Report
Share
Download to read offline
Recommended
1. Smart Labels are computerized, bar-coded label that is used to pr.pdf
1. Smart Labels are computerized, bar-coded label that is used to provide shipping information.
They play a large role in UPS operations because they provide customer information for shipping
which makes for faster and more reliable shipping. The label is scanned at each part of shipment
and delivery so the customer can track their packages and it\'s accounted for from start to finish.
2.
The driver picks up the package from the customer and creates a smart label that contains all
shipping information and a tracking number, this information is sent to the warehouse. The
driver will always carry a DIAD which is basically a hand held computer that will show the
driver which routes to take and his stops for the day. The driver will scan the package
information into the DIAD so that the information is available to the Data Center and to the
customer. All packages will be process through a UPS package center.
3.
Wireless communication allows the UPS drivers to have complete flexibility in the field and upto
date information about the changes to their routes or new pick-ups. This makes mileage which is
really important these days.
4.
Information technology definitely revolutionized the business of UPS. It enabled UPS to reach
out to countries and territories and expand their business. UPS operations has become more
efficient, expedient and more up-to-date.
5.
By putting investment in improving IT facility the company was able to optimally drive the
operations of the company. It further enhanced its customer representation and relation and
expands its business portfolio
Solution
1. Smart Labels are computerized, bar-coded label that is used to provide shipping information.
They play a large role in UPS operations because they provide customer information for shipping
which makes for faster and more reliable shipping. The label is scanned at each part of shipment
and delivery so the customer can track their packages and it\'s accounted for from start to finish.
2.
The driver picks up the package from the customer and creates a smart label that contains all
shipping information and a tracking number, this information is sent to the warehouse. The
driver will always carry a DIAD which is basically a hand held computer that will show the
driver which routes to take and his stops for the day. The driver will scan the package
information into the DIAD so that the information is available to the Data Center and to the
customer. All packages will be process through a UPS package center.
3.
Wireless communication allows the UPS drivers to have complete flexibility in the field and upto
date information about the changes to their routes or new pick-ups. This makes mileage which is
really important these days.
4.
Information technology definitely revolutionized the business of UPS. It enabled UPS to reach
out to countries and territories and expand their business. UPS operations has become more
efficient, expedient and more up-to-date.
5.
By putting investment.
1 ) a) Android Security One of the biggest challange is android sm.pdf
1 ) a) Android Security : One of the biggest challange is android smart phone security as once
user had given permission control, the app has the ability to display popups and install the
adware and inject virusesm also.
b) Hackers era : In todays digital era these bad guys use coordinated attack and get from us
whatever they want from us.
and many more countless number of trend like IOS malware attack is still a hot issue in the
market these days.
Steps to Take to Minimize the Possibility of an Attack on a System :
2) a) You should create strong passwords :Store your passwords safely. Don’t use sticky notes to
write or store them if you want to store them manualy, file passwords away from your
computers. Write clues rather than the actual password as a protective measure.
b) Account Log-off : Its a simple step that can prevent you from cyber attack if you always log
off your account.
c) Install System and software updates : As we know that daily new number of malwares and
viruses are being created so update system and software to get rid of those attacks
3) Trojan , Adwares , malwares , Spy Wares , Backdoor , wabbits ,Exploit , Phishing are some of
the wellknown Threats That Exist for Computers and Networks.
4) Number of Crimes happened in recent times , Some are :
a) Hackers published 13GB of Ashley Madison data
b)Credit card data hacked from Mandarin Oriental
Solution
1 ) a) Android Security : One of the biggest challange is android smart phone security as once
user had given permission control, the app has the ability to display popups and install the
adware and inject virusesm also.
b) Hackers era : In todays digital era these bad guys use coordinated attack and get from us
whatever they want from us.
and many more countless number of trend like IOS malware attack is still a hot issue in the
market these days.
Steps to Take to Minimize the Possibility of an Attack on a System :
2) a) You should create strong passwords :Store your passwords safely. Don’t use sticky notes to
write or store them if you want to store them manualy, file passwords away from your
computers. Write clues rather than the actual password as a protective measure.
b) Account Log-off : Its a simple step that can prevent you from cyber attack if you always log
off your account.
c) Install System and software updates : As we know that daily new number of malwares and
viruses are being created so update system and software to get rid of those attacks
3) Trojan , Adwares , malwares , Spy Wares , Backdoor , wabbits ,Exploit , Phishing are some of
the wellknown Threats That Exist for Computers and Networks.
4) Number of Crimes happened in recent times , Some are :
a) Hackers published 13GB of Ashley Madison data
b)Credit card data hacked from Mandarin Oriental.
1) front to back2) smart dimension or noteSolution1) front t.pdf
1) front to back
2) smart dimension or note
Solution
1) front to back
2) smart dimension or note.
Noonan syndrome is a developmental disorder. Noonan syndrome is a ge.pdf
Noonan syndrome is a developmental disorder. Noonan syndrome is a genetical disorder.
Nuerofibromatosis 1 (NF1), Noonan Syndrome (NS), Costello syndrome, cardiofaciocutaneous
syndrome(CSC) are autosomal dominant disorders. All these diseases leads to mental disorders.
As these are autosomal dominant disorders, one mutated copy of gene is enough to get the
disease. One of the parent might be a carrier of the gene, get affected with the disease and pass
the gene to the children. Both males and females get affected equally as one gene is enough to
get the disease. Due to a sudden mutation in any gene also cause the autosomal dominant
disorder.
Solution
Noonan syndrome is a developmental disorder. Noonan syndrome is a genetical disorder.
Nuerofibromatosis 1 (NF1), Noonan Syndrome (NS), Costello syndrome, cardiofaciocutaneous
syndrome(CSC) are autosomal dominant disorders. All these diseases leads to mental disorders.
As these are autosomal dominant disorders, one mutated copy of gene is enough to get the
disease. One of the parent might be a carrier of the gene, get affected with the disease and pass
the gene to the children. Both males and females get affected equally as one gene is enough to
get the disease. Due to a sudden mutation in any gene also cause the autosomal dominant
disorder..
Blastomyces dermatitidis, coccidiodies immitis and histoplasma capsu.pdf
Blastomyces dermatitidis, coccidiodies immitis and histoplasma capsulatum are the pathogenic
fungi that exhibit fungal dimorphism. they are thermally dimorphic fungi that convert from
nonpathogenic as filamentous mold with asexual spores(conidia) in soil to pathogenic budding
yeast (or spherule in c.immitis) after infectious spores are inhaled into lungs of human.This
morphological conversion is required for virulence. conversion to yeast form provide them
protection against killing by neutrophils, monocytes and macophages. The conversion from mold
to yeast results in a change in cell shape, composition of cell wall,presence of antigenic
molecules and expression of viulence traits. Dimorphic fungi other phase specific products that
are essential for virulence for eg;In B. dermatitidis only yeast phase expresses immunoreactive
protein antigen BAD1, which is responsible for virulence. In H.capsulatum only pathogenic
yeast phase expressa released calcium binding protein (CBP1) which is essential for growth in
macrophages. laboratory Conditions that are necessary to elicit the coversion and gowth of
fungus are temperature, carbon dioxide, cysteine, and estradiol. Binding of estradiol to cell
surface receptors accelerates the gowth of coccidioides spherules at 370c. Dimorphism in
H.Capsulatum occurs in 3 stages. In first stage, induced by increase in temperature, respiration
ceases and level of cytochrome decreases. in second stage conversion from mycelia to yeast
occurs for which cysteine and sulfhydryl containing compounds are required. cysteine is required
for yeast form to grow. final stage characteized by normal cytochrome level and respiration as
yeast grows. B.dermatitidis produces hyphae and one celled conidia at 250c and budding yeast
cells at 370c.
Solution
Blastomyces dermatitidis, coccidiodies immitis and histoplasma capsulatum are the pathogenic
fungi that exhibit fungal dimorphism. they are thermally dimorphic fungi that convert from
nonpathogenic as filamentous mold with asexual spores(conidia) in soil to pathogenic budding
yeast (or spherule in c.immitis) after infectious spores are inhaled into lungs of human.This
morphological conversion is required for virulence. conversion to yeast form provide them
protection against killing by neutrophils, monocytes and macophages. The conversion from mold
to yeast results in a change in cell shape, composition of cell wall,presence of antigenic
molecules and expression of viulence traits. Dimorphic fungi other phase specific products that
are essential for virulence for eg;In B. dermatitidis only yeast phase expresses immunoreactive
protein antigen BAD1, which is responsible for virulence. In H.capsulatum only pathogenic
yeast phase expressa released calcium binding protein (CBP1) which is essential for growth in
macrophages. laboratory Conditions that are necessary to elicit the coversion and gowth of
fungus are temperature, carbon dioxide, cysteine, and estradiol. Binding of estradiol to c.
When the compound is labeled as a liquid or aqueo.pdf
When the compound is labeled as a liquid or aqueous it is considered wet.
Solution
When the compound is labeled as a liquid or aqueous it is considered wet..
They represent exception as they have more than 8.pdf
They represent exception as they have more than 8 electrons participating in their
bonding.. this occurs as they have vaccant d orbitals where bonding can occur..
Solution
They represent exception as they have more than 8 electrons participating in their
bonding.. this occurs as they have vaccant d orbitals where bonding can occur...
The elements of inner transition elements differ .pdf
The elements of inner transition elements differ from, transition having just one
difference that transition have single d-orbital in their shells, but the inner transition elements
have two or more d- orbital so they are more complex and tough to study . Hence they are
distinguished. co-ordination compounds are result of the lone pairs in the d-orbitals only. o.k.
nothing else.
Solution
The elements of inner transition elements differ from, transition having just one
difference that transition have single d-orbital in their shells, but the inner transition elements
have two or more d- orbital so they are more complex and tough to study . Hence they are
distinguished. co-ordination compounds are result of the lone pairs in the d-orbitals only. o.k.
nothing else..
Recommended
1. Smart Labels are computerized, bar-coded label that is used to pr.pdf
1. Smart Labels are computerized, bar-coded label that is used to provide shipping information.
They play a large role in UPS operations because they provide customer information for shipping
which makes for faster and more reliable shipping. The label is scanned at each part of shipment
and delivery so the customer can track their packages and it\'s accounted for from start to finish.
2.
The driver picks up the package from the customer and creates a smart label that contains all
shipping information and a tracking number, this information is sent to the warehouse. The
driver will always carry a DIAD which is basically a hand held computer that will show the
driver which routes to take and his stops for the day. The driver will scan the package
information into the DIAD so that the information is available to the Data Center and to the
customer. All packages will be process through a UPS package center.
3.
Wireless communication allows the UPS drivers to have complete flexibility in the field and upto
date information about the changes to their routes or new pick-ups. This makes mileage which is
really important these days.
4.
Information technology definitely revolutionized the business of UPS. It enabled UPS to reach
out to countries and territories and expand their business. UPS operations has become more
efficient, expedient and more up-to-date.
5.
By putting investment in improving IT facility the company was able to optimally drive the
operations of the company. It further enhanced its customer representation and relation and
expands its business portfolio
Solution
1. Smart Labels are computerized, bar-coded label that is used to provide shipping information.
They play a large role in UPS operations because they provide customer information for shipping
which makes for faster and more reliable shipping. The label is scanned at each part of shipment
and delivery so the customer can track their packages and it\'s accounted for from start to finish.
2.
The driver picks up the package from the customer and creates a smart label that contains all
shipping information and a tracking number, this information is sent to the warehouse. The
driver will always carry a DIAD which is basically a hand held computer that will show the
driver which routes to take and his stops for the day. The driver will scan the package
information into the DIAD so that the information is available to the Data Center and to the
customer. All packages will be process through a UPS package center.
3.
Wireless communication allows the UPS drivers to have complete flexibility in the field and upto
date information about the changes to their routes or new pick-ups. This makes mileage which is
really important these days.
4.
Information technology definitely revolutionized the business of UPS. It enabled UPS to reach
out to countries and territories and expand their business. UPS operations has become more
efficient, expedient and more up-to-date.
5.
By putting investment.
1 ) a) Android Security One of the biggest challange is android sm.pdf
1 ) a) Android Security : One of the biggest challange is android smart phone security as once
user had given permission control, the app has the ability to display popups and install the
adware and inject virusesm also.
b) Hackers era : In todays digital era these bad guys use coordinated attack and get from us
whatever they want from us.
and many more countless number of trend like IOS malware attack is still a hot issue in the
market these days.
Steps to Take to Minimize the Possibility of an Attack on a System :
2) a) You should create strong passwords :Store your passwords safely. Don’t use sticky notes to
write or store them if you want to store them manualy, file passwords away from your
computers. Write clues rather than the actual password as a protective measure.
b) Account Log-off : Its a simple step that can prevent you from cyber attack if you always log
off your account.
c) Install System and software updates : As we know that daily new number of malwares and
viruses are being created so update system and software to get rid of those attacks
3) Trojan , Adwares , malwares , Spy Wares , Backdoor , wabbits ,Exploit , Phishing are some of
the wellknown Threats That Exist for Computers and Networks.
4) Number of Crimes happened in recent times , Some are :
a) Hackers published 13GB of Ashley Madison data
b)Credit card data hacked from Mandarin Oriental
Solution
1 ) a) Android Security : One of the biggest challange is android smart phone security as once
user had given permission control, the app has the ability to display popups and install the
adware and inject virusesm also.
b) Hackers era : In todays digital era these bad guys use coordinated attack and get from us
whatever they want from us.
and many more countless number of trend like IOS malware attack is still a hot issue in the
market these days.
Steps to Take to Minimize the Possibility of an Attack on a System :
2) a) You should create strong passwords :Store your passwords safely. Don’t use sticky notes to
write or store them if you want to store them manualy, file passwords away from your
computers. Write clues rather than the actual password as a protective measure.
b) Account Log-off : Its a simple step that can prevent you from cyber attack if you always log
off your account.
c) Install System and software updates : As we know that daily new number of malwares and
viruses are being created so update system and software to get rid of those attacks
3) Trojan , Adwares , malwares , Spy Wares , Backdoor , wabbits ,Exploit , Phishing are some of
the wellknown Threats That Exist for Computers and Networks.
4) Number of Crimes happened in recent times , Some are :
a) Hackers published 13GB of Ashley Madison data
b)Credit card data hacked from Mandarin Oriental.
1) front to back2) smart dimension or noteSolution1) front t.pdf
1) front to back
2) smart dimension or note
Solution
1) front to back
2) smart dimension or note.
Noonan syndrome is a developmental disorder. Noonan syndrome is a ge.pdf
Noonan syndrome is a developmental disorder. Noonan syndrome is a genetical disorder.
Nuerofibromatosis 1 (NF1), Noonan Syndrome (NS), Costello syndrome, cardiofaciocutaneous
syndrome(CSC) are autosomal dominant disorders. All these diseases leads to mental disorders.
As these are autosomal dominant disorders, one mutated copy of gene is enough to get the
disease. One of the parent might be a carrier of the gene, get affected with the disease and pass
the gene to the children. Both males and females get affected equally as one gene is enough to
get the disease. Due to a sudden mutation in any gene also cause the autosomal dominant
disorder.
Solution
Noonan syndrome is a developmental disorder. Noonan syndrome is a genetical disorder.
Nuerofibromatosis 1 (NF1), Noonan Syndrome (NS), Costello syndrome, cardiofaciocutaneous
syndrome(CSC) are autosomal dominant disorders. All these diseases leads to mental disorders.
As these are autosomal dominant disorders, one mutated copy of gene is enough to get the
disease. One of the parent might be a carrier of the gene, get affected with the disease and pass
the gene to the children. Both males and females get affected equally as one gene is enough to
get the disease. Due to a sudden mutation in any gene also cause the autosomal dominant
disorder..
Blastomyces dermatitidis, coccidiodies immitis and histoplasma capsu.pdf
Blastomyces dermatitidis, coccidiodies immitis and histoplasma capsulatum are the pathogenic
fungi that exhibit fungal dimorphism. they are thermally dimorphic fungi that convert from
nonpathogenic as filamentous mold with asexual spores(conidia) in soil to pathogenic budding
yeast (or spherule in c.immitis) after infectious spores are inhaled into lungs of human.This
morphological conversion is required for virulence. conversion to yeast form provide them
protection against killing by neutrophils, monocytes and macophages. The conversion from mold
to yeast results in a change in cell shape, composition of cell wall,presence of antigenic
molecules and expression of viulence traits. Dimorphic fungi other phase specific products that
are essential for virulence for eg;In B. dermatitidis only yeast phase expresses immunoreactive
protein antigen BAD1, which is responsible for virulence. In H.capsulatum only pathogenic
yeast phase expressa released calcium binding protein (CBP1) which is essential for growth in
macrophages. laboratory Conditions that are necessary to elicit the coversion and gowth of
fungus are temperature, carbon dioxide, cysteine, and estradiol. Binding of estradiol to cell
surface receptors accelerates the gowth of coccidioides spherules at 370c. Dimorphism in
H.Capsulatum occurs in 3 stages. In first stage, induced by increase in temperature, respiration
ceases and level of cytochrome decreases. in second stage conversion from mycelia to yeast
occurs for which cysteine and sulfhydryl containing compounds are required. cysteine is required
for yeast form to grow. final stage characteized by normal cytochrome level and respiration as
yeast grows. B.dermatitidis produces hyphae and one celled conidia at 250c and budding yeast
cells at 370c.
Solution
Blastomyces dermatitidis, coccidiodies immitis and histoplasma capsulatum are the pathogenic
fungi that exhibit fungal dimorphism. they are thermally dimorphic fungi that convert from
nonpathogenic as filamentous mold with asexual spores(conidia) in soil to pathogenic budding
yeast (or spherule in c.immitis) after infectious spores are inhaled into lungs of human.This
morphological conversion is required for virulence. conversion to yeast form provide them
protection against killing by neutrophils, monocytes and macophages. The conversion from mold
to yeast results in a change in cell shape, composition of cell wall,presence of antigenic
molecules and expression of viulence traits. Dimorphic fungi other phase specific products that
are essential for virulence for eg;In B. dermatitidis only yeast phase expresses immunoreactive
protein antigen BAD1, which is responsible for virulence. In H.capsulatum only pathogenic
yeast phase expressa released calcium binding protein (CBP1) which is essential for growth in
macrophages. laboratory Conditions that are necessary to elicit the coversion and gowth of
fungus are temperature, carbon dioxide, cysteine, and estradiol. Binding of estradiol to c.
When the compound is labeled as a liquid or aqueo.pdf
When the compound is labeled as a liquid or aqueous it is considered wet.
Solution
When the compound is labeled as a liquid or aqueous it is considered wet..
They represent exception as they have more than 8.pdf
They represent exception as they have more than 8 electrons participating in their
bonding.. this occurs as they have vaccant d orbitals where bonding can occur..
Solution
They represent exception as they have more than 8 electrons participating in their
bonding.. this occurs as they have vaccant d orbitals where bonding can occur...
The elements of inner transition elements differ .pdf
The elements of inner transition elements differ from, transition having just one
difference that transition have single d-orbital in their shells, but the inner transition elements
have two or more d- orbital so they are more complex and tough to study . Hence they are
distinguished. co-ordination compounds are result of the lone pairs in the d-orbitals only. o.k.
nothing else.
Solution
The elements of inner transition elements differ from, transition having just one
difference that transition have single d-orbital in their shells, but the inner transition elements
have two or more d- orbital so they are more complex and tough to study . Hence they are
distinguished. co-ordination compounds are result of the lone pairs in the d-orbitals only. o.k.
nothing else..
PV = n RT 15 x 0.068 x 1.01 = 0.75M x 0.0821 x .pdf
PV = n RT 15 x 0.068 x 1.01 = 0.75/M x 0.0821 x (273+25) Molar mass = M =
17.811 gms
Solution
PV = n RT 15 x 0.068 x 1.01 = 0.75/M x 0.0821 x (273+25) Molar mass = M =
17.811 gms.
HOkay. So. Youve got Ruthenium(II)..this is y.pdf
HOkay. So. You\'ve got: Ruthenium(II)..this is your transition metal Ammonia..a
common ligand (you\'ve just go to befamiliar with common ligands and their charges if they
have them) Chloride..this is your cation now. from this information you must deduce:
Ruthenium (R) has a charge of +2 which (because thequestion asks for a coordination
compound) binds with ammonia (NH3) (remembering that ammonia is a common ligand who
donates it\'selections to the metal..in other words it is a Lewisbase..). Ammonia ligands do not
have any charge(like I said before, you just have to know that,there is a chart in your text book
that will give you thisinformation). About the coordination number 6.. I\'ll get tothat in a second.
Notating a complex ion we use brackets like this: [complexion]. Butremember, R has a +2
charge. so the complex ion is an cationiccomplexion like this: [complexion]+2. so thisis were
chlorine comes in. chlorine is an anion meaning it has anegative charge, chlorine has a -1 charge.
so in order to neutralout the +2 charge that R gives the [complexion] we need two of them(Cl2).
so now we have everything to write this out in complex ionnotation. [R(NH3)6]Cl2 and now a
word about the coordination number 6. you don\'t really know that there are 6 ligands present but
it\'s agood assumption because 6 happens to be a very common coordinationfor ammonia
ligands. don\'t worry about this too much, just know howto count coordination numbers when
given the complex ion. Okay, NOW about naming this ridiculous thing. name the ligand first!
and then name other ligands in alphabeticalorder! ammonia in the form of a ligand is called
anammine...(you just have to memorize that table ofligand names which I\'m sure is in the book
as well). 6 ammines = hexamine. that\'s prettystraight forward unless you need to also memorize
numberprefixes. and then you name the metal it\'s boundtoo,Ruthenium(II). here is the tricky
part. add ate to the end of themetal IF and only IF the charge on the complex ion (before
itbecame neutral) was negative. (and in this case thepositive ion that would bind to the [complex
ion notation] would gobefore it, like lets say it was potassium, then you would writeK[complex
ion] and not [complex ion]K...oy...) but in thecase of our Ruthenium(II) we don\'t need to add
ate, because thecharge was originally positive. so now we have hexamineruthenium(II) and now
we add our little chloride... hexamineruthenium(II) chloride. (you don\'t have to say
dichloride,it\'s deducible that there are two..clearly, because i explained howto figure it out up at
the top, but hopefully my explanation wasclear too) (and in this same manner, if the [complex
ion] wasnegative at first and we bound a positive potassium to it we\'d putthe name potassium
before the complex ion like: potassiumwhateverwhatever(roman numeral).)
Solution
HOkay. So. You\'ve got: Ruthenium(II)..this is your transition metal Ammonia..a
common ligand (you\'ve just go to befamiliar with common l.
enthalpy of the given reaction is 4721 KJ since 4.pdf
enthalpy of the given reaction is 4721 KJ since 4 moles of boron are used enthalpy
change / mole boron = 4721/4 = 1180 KJ/mol
Solution
enthalpy of the given reaction is 4721 KJ since 4 moles of boron are used enthalpy
change / mole boron = 4721/4 = 1180 KJ/mol.
in.txt ( Suppose if we store the text file in DDrive then the path .pdf
in.txt ( Suppose if we store the text file in D:Drive then the path of the file pointing to this file is
D:\\\\in.txt )
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
3
38
39
40
41
42
43
44
45
46
45
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
___________________________________________________________________
ReadFile.java
import java.io.*;
import java.util.*;
public class ReadFile {
public static void main(String[] args) {
//Creating the Scanner class reference.
Scanner scanner=null;
try {
//Creating Scanner object by passing the file Object as input
scanner = new Scanner(new File(\"D:\\\\in.txt\"));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
//Creating the ArrayList Obkect
ArrayList integers = new ArrayList();
while (scanner.hasNext()) {
if (scanner.hasNextInt()) {
//Adding each element in the file to the ArrayList.
integers.add(scanner.nextInt());
}
}
//Create the Iterator Object.
Iterator i=integers.iterator();
//Iterator over the ArrayList
while(i.hasNext())
{
//Displaying the contents in the ArrayList.
System.out.println(i.next());
}
}
}
______________________________________________________________________
Output:
Enter File Name :D:\\\\in.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
3
38
39
40
41
42
43
44
45
46
4
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
_____________________________________________________________________
Solution
in.txt ( Suppose if we store the text file in D:Drive then the path of the file pointing to this file is
D:\\\\in.txt )
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
3
38
39
40
41
42
43
44
45
46
45
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
___________________________________________________________________
ReadFile.java
import java.io.*;
import java.util.*;
public class ReadFile {
public static void main(String[] args) {
//Creating the Scanner class reference.
Scanner scanner=null;
try {
//Creating Scanner object by passing the file Object as input
scanner = new Scanner(new File(\"D:\\\\in.txt\"));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
//Creating the ArrayList Obkect
ArrayList integers = new ArrayList();
while (scanner.hasNext()) {
if (scanner.hasNextInt()) {
//Adding each element in the file to the ArrayList.
integers.add(scanner.nextInt());
}
}
//Create the Iterator Object.
Iterator i=integers.iterator();
//Iterator over the ArrayList
while(i.hasNext())
{
//Displaying the contents in the ArrayList.
System.out.println(i.next());
}
}
}
_____________________.
The matrix, the dominant component in the landscape, is the most ext.pdf
The matrix, the dominant component in the landscape, is the most extensive and connected
landscape type, and it plays the dominant role in landscape functioning. A landscape is
composed typically of several types of landscape elements (usually patches). Of these, the matrix
is the most extensive and most connected landscape element type, and therefore plays the
dominant role in the functioning of the landscape .The characteristics of matrix structure are the
density of the patches (porosity), boundary shape, networks, and heterogeneity . For example, in
a large contiguous area of mature forest embedded with numerous small disturbance patches
(e.g., timber harvest patches), the mature forest constitutes the matrix element type because it is
greatest in areal extent, is mostly connected, and exerts a dominant influence on the area flora
and fauna and ecological processes.
Effect of matrix on metapopulation:
The core conceptual framework of metapopulation theory (Levins 1969) is that with-in-patch
dynamics are influenced not only by local extinction but also by colonization from elsewhere in
the landscape. When the landscape becomes more fragmented, colonization rates are reduced.
Then, within-patch species richness will decline and extinction risk at the scale of the entire
metapopulation will increase.
Matrix affects its communities by the following ways.
(1) altering dispersal and colonization rates, which may be reduced or enhanced, depending on
the characteristics of each species
(2) providing alternative habitat to existing species (for some species matrix habitat may be of
lower quality than the original habitat, while for others it may be of higher quality)
(3) as a source of novel invading species for fragments, as the matrix may provide habitat for
new species.
Holt a researcher studied \"Potential effects of the matrix on communities in patchy
environments \". He predicted that species that were abundant in the common habitat (matrix)
had potential to become more common members of local communities in the sparser habitat
(fragments) than species that did not inhabit the common habitat. This spillover effect had
potential to be important in determining which species make up local communities.
Solution
The matrix, the dominant component in the landscape, is the most extensive and connected
landscape type, and it plays the dominant role in landscape functioning. A landscape is
composed typically of several types of landscape elements (usually patches). Of these, the matrix
is the most extensive and most connected landscape element type, and therefore plays the
dominant role in the functioning of the landscape .The characteristics of matrix structure are the
density of the patches (porosity), boundary shape, networks, and heterogeneity . For example, in
a large contiguous area of mature forest embedded with numerous small disturbance patches
(e.g., timber harvest patches), the mature forest constitutes the matrix element type because it i.
The non-disjunctioning had occur in her mother.Turners syndrome .pdf
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia.
Solution
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia..
The answer isB. HNO2, NO2+Conjugate acids have an extra H+ comp.pdf
The answer is:
B. HNO2, NO2+
Conjugate acids have an extra H+ compared to a comjugate base.
Solution
The answer is:
B. HNO2, NO2+
Conjugate acids have an extra H+ compared to a comjugate base..
Team 1 should start with Rohnda because of her no loss in fight 1 an.pdf
Team 1 should start with Rohnda because of her no loss in fight 1 and Cat for 2 fight
Team 2 should start with Holly because of her no loss in fight 1 and Miesha for 2 fight
And team 2 has advantage over team 1
Solution
Team 1 should start with Rohnda because of her no loss in fight 1 and Cat for 2 fight
Team 2 should start with Holly because of her no loss in fight 1 and Miesha for 2 fight
And team 2 has advantage over team 1.
Rods are responsible for dim light vision and cones are responsible .pdf
Rods are responsible for dim light vision and cones are responsible for color vision. Cones are of
three types( three wavelengths) Red , green and blue. A severe color deficiency or color blind is
present when one or more of the cones light sensitive pigments is really wrong
Solution
Rods are responsible for dim light vision and cones are responsible for color vision. Cones are of
three types( three wavelengths) Red , green and blue. A severe color deficiency or color blind is
present when one or more of the cones light sensitive pigments is really wrong.
Navigation DesignBecause of navigation component present in the int.pdf
Navigation Design:Because of navigation component present in the interface we are able to
navigate through the system.How this navigation component does this navigation means it takes
the input as comand from the user and perform corresponding action and review the information.
Navigation component enable users know about their success or failture of their actions by
messages.The main goal of any system is providing user ease of access.That ease of access is
given by this navigation component.
A good navigation component does the things as the way user expecting actions from the system.
Basic principles of Navigation Design:
1)Prevent mistakes:Designing navigation controls is to prevent from the user making
mistakes.This mistake costs of time as well as frustation.Making continuous mistakes by user
discard the system.
2)Simplify recovery from mistakes:The system have Undo button that do the mistakes override
simply.
3)Use Consistent Grammar Order:
Input Design:It is all about designing screens so that user can enter information.We can say this
in different manner also that is providing forms so that can write or enter information.
Principles of input design:
1)Capture data at the source:capturing data in an electronic format.
2)Online versus Batch Processing:Online means enter the data individually while tranction is
going on
Batch Processing:Collecting or gathering data over some time and enter the data into the
system.
3)Minimize keystrokes:Whatever the system we design It is not allowed user to enter
information via keyboard
Because it costs time and also money.So,minimize this by providing drop down or
involving database calculation.
Output Design:Outputs are the reprots that system produces on the screen or paper or in web.
1. Manage information load:Output motive is not about providing all the information.So,we
have to mangage the information so that it is intended about that particular task or input.
2.Minimize Bias:
Solution
Navigation Design:Because of navigation component present in the interface we are able to
navigate through the system.How this navigation component does this navigation means it takes
the input as comand from the user and perform corresponding action and review the information.
Navigation component enable users know about their success or failture of their actions by
messages.The main goal of any system is providing user ease of access.That ease of access is
given by this navigation component.
A good navigation component does the things as the way user expecting actions from the system.
Basic principles of Navigation Design:
1)Prevent mistakes:Designing navigation controls is to prevent from the user making
mistakes.This mistake costs of time as well as frustation.Making continuous mistakes by user
discard the system.
2)Simplify recovery from mistakes:The system have Undo button that do the mistakes override
simply.
3)Use Consistent Grammar Order:
Input Design:It is all about designing screens so that user can ent.
One lone pair on each element.C OSolutionOne lone .pdf
One lone pair on each element.
:C
|||
:O
Solution
One lone pair on each element.
:C
|||
:O.
LHS = ( sin x - cos x )^2= sin^2 x + cos^2 x -2 sinx cos xBut si.pdf
LHS = ( sin x - cos x )^2
= sin^2 x + cos^2 x -2 sinx cos x
But sin^2 x + cos^2 x = 1 and 2 sin x cos x = sin 2 x
So
LHS = 1 - 2sin x cos x
= 1 - sin 2x = RHS
Hence pROved
Solution
LHS = ( sin x - cos x )^2
= sin^2 x + cos^2 x -2 sinx cos x
But sin^2 x + cos^2 x = 1 and 2 sin x cos x = sin 2 x
So
LHS = 1 - 2sin x cos x
= 1 - sin 2x = RHS
Hence pROved.
I think since its 9F which is 12 degrees above the all-time low. J.pdf
I think since its 9F which is 12 degrees /above/ the all-time low. Just do 9-12= Your answer and
then
Solution
I think since its 9F which is 12 degrees /above/ the all-time low. Just do 9-12= Your answer and
then.
FuelGauge class class FuelGauge { private int fuel; public.pdf
//FuelGauge class
class FuelGauge
{
private int fuel;
public static int max_capacity = 15;
public FuelGauge()
{
fuel = 0;
}
// parameterized constructor
public FuelGauge(int oil)
{
if (oil <= max_capacity)
{
fuel = oil;
}
else
{
fuel = max_capacity;
}
}
// return amount of fuel
public int getfuel()
{
return fuel;
}
// add fuel till max capacity is reached
public void addfuel()
{
if (fuel < max_capacity)
{
fuel++;
}
else
{
System.out.println(\"Maximum capacity reached, Burn some fuel\ \");
}
}
// burn the fuel
public void burnFuel()
{
if (fuel > 0)
{
fuel--;
}
else
{
System.out.println(\"No fuel left, Add some fuel\ \");
}
}
}
//Odometer class
class Odometer
{
// maximum mileage
public int max_mileage = 999999;
// miles per gallon
public int fuel_economy = 22;
private int startmileage;
private int mileage;
// FuelGauge object
private FuelGauge fuelgauge;
public Odometer(int mlg, FuelGauge fg) {
startmileage = mlg;
mileage = mlg;
fuelgauge = fg;
}
public int getMileage()
{
return mileage;
}
public void addMileage()
{
if (mileage < max_mileage)
{
mileage++;
}
else
{
mileage = 0;
}
int d = startmileage - mileage;
if (d % fuel_economy == 0)
{
fuelgauge.burnFuel();
}
}
}
// Driver class
public class Driver
{
public static void main(String[] args)
{
FuelGauge fuel = new FuelGauge();
Odometer odometer = new Odometer(0, fuel);
for (int i = 0; i < FuelGauge.max_capacity; i++)
{
fuel.addfuel();
}
// while there is fuel left
while (fuel.getfuel() > 0)
{
odometer.addMileage();
System.out.println(\"Mileage: \" + odometer.getMileage());
// amount of fuel left
System.out.println(\"Fuel left: \" + fuel.getfuel() + \" Gallons\ \");
}
}
}
/*
output:
Mileage: 1
Fuel left: 15 Gallons
Mileage: 2
Fuel left: 15 Gallons
Mileage: 3
Fuel left: 15 Gallons
Mileage: 4
Fuel left: 15 Gallons
Mileage: 5
Fuel left: 15 Gallons
Mileage: 6
Fuel left: 15 Gallons
Mileage: 7
Fuel left: 15 Gallons
Mileage: 8
Fuel left: 15 Gallons
Mileage: 9
Fuel left: 15 Gallons
Mileage: 10
Fuel left: 15 Gallons
Mileage: 11
Fuel left: 15 Gallons
Mileage: 12
Fuel left: 15 Gallons
Mileage: 13
Fuel left: 15 Gallons
Mileage: 14
Fuel left: 15 Gallons
Mileage: 15
Fuel left: 15 Gallons
Mileage: 16
Fuel left: 15 Gallons
Mileage: 17
Fuel left: 15 Gallons
Mileage: 18
Fuel left: 15 Gallons
Mileage: 19
Fuel left: 15 Gallons
Mileage: 20
Fuel left: 15 Gallons
Mileage: 21
Fuel left: 15 Gallons
Mileage: 22
Fuel left: 14 Gallons
Mileage: 23
Fuel left: 14 Gallons
Mileage: 24
Fuel left: 14 Gallons
Mileage: 25
Fuel left: 14 Gallons
Mileage: 26
Fuel left: 14 Gallons
Mileage: 27
Fuel left: 14 Gallons
Mileage: 28
Fuel left: 14 Gallons
Mileage: 29
Fuel left: 14 Gallons
Mileage: 30
Fuel left: 14 Gallons
Mileage: 31
Fuel left: 14 Gallons
Mileage: 32
Fuel left: 14 Gallons
Mileage: 33
Fuel left: 14 Gallons
Mileage: 34
Fuel left: 14 Gallons
Mileage: 35
Fuel left: 14 Gallons
Mileage: 36
Fuel left: 14 Gallons
Mileage: 37
Fuel left: 14 Gallons
Mileage: 38
Fuel left: 14 Gallons
Mileage: 39
Fuel left: 14 Gallons
Mileage: 40
Fuel lef.
More Related Content
More from angelnxcom
PV = n RT 15 x 0.068 x 1.01 = 0.75M x 0.0821 x .pdf
PV = n RT 15 x 0.068 x 1.01 = 0.75/M x 0.0821 x (273+25) Molar mass = M =
17.811 gms
Solution
PV = n RT 15 x 0.068 x 1.01 = 0.75/M x 0.0821 x (273+25) Molar mass = M =
17.811 gms.
HOkay. So. Youve got Ruthenium(II)..this is y.pdf
HOkay. So. You\'ve got: Ruthenium(II)..this is your transition metal Ammonia..a
common ligand (you\'ve just go to befamiliar with common ligands and their charges if they
have them) Chloride..this is your cation now. from this information you must deduce:
Ruthenium (R) has a charge of +2 which (because thequestion asks for a coordination
compound) binds with ammonia (NH3) (remembering that ammonia is a common ligand who
donates it\'selections to the metal..in other words it is a Lewisbase..). Ammonia ligands do not
have any charge(like I said before, you just have to know that,there is a chart in your text book
that will give you thisinformation). About the coordination number 6.. I\'ll get tothat in a second.
Notating a complex ion we use brackets like this: [complexion]. Butremember, R has a +2
charge. so the complex ion is an cationiccomplexion like this: [complexion]+2. so thisis were
chlorine comes in. chlorine is an anion meaning it has anegative charge, chlorine has a -1 charge.
so in order to neutralout the +2 charge that R gives the [complexion] we need two of them(Cl2).
so now we have everything to write this out in complex ionnotation. [R(NH3)6]Cl2 and now a
word about the coordination number 6. you don\'t really know that there are 6 ligands present but
it\'s agood assumption because 6 happens to be a very common coordinationfor ammonia
ligands. don\'t worry about this too much, just know howto count coordination numbers when
given the complex ion. Okay, NOW about naming this ridiculous thing. name the ligand first!
and then name other ligands in alphabeticalorder! ammonia in the form of a ligand is called
anammine...(you just have to memorize that table ofligand names which I\'m sure is in the book
as well). 6 ammines = hexamine. that\'s prettystraight forward unless you need to also memorize
numberprefixes. and then you name the metal it\'s boundtoo,Ruthenium(II). here is the tricky
part. add ate to the end of themetal IF and only IF the charge on the complex ion (before
itbecame neutral) was negative. (and in this case thepositive ion that would bind to the [complex
ion notation] would gobefore it, like lets say it was potassium, then you would writeK[complex
ion] and not [complex ion]K...oy...) but in thecase of our Ruthenium(II) we don\'t need to add
ate, because thecharge was originally positive. so now we have hexamineruthenium(II) and now
we add our little chloride... hexamineruthenium(II) chloride. (you don\'t have to say
dichloride,it\'s deducible that there are two..clearly, because i explained howto figure it out up at
the top, but hopefully my explanation wasclear too) (and in this same manner, if the [complex
ion] wasnegative at first and we bound a positive potassium to it we\'d putthe name potassium
before the complex ion like: potassiumwhateverwhatever(roman numeral).)
Solution
HOkay. So. You\'ve got: Ruthenium(II)..this is your transition metal Ammonia..a
common ligand (you\'ve just go to befamiliar with common l.
enthalpy of the given reaction is 4721 KJ since 4.pdf
enthalpy of the given reaction is 4721 KJ since 4 moles of boron are used enthalpy
change / mole boron = 4721/4 = 1180 KJ/mol
Solution
enthalpy of the given reaction is 4721 KJ since 4 moles of boron are used enthalpy
change / mole boron = 4721/4 = 1180 KJ/mol.
in.txt ( Suppose if we store the text file in DDrive then the path .pdf
in.txt ( Suppose if we store the text file in D:Drive then the path of the file pointing to this file is
D:\\\\in.txt )
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
3
38
39
40
41
42
43
44
45
46
45
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
___________________________________________________________________
ReadFile.java
import java.io.*;
import java.util.*;
public class ReadFile {
public static void main(String[] args) {
//Creating the Scanner class reference.
Scanner scanner=null;
try {
//Creating Scanner object by passing the file Object as input
scanner = new Scanner(new File(\"D:\\\\in.txt\"));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
//Creating the ArrayList Obkect
ArrayList integers = new ArrayList();
while (scanner.hasNext()) {
if (scanner.hasNextInt()) {
//Adding each element in the file to the ArrayList.
integers.add(scanner.nextInt());
}
}
//Create the Iterator Object.
Iterator i=integers.iterator();
//Iterator over the ArrayList
while(i.hasNext())
{
//Displaying the contents in the ArrayList.
System.out.println(i.next());
}
}
}
______________________________________________________________________
Output:
Enter File Name :D:\\\\in.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
3
38
39
40
41
42
43
44
45
46
4
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
_____________________________________________________________________
Solution
in.txt ( Suppose if we store the text file in D:Drive then the path of the file pointing to this file is
D:\\\\in.txt )
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
3
38
39
40
41
42
43
44
45
46
45
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
___________________________________________________________________
ReadFile.java
import java.io.*;
import java.util.*;
public class ReadFile {
public static void main(String[] args) {
//Creating the Scanner class reference.
Scanner scanner=null;
try {
//Creating Scanner object by passing the file Object as input
scanner = new Scanner(new File(\"D:\\\\in.txt\"));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
//Creating the ArrayList Obkect
ArrayList integers = new ArrayList();
while (scanner.hasNext()) {
if (scanner.hasNextInt()) {
//Adding each element in the file to the ArrayList.
integers.add(scanner.nextInt());
}
}
//Create the Iterator Object.
Iterator i=integers.iterator();
//Iterator over the ArrayList
while(i.hasNext())
{
//Displaying the contents in the ArrayList.
System.out.println(i.next());
}
}
}
_____________________.
The matrix, the dominant component in the landscape, is the most ext.pdf
The matrix, the dominant component in the landscape, is the most extensive and connected
landscape type, and it plays the dominant role in landscape functioning. A landscape is
composed typically of several types of landscape elements (usually patches). Of these, the matrix
is the most extensive and most connected landscape element type, and therefore plays the
dominant role in the functioning of the landscape .The characteristics of matrix structure are the
density of the patches (porosity), boundary shape, networks, and heterogeneity . For example, in
a large contiguous area of mature forest embedded with numerous small disturbance patches
(e.g., timber harvest patches), the mature forest constitutes the matrix element type because it is
greatest in areal extent, is mostly connected, and exerts a dominant influence on the area flora
and fauna and ecological processes.
Effect of matrix on metapopulation:
The core conceptual framework of metapopulation theory (Levins 1969) is that with-in-patch
dynamics are influenced not only by local extinction but also by colonization from elsewhere in
the landscape. When the landscape becomes more fragmented, colonization rates are reduced.
Then, within-patch species richness will decline and extinction risk at the scale of the entire
metapopulation will increase.
Matrix affects its communities by the following ways.
(1) altering dispersal and colonization rates, which may be reduced or enhanced, depending on
the characteristics of each species
(2) providing alternative habitat to existing species (for some species matrix habitat may be of
lower quality than the original habitat, while for others it may be of higher quality)
(3) as a source of novel invading species for fragments, as the matrix may provide habitat for
new species.
Holt a researcher studied \"Potential effects of the matrix on communities in patchy
environments \". He predicted that species that were abundant in the common habitat (matrix)
had potential to become more common members of local communities in the sparser habitat
(fragments) than species that did not inhabit the common habitat. This spillover effect had
potential to be important in determining which species make up local communities.
Solution
The matrix, the dominant component in the landscape, is the most extensive and connected
landscape type, and it plays the dominant role in landscape functioning. A landscape is
composed typically of several types of landscape elements (usually patches). Of these, the matrix
is the most extensive and most connected landscape element type, and therefore plays the
dominant role in the functioning of the landscape .The characteristics of matrix structure are the
density of the patches (porosity), boundary shape, networks, and heterogeneity . For example, in
a large contiguous area of mature forest embedded with numerous small disturbance patches
(e.g., timber harvest patches), the mature forest constitutes the matrix element type because it i.
The non-disjunctioning had occur in her mother.Turners syndrome .pdf
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia.
Solution
The non-disjunctioning had occur in her mother.
Turner\'s syndrome is caused by incomplete inheritance of or missing X chromosome, which
generally affects females. The poor growth and development of sexual organs can be seen early.
Haemophilia is an X-lined recessive disease. Females inherit one X-chromosome from the father
and another from the mother. The inheritance of haemophilia from the father ensures the
inheritance of X-chromosomes from the father. So, the nondisjunction occurred in females
during meiosis-I. The daughter received one X-chromosome from the father and no
chromosomes from the mother, resulting in 45, X karyotype (Turner\'s syndrome) and
haemophilia..
The answer isB. HNO2, NO2+Conjugate acids have an extra H+ comp.pdf
The answer is:
B. HNO2, NO2+
Conjugate acids have an extra H+ compared to a comjugate base.
Solution
The answer is:
B. HNO2, NO2+
Conjugate acids have an extra H+ compared to a comjugate base..
Team 1 should start with Rohnda because of her no loss in fight 1 an.pdf
Team 1 should start with Rohnda because of her no loss in fight 1 and Cat for 2 fight
Team 2 should start with Holly because of her no loss in fight 1 and Miesha for 2 fight
And team 2 has advantage over team 1
Solution
Team 1 should start with Rohnda because of her no loss in fight 1 and Cat for 2 fight
Team 2 should start with Holly because of her no loss in fight 1 and Miesha for 2 fight
And team 2 has advantage over team 1.
Rods are responsible for dim light vision and cones are responsible .pdf
Rods are responsible for dim light vision and cones are responsible for color vision. Cones are of
three types( three wavelengths) Red , green and blue. A severe color deficiency or color blind is
present when one or more of the cones light sensitive pigments is really wrong
Solution
Rods are responsible for dim light vision and cones are responsible for color vision. Cones are of
three types( three wavelengths) Red , green and blue. A severe color deficiency or color blind is
present when one or more of the cones light sensitive pigments is really wrong.
Navigation DesignBecause of navigation component present in the int.pdf
Navigation Design:Because of navigation component present in the interface we are able to
navigate through the system.How this navigation component does this navigation means it takes
the input as comand from the user and perform corresponding action and review the information.
Navigation component enable users know about their success or failture of their actions by
messages.The main goal of any system is providing user ease of access.That ease of access is
given by this navigation component.
A good navigation component does the things as the way user expecting actions from the system.
Basic principles of Navigation Design:
1)Prevent mistakes:Designing navigation controls is to prevent from the user making
mistakes.This mistake costs of time as well as frustation.Making continuous mistakes by user
discard the system.
2)Simplify recovery from mistakes:The system have Undo button that do the mistakes override
simply.
3)Use Consistent Grammar Order:
Input Design:It is all about designing screens so that user can enter information.We can say this
in different manner also that is providing forms so that can write or enter information.
Principles of input design:
1)Capture data at the source:capturing data in an electronic format.
2)Online versus Batch Processing:Online means enter the data individually while tranction is
going on
Batch Processing:Collecting or gathering data over some time and enter the data into the
system.
3)Minimize keystrokes:Whatever the system we design It is not allowed user to enter
information via keyboard
Because it costs time and also money.So,minimize this by providing drop down or
involving database calculation.
Output Design:Outputs are the reprots that system produces on the screen or paper or in web.
1. Manage information load:Output motive is not about providing all the information.So,we
have to mangage the information so that it is intended about that particular task or input.
2.Minimize Bias:
Solution
Navigation Design:Because of navigation component present in the interface we are able to
navigate through the system.How this navigation component does this navigation means it takes
the input as comand from the user and perform corresponding action and review the information.
Navigation component enable users know about their success or failture of their actions by
messages.The main goal of any system is providing user ease of access.That ease of access is
given by this navigation component.
A good navigation component does the things as the way user expecting actions from the system.
Basic principles of Navigation Design:
1)Prevent mistakes:Designing navigation controls is to prevent from the user making
mistakes.This mistake costs of time as well as frustation.Making continuous mistakes by user
discard the system.
2)Simplify recovery from mistakes:The system have Undo button that do the mistakes override
simply.
3)Use Consistent Grammar Order:
Input Design:It is all about designing screens so that user can ent.
One lone pair on each element.C OSolutionOne lone .pdf
One lone pair on each element.
:C
|||
:O
Solution
One lone pair on each element.
:C
|||
:O.
LHS = ( sin x - cos x )^2= sin^2 x + cos^2 x -2 sinx cos xBut si.pdf
LHS = ( sin x - cos x )^2
= sin^2 x + cos^2 x -2 sinx cos x
But sin^2 x + cos^2 x = 1 and 2 sin x cos x = sin 2 x
So
LHS = 1 - 2sin x cos x
= 1 - sin 2x = RHS
Hence pROved
Solution
LHS = ( sin x - cos x )^2
= sin^2 x + cos^2 x -2 sinx cos x
But sin^2 x + cos^2 x = 1 and 2 sin x cos x = sin 2 x
So
LHS = 1 - 2sin x cos x
= 1 - sin 2x = RHS
Hence pROved.
I think since its 9F which is 12 degrees above the all-time low. J.pdf
I think since its 9F which is 12 degrees /above/ the all-time low. Just do 9-12= Your answer and
then
Solution
I think since its 9F which is 12 degrees /above/ the all-time low. Just do 9-12= Your answer and
then.
FuelGauge class class FuelGauge { private int fuel; public.pdf
//FuelGauge class
class FuelGauge
{
private int fuel;
public static int max_capacity = 15;
public FuelGauge()
{
fuel = 0;
}
// parameterized constructor
public FuelGauge(int oil)
{
if (oil <= max_capacity)
{
fuel = oil;
}
else
{
fuel = max_capacity;
}
}
// return amount of fuel
public int getfuel()
{
return fuel;
}
// add fuel till max capacity is reached
public void addfuel()
{
if (fuel < max_capacity)
{
fuel++;
}
else
{
System.out.println(\"Maximum capacity reached, Burn some fuel\ \");
}
}
// burn the fuel
public void burnFuel()
{
if (fuel > 0)
{
fuel--;
}
else
{
System.out.println(\"No fuel left, Add some fuel\ \");
}
}
}
//Odometer class
class Odometer
{
// maximum mileage
public int max_mileage = 999999;
// miles per gallon
public int fuel_economy = 22;
private int startmileage;
private int mileage;
// FuelGauge object
private FuelGauge fuelgauge;
public Odometer(int mlg, FuelGauge fg) {
startmileage = mlg;
mileage = mlg;
fuelgauge = fg;
}
public int getMileage()
{
return mileage;
}
public void addMileage()
{
if (mileage < max_mileage)
{
mileage++;
}
else
{
mileage = 0;
}
int d = startmileage - mileage;
if (d % fuel_economy == 0)
{
fuelgauge.burnFuel();
}
}
}
// Driver class
public class Driver
{
public static void main(String[] args)
{
FuelGauge fuel = new FuelGauge();
Odometer odometer = new Odometer(0, fuel);
for (int i = 0; i < FuelGauge.max_capacity; i++)
{
fuel.addfuel();
}
// while there is fuel left
while (fuel.getfuel() > 0)
{
odometer.addMileage();
System.out.println(\"Mileage: \" + odometer.getMileage());
// amount of fuel left
System.out.println(\"Fuel left: \" + fuel.getfuel() + \" Gallons\ \");
}
}
}
/*
output:
Mileage: 1
Fuel left: 15 Gallons
Mileage: 2
Fuel left: 15 Gallons
Mileage: 3
Fuel left: 15 Gallons
Mileage: 4
Fuel left: 15 Gallons
Mileage: 5
Fuel left: 15 Gallons
Mileage: 6
Fuel left: 15 Gallons
Mileage: 7
Fuel left: 15 Gallons
Mileage: 8
Fuel left: 15 Gallons
Mileage: 9
Fuel left: 15 Gallons
Mileage: 10
Fuel left: 15 Gallons
Mileage: 11
Fuel left: 15 Gallons
Mileage: 12
Fuel left: 15 Gallons
Mileage: 13
Fuel left: 15 Gallons
Mileage: 14
Fuel left: 15 Gallons
Mileage: 15
Fuel left: 15 Gallons
Mileage: 16
Fuel left: 15 Gallons
Mileage: 17
Fuel left: 15 Gallons
Mileage: 18
Fuel left: 15 Gallons
Mileage: 19
Fuel left: 15 Gallons
Mileage: 20
Fuel left: 15 Gallons
Mileage: 21
Fuel left: 15 Gallons
Mileage: 22
Fuel left: 14 Gallons
Mileage: 23
Fuel left: 14 Gallons
Mileage: 24
Fuel left: 14 Gallons
Mileage: 25
Fuel left: 14 Gallons
Mileage: 26
Fuel left: 14 Gallons
Mileage: 27
Fuel left: 14 Gallons
Mileage: 28
Fuel left: 14 Gallons
Mileage: 29
Fuel left: 14 Gallons
Mileage: 30
Fuel left: 14 Gallons
Mileage: 31
Fuel left: 14 Gallons
Mileage: 32
Fuel left: 14 Gallons
Mileage: 33
Fuel left: 14 Gallons
Mileage: 34
Fuel left: 14 Gallons
Mileage: 35
Fuel left: 14 Gallons
Mileage: 36
Fuel left: 14 Gallons
Mileage: 37
Fuel left: 14 Gallons
Mileage: 38
Fuel left: 14 Gallons
Mileage: 39
Fuel left: 14 Gallons
Mileage: 40
Fuel lef.
More from angelnxcom (20)
PV = n RT 15 x 0.068 x 1.01 = 0.75M x 0.0821 x .pdf
PV = n RT 15 x 0.068 x 1.01 = 0.75M x 0.0821 x .pdf
enthalpy of the given reaction is 4721 KJ since 4.pdf
enthalpy of the given reaction is 4721 KJ since 4.pdf
in.txt ( Suppose if we store the text file in DDrive then the path .pdf
in.txt ( Suppose if we store the text file in DDrive then the path .pdf
The matrix, the dominant component in the landscape, is the most ext.pdf
The matrix, the dominant component in the landscape, is the most ext.pdf
The non-disjunctioning had occur in her mother.Turners syndrome .pdf
The non-disjunctioning had occur in her mother.Turners syndrome .pdf
The answer isB. HNO2, NO2+Conjugate acids have an extra H+ comp.pdf
The answer isB. HNO2, NO2+Conjugate acids have an extra H+ comp.pdf
Team 1 should start with Rohnda because of her no loss in fight 1 an.pdf
Team 1 should start with Rohnda because of her no loss in fight 1 an.pdf
Rods are responsible for dim light vision and cones are responsible .pdf
Rods are responsible for dim light vision and cones are responsible .pdf
Navigation DesignBecause of navigation component present in the int.pdf
Navigation DesignBecause of navigation component present in the int.pdf
One lone pair on each element.C OSolutionOne lone .pdf
One lone pair on each element.C OSolutionOne lone .pdf
LHS = ( sin x - cos x )^2= sin^2 x + cos^2 x -2 sinx cos xBut si.pdf
LHS = ( sin x - cos x )^2= sin^2 x + cos^2 x -2 sinx cos xBut si.pdf
I think since its 9F which is 12 degrees above the all-time low. J.pdf
I think since its 9F which is 12 degrees above the all-time low. J.pdf
FuelGauge class class FuelGauge { private int fuel; public.pdf
FuelGauge class class FuelGauge { private int fuel; public.pdf
Recently uploaded
أفضل 11 موقع لعمل اختبارات إلكترونية (Slide Decks).pdf
مع تطور التكنولوجيا، أصبحت أنظمة وأدوات الامتحانات الإلكترونية جزءاً أساسياً من التعليم الحديث. في هذا العرض، سنستعرض أفضل الأنظمة والأدوات التي تساعد المؤسسات التعليمية على تحسين عمليات الامتحان وتقديم تجربة تعليمية متميزة.
النزاع-ا لعرقي-والقبلي-في-دولة-جنوب-السودان-وانعكاساته-على-بناء-الدولة-4.pdf
النزاع-العرقي-والقبلي-في-دولة-جنوب-السودان-وانعكاساته-على-بناء-الدولة-4.pdf
Recently uploaded (6)
Panduan Pelaksanaan KCJ .pdf Kementerian Pendidikan Malaysia
Panduan Pelaksanaan KCJ .pdf Kementerian Pendidikan Malaysia
أفضل 11 موقع لعمل اختبارات إلكترونية (Slide Decks).pdf
أفضل 11 موقع لعمل اختبارات إلكترونية (Slide Decks).pdf
النزاع-ا لعرقي-والقبلي-في-دولة-جنوب-السودان-وانعكاساته-على-بناء-الدولة-4.pdf
النزاع-ا لعرقي-والقبلي-في-دولة-جنوب-السودان-وانعكاساته-على-بناء-الدولة-4.pdf