The table below gives approximate minimal values for the constant k in the Erdős bound for ${\displaystyle X=10^{n}}$ as n grows:
${\displaystyle n}$ k 4 6 8 10 12 14 16 18 20 21 2.19547 1.97946 1.90495 1.8687 1.86377 1.86293 1.86406 1.86522 1.86598 1.86619
In the other direction, Alford, Granville and Pomerance proved in 1994[11] that for sufficiently large X,
${\displaystyle C(X)>X^{\frac {2}{7}}.}$
In 2005, this bound was further improved by Harman[14] to
${\displaystyle C(X)>X^{0.332}}$
who subsequently improved the exponent to ${\displaystyle 0.7039\cdot 0.4736=0.33336704>1/3}$. [15]
Regarding the asymptotic distribution of Carmichael numbers, there have been several conjectures. In 1956, Erdős[13] conjectured that there were ${\displaystyle X^{1-o(1)}}$ Carmichael numbers for X sufficiently large. In 1981, Pomerance[16] sharpened Erdős' heuristic arguments to conjecture that there are
${\displaystyle X^{1-{\frac {\{1+o(1)\}\log \log \log X}{\log \log X}}}}$
Carmichael numbers up to X. However, inside current computational ranges (such as the counts of Carmichael numbers performed by Pinch[4] up to 1021), these conjectures are not yet borne out by the data.
## Generalizations
The notion of Carmichael number generalizes to a Carmichael ideal in any number field K. For any nonzero prime ideal ${\displaystyle {\mathfrak {p}}}$ in ${\displaystyle {\mathcal {O}}_{K}}$, we have ${\displaystyle \alpha ^{{\rm {N}}({\mathfrak {p}})}\equiv \alpha {\bmod {\mathfrak {p}}}}$ for all ${\displaystyle \alpha }$ in ${\displaystyle {\mathcal {O}}_{K}}$, where ${\displaystyle {\rm {N}}({\mathfrak {p}})}$ is the norm of the ideal ${\displaystyle {\mathfrak {p}}}$. (This generalizes Fermat's little theorem, that ${\displaystyle m^{p}\equiv m{\bmod {p}}}$ for all integers m when p is prime.) Call a nonzero ideal ${\displaystyle {\mathfrak {a}}}$ in ${\displaystyle {\mathcal {O}}_{K}}$ Carmichael if it is not a prime ideal and ${\displaystyle \alpha ^{{\rm {N}}({\mathfrak {a}})}\equiv \alpha {\bmod {\mathfrak {a}}}}$ for all ${\displaystyle \alpha \in {\mathcal {O}}_{K}}$, where ${\displaystyle {\rm {N}}({\mathfrak {a}})}$ is the norm of the ideal ${\displaystyle {\mathfrak {a}}}$. When K is ${\displaystyle \mathbf {Q} }$, the ideal ${\displaystyle {\mathfrak {a}}}$ is principal, and if we let a be its positive generator then the ideal ${\displaystyle {\mathfrak {a}}=(a)}$ is Carmichael exactly when a is a Carmichael number in the usual sense.
When K is larger than the rationals it is easy to write down Carmichael ideals in ${\displaystyle {\mathcal {O}}_{K}}$: for any prime number p that splits completely in K, the principal ideal ${\displaystyle p{\mathcal {O}}_{K}}$ is a Carmichael ideal. Since infinitely many prime numbers split completely in any number field, there are infinitely many Carmichael ideals in ${\displaystyle {\mathcal {O}}_{K}}$. For example, if p is any prime number that is 1 mod 4, the ideal (p) in the Gaussian integers Z[i] is a Carmichael ideal.
Both prime and Carmichael numbers satisfy the following equality:
${\displaystyle \gcd \left(\sum _{x=1}^{n-1}x^{n-1},n\right)=1}$
## Higher-order Carmichael numbers
Carmichael numbers can be generalized using concepts of abstract algebra.
The above definition states that a composite integer n is Carmichael precisely when the nth-power-raising function pn from the ring Zn of integers modulo n to itself is the identity function. The identity is the only Zn-algebra endomorphism on Zn so we can restate the definition as asking that pn be an algebra endomorphism of Zn. As above, pn satisfies the same property whenever n is prime.
The nth-power-raising function pn is also defined on any Zn-algebra A. A theorem states that n is prime if and only if all such functions pn are algebra endomorphisms.
In-between these two conditions lies the definition of Carmichael number of order m for any positive integer m as any composite number n such that pn is an endomorphism on every Zn-algebra that can be generated as Zn-module by m elements. Carmichael numbers of order 1 are just the ordinary Carmichael numbers.
### An order 2 Carmichael number
According to Howe, 17 · 31 · 41 · 43 · 89 · 97 · 167 · 331 is an order 2 Carmichael number. This product is equal to 443,372,888,629,441.[17]
### Properties
Korselt's criterion can be generalized to higher-order Carmichael numbers, as shown by Howe.
A heuristic argument, given in the same paper, appears to suggest that there are infinitely many Carmichael numbers of order m, for any m. However, not a single Carmichael number of order 3 or above is known.
## Notes
1. ^ Riesel, Hans (1994). Prime Numbers and Computer Methods for Factorization. Progress in Mathematics. 126 (second ed.). Boston, MA: Birkhäuser. ISBN 0-8176-3743-5. Zbl 0821.11001.
2. ^ Crandall, Richard; Pomerance, Carl (2005). Prime Numbers: A Computational Perspective (second ed.). New York: Springer. p. 133. ISBN 978-0387-25282-7.
3. ^ D. H. Lehmer (1976). "Strong Carmichael numbers". J. Austral. Math. Soc. 21: 508–510. doi:10.1017/s1446788700019364. Lehmer proved that no Carmichael number is an Euler-Jacobi pseudoprime to every base relatively prime to it. He used the term strong pseudoprime, but the terminology has changed since then. Strong pseudoprimes are a subset of Euler-Jacobi pseudoprimes. Therefore, no Carmichael number is a strong pseudoprime to every base relatively prime to it.
4. ^ a b c Pinch, Richard (December 2007). Anne-Maria Ernvall-Hytönen, ed. The Carmichael numbers up to 1021 (PDF). Proceedings of Conference on Algorithmic Number Theory. 46. Turku, Finland: Turku Centre for Computer Science. pp. 129–131. Retrieved 2017-06-26.
5. ^ Carmichael Multiples of Odd Cyclic Numbers "Any divisor of a Carmichael number must be an odd cyclic number"
6. ^ Proof sketch: If ${\displaystyle n}$ is square-free but not cyclic, ${\displaystyle p_{i}\mid p_{j}-1}$ for two prime factors ${\displaystyle p_{i}}$ and ${\displaystyle p_{j}}$ of ${\displaystyle n}$. But if ${\displaystyle n}$ satisfies Korselt then ${\displaystyle p_{j}-1\mid n-1}$, so by transitivity of the "divides" relation ${\displaystyle p_{i}\mid n-1}$. But ${\displaystyle p_{i}}$ is also a factor of ${\displaystyle n}$, a contradiction.
7. ^ R. D. Carmichael (1910). "Note on a new number theory function". Bulletin of the American Mathematical Society. 16 (5): 232–238. doi:10.1090/s0002-9904-1910-01892-9.
8. ^ V. Šimerka (1885). "Zbytky z arithmetické posloupnosti (On the remainders of an arithmetic progression)". Časopis pro pěstování matematiky a fysiky. 14 (5): 221–225.
9. ^ Lemmermeyer, F. (2013). "Václav Šimerka: quadratic forms and factorization". LMS Journal of Computation and Mathematics. 16: 118–129. doi:10.1112/S1461157013000065.
10. ^ Chernick, J. (1939). "On Fermat's simple theorem" (PDF). Bull. Amer. Math. Soc. 45: 269–274. doi:10.1090/S0002-9904-1939-06953-X.
11. ^ a b W. R. Alford; Andrew Granville; Carl Pomerance (1994). "There are Infinitely Many Carmichael Numbers" (PDF). Annals of Mathematics. 139: 703–722. doi:10.2307/2118576.
12. ^ Wright, Thomas (2016-06-01). "Factors of Carmichael numbers and a weak k-tuples conjecture". Journal of the Australian Mathematical Society. Australian Mathematical Publishing Association Inc. 100 (3): 421–429. doi:10.1017/S1446788715000427. Retrieved 2016-08-13.
13. ^ a b Erdős, P. (1956). "On pseudoprimes and Carmichael numbers" (PDF). Publ. Math. Debrecen. 4: 201–206. MR 0079031.
14. ^ Glyn Harman (2005). "On the number of Carmichael numbers up to x". Bulletin of the London Mathematical Society. 37: 641–650. doi:10.1112/S0024609305004686.
15. ^ Harman, Glyn (2008). "Watt's mean value theorem and Carmichael numbers". International Journal of Number Theory. 4 (2): 242, 243. doi:10.1142/S1793042108001316. MR 2404800.
16. ^ Pomerance, C. (1981). "On the distribution of pseudoprimes". Math. Comp. 37: 587–593. doi:10.1090/s0025-5718-1981-0628717-0. JSTOR 2007448.
17. ^ Everett W. Howe (October 2000). "Higher-order Carmichael numbers". Mathematics of Computation. 69 (232): 1711–1719. arXiv:math.NT/9812089. doi:10.1090/s0025-5718-00-01225-4. JSTOR 2585091. | HuggingFaceTB/finemath |
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