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f(x) = cos(x) Solution f(x) = cos(x).

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Matter is made up of electrically charge particle but the Constituents of objects have opposite charges, adding up to electric neutrality overall. Think of it in this way that two bring two nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough task. But in case of atoms or molecules that are the smallest entity of matter that can individually exist; it always contain equal number of proton and electron and hence are electrically neutral on a whole. Atom being neutral on whole there no force on other atoms Solution Matter is made up of electrically charge particle but the Constituents of objects have opposite charges, adding up to electric neutrality overall. Think of it in this way that two bring two nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough task. But in case of atoms or molecules that are the smallest entity of matter that can individually exist; it always contain equal number of proton and electron and hence are electrically neutral on a whole. Atom being neutral on whole there no force on other atoms.

Matter is made up of electrically charge particle but the Constituen.pdf

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it is a polynomial of degree 3Solutionit is a polynomial of de.pdf

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JUnit is a Regression Testing Framework used by developers to implement unit testing in Java, and accelerate programming speed and increase the quality of code. JUnit Framework can be easily integrated with either of the following Features of JUnit Test Framework JUnit test framework provides the following important features Fixtures Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a test fixture is to ensure that there is a well-known and fixed environment in which tests are run so that results are repeatable Test Suites A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and @Suite annotation are used to run the suite test. Test Runners Test runner is used for executing the test cases. Here is an example that assumes the test class TestJunit already exists. JUnit Classes JUnit classes are important classes, used in writing and testing JUnits. Some of the important classes are Assert Contains a set of assert methods. TestCase Contains a test case that defines the fixture to run multiple tests. TestResult Contains methods to collect the results of executing a test case. Solution JUnit is a Regression Testing Framework used by developers to implement unit testing in Java, and accelerate programming speed and increase the quality of code. JUnit Framework can be easily integrated with either of the following Features of JUnit Test Framework JUnit test framework provides the following important features Fixtures Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a test fixture is to ensure that there is a well-known and fixed environment in which tests are run so that results are repeatable Test Suites A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and @Suite annotation are used to run the suite test. Test Runners Test runner is used for executing the test cases. Here is an example that assumes the test class TestJunit already exists. JUnit Classes JUnit classes are important classes, used in writing and testing JUnits. Some of the important classes are Assert Contains a set of assert methods. TestCase Contains a test case that defines the fixture to run multiple tests. TestResult Contains methods to collect the results of executing a test case..

JUnit is a Regression Testing Framework used by developers to implem.pdf

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Intitially take a graph sheet and plot the points and the whole figure as it is. Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical representation. Apply corialis acceleration component at link 6. Solution Intitially take a graph sheet and plot the points and the whole figure as it is. Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical representation. Apply corialis acceleration component at link 6..

Intitially take a graph sheet and plot the points and the whole figu.pdf

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Incapsula Enterprise is the best mitigation service provider with the gold award winner and the other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud respectively With DDoS attacks growing in complexity and size daily, you need a DDoS protection service with a robust network and variety of mitigation techniques to thwart any attacks directed at your site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud offered the best protection. Incapsula’s growing global network and scrubbing capacities equip it with the size needed to prevent large-scale volumetric attacks,27 data centers are connected which are located all over the world, the service is prepared for even the most advanced DDoS attacks. In addition to enormous network capacity and a high number of data centers, Incapsula hosts a variety of mitigation techniques that intercept application layer, volumetric and protocol attacks. With the recent acquisition of Defense.net, F5 has adjusted its service to offer cloud-based mitigation technique, on-site and hybrid deployment methods,both hardware and the cloud is used, F5 Silverline Protection is equipped for sophisticated DDoS attacks. This service surpasses Incapsula in scrubbing capacity, it provides 2TB of bandwidth to filter, clean and remove infected traffic. While the service does not offer unlimited mitigation, this protection service can be used on-demand or always on, customizing it to fit the needs of any company. Third one comes Arbor Cloud which offers a large network and scrubbing capacity and have 1.14TB of bandwidth, this server prepare the system for specialised volumetric attacks.variety of protection techniques are provided by this service including IP blocking, rate limiting, automatic bot discernment and more. While the cloud-based mitigation techniques are impressive, Arbor is known for its hardware. The service is in a unique position, it\'s hardware is the challenge for it\'s competitors similar mitigation techniques is provided by all DDOS mitigation service providers to keep an eye on any attacks and constant monitoring, they differ in price, deployment methods and network size.It is not necessary that bigger mean better, the services that offer larger networks and more mitigation technologies allow for more customization that you employ the best protection service for your IT infrastructure. Solution Incapsula Enterprise is the best mitigation service provider with the gold award winner and the other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud respectively With DDoS attacks growing in complexity and size daily, you need a DDoS protection service with a robust network and variety of mitigation techniques to thwart any attacks directed at your site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud offered the best protection. Incapsula’s growing global network and scrubbing capacit.

Incapsula Enterprise is the best mitigation service provider with th.pdf

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Go with the most obvious one first... Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area (aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be spectrum 4. Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere near as broad as the acid. It just looks like a really fat absorption line, rather than a well... Anyway, the characteristic spectrum for the alcohol is spectrum 5. Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about the small absorption in the 1500-1700cm-1 area. Now for the fine details... The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester. Summary: Spectrum 1 = structure B Spectrum 2 = structure E Spectrum 3 = structure C Spectrum 4 = structure D Spectrum 5 = structure A Solution Go with the most obvious one first... Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area (aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be spectrum 4. Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere near as broad as the acid. It just looks like a really fat absorption line, rather than a well... Anyway, the characteristic spectrum for the alcohol is spectrum 5. Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about the small absorption in the 1500-1700cm-1 area. Now for the fine details... The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester. Summary: Spectrum 1 = structure B Spectrum 2 = structure E Spectrum 3 = structure C Spectrum 4 = structure D Spectrum 5 = structure A.

Go with the most obvious one first...Carboxylic Acids will have a.pdf

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Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health insurance and retirement benefits. The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts. Under the cafeteria plan, employees have the option of choosing from several different benefit packages. Employees can select the option of receiving some or all of the employer\'s nontaxable benefits or receiving cash or other taxable benefits. The benefits offered under cafeteria plans are health and group insurance, medical reimbursement schemes for non-insured expenses and vacation days. Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax deferred savings account. These accounts are formed by employers and helps the employee in meeting certain medical expenses that are not a part of employer\'s insurance plan. Solution Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health insurance and retirement benefits. The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts. Under the cafeteria plan, employees have the option of choosing from several different benefit packages. Employees can select the option of receiving some or all of the employer\'s nontaxable benefits or receiving cash or other taxable benefits. The benefits offered under cafeteria plans are health and group insurance, medical reimbursement schemes for non-insured expenses and vacation days. Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax deferred savings account. These accounts are formed by employers and helps the employee in meeting certain medical expenses that are not a part of employer\'s insurance plan..

Flexible benefit plans gives employees a choice between qualified be.pdf

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Environment: Let assume the environment is a grid of colored tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has a(X,Y) Position We use a function getTileColor(x,y) to know color of current tile(x,y position). ALGORITHM: TRESUREHUNT(x,y) 1. color=getColor(x,y) 2. IF color=YELLOW 3. Return \"tiles\"+x+y+\" Is a tresure\" 4. ELSE IF color=WHITE 5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1 6 ELSE IF color=BLUE 7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1 8. ELSE IF color=GREEN 9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1 10. ELSE IF color=BLACK 11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2 12. ELSE 13. Return \"Unsucessfull\" 12.END iF C++ CODE //for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains arbitary color values,0-BLACK //1-BLUE,2;GREEN,14;YELLOW,15-WHITE //accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile. // so x=x-1 and y=y #include #include void tresurehunt(int x,int y); int grid[][7]={ {1, 1, 1, 14,2, 0, 15}, {0, 1, 1, 1, 0, 0, 2}, {2, 2 ,1, 0, 15,14,1}, {15,15,14,0, 1, 2, 1}, {15,0, 0, 1, 2, 2, 14}, {14,14,14,15,0, 2, 2}, {15,15,15,1, 1, 2, 1}}; int main() { tresurehunt(4,4); return 0; } void tresurehunt(int x,int y) { //int color=getpixel(x,y); int color=grid[x][y]; if(color==YELLOW) { cout<<\"Tresure is in:\"< Solution Environment: Let assume the environment is a grid of colored tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has a(X,Y) Position We use a function getTileColor(x,y) to know color of current tile(x,y position). ALGORITHM: TRESUREHUNT(x,y) 1. color=getColor(x,y) 2. IF color=YELLOW 3. Return \"tiles\"+x+y+\" Is a tresure\" 4. ELSE IF color=WHITE 5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1 6 ELSE IF color=BLUE 7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1 8. ELSE IF color=GREEN 9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1 10. ELSE IF color=BLACK 11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2 12. ELSE 13. Return \"Unsucessfull\" 12.END iF C++ CODE //for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains arbitary color values,0-BLACK //1-BLUE,2;GREEN,14;YELLOW,15-WHITE //accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile. // so x=x-1 and y=y #include #include void tresurehunt(int x,int y); int grid[][7]={ {1, 1, 1, 14,2, 0, 15}, {0, 1, 1, 1, 0, 0, 2}, {2, 2 ,1, 0, 15,14,1}, {15,15,14,0, 1, 2, 1}, {15,0, 0, 1, 2, 2, 14}, {14,14,14,15,0, 2, 2}, {15,15,15,1, 1, 2, 1}}; int main() { tresurehunt(4,4); return 0; } void tresurehunt(int x,int y) { //int color=getpixel(x,y); int color=grid[x][y]; if(color==YELLOW) { cout<<\"Tresure is in:\"<.

EnvironmentLet assume the environment is a grid of colored tiles(.pdf

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Classification of organisms is based upon a number of physical and physiological features. The structure and organization of reproductive organs remains in important feature usually used to distinguish organisms. Some of the features can be discussed as below: 1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body is present or not. Presence of a fleshy or modified cover around the seed is a characteristic feature of angiosperms whereas a naked seed located directly above a vegetative structure of a plant is a feature of gymnosperms. 2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the leaf blades of lower and higher ferns. Their arrangement and mode of release determines the nature, frequency and ease of sexual reproduction in these lower plants. 3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for sexual reproduction. The mode of transfer of pollen determines the geographical range of pollination in plants. Sulphur shower is a critical example for sexual reproduction in pines (gymnosperms) which distinguishes them from other trees. 4. Sporocarp: A sporocarp or a fruiting body is a bulging of the vegetative extension from the main structure of a lower fungi which contains a group of cells which either follow meiotic division or mitotic division and serve as the reproductive structures. Their growth and nature can be monitored microscopically and the fungus can be thus classified appropriately. Some algae also contain a unique localization and arrangement of haploid cells called spores which can be present in singles or doublets and thus represent a very unique discrimination from other organisms. These, these set of information provide an insight into usability of features of a reproductive structure for classification of organisms. Solution Classification of organisms is based upon a number of physical and physiological features. The structure and organization of reproductive organs remains in important feature usually used to distinguish organisms. Some of the features can be discussed as below: 1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body is present or not. Presence of a fleshy or modified cover around the seed is a characteristic feature of angiosperms whereas a naked seed located directly above a vegetative structure of a plant is a feature of gymnosperms. 2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the leaf blades of lower and higher ferns. Their arrangement and mode of release determines the nature, frequency and ease of sexual reproduction in these lower plants. 3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for sexual reproduction. The mode of transfer of pollen determines the geographical range of pollination in plants. Sulphur shower is a critical example for.

Classification of organisms is based upon a number of physical and p.pdf

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Assets=Liabilities+Paid in capital+Retained earnings1..pdf

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Hypothesis Test for population variance If we have a sample fro.pdf

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ANSWER: Fundamental configurations of Windows printer Deployments: The Following sections describe four fundamental configuraions that are the basis of most Windows Printer Deployments 1.Direct Printing 2.Locally Attached printer Sharing 3.Network-attached Printing 4.Network-Attached printer sharing you can scale these configurations up to accommodate a network of virtually any size. 1.DIRECT PRINTING: The simplest print architecture consists of one print device connected to one computer, also known as a locally attached print device, as shown in Figure 2-13. When you connect a print device directly to a Windows Server 2012 R2 computer and print from an application running on that system, the computer supplies the printer, printer driver, and print server functions. 2.LOCALLY ATTACHED PRINTER SHARING: In addition to printing from an application running on that computer, you can also share the printer (and the print device) with other users on the same network. In this arrangement, the computer with the locally attached print device functions as a print server. Figure 2-14 shows the other computers on the network, which are known as the print clients. In the default Windows Server 2012 R2 printer-sharing configuration, each client uses its own printer and printer driver. As before, the application running on the client computer sends the print job to the printer and the printer driver renders the job, based on the capabilities of the print device. The main advantage of this printing arrangement is that multiple users, located anywhere on the network, can send jobs to a single print device connected to a computer functioning as a print server. The downside is that processing the print jobs for many users can impose a significant burden on the print server. Although any Windows computer can function as a print server, you should use a workstation for this purpose only when you have no more than a handful of print clients to support or you have a very light printing volume. 3.NETWORK-ATTACHED PRINTING: The printing solutions discussed thus far involve print devices connected directly to a computer using a USB or other port. Print devices do not necessarily have to be attached to computers, however. You can connect a print device directly to the network instead. Many print device models are equipped with network interface adapters, enabling you to attach a standard network cable. Some print devices have expansion slots into which you can install a network printing adapter you have purchased separately. Finally, for print devices with no networking capabilities, standalone network print servers are available, which connect to the network and enable you to attach one or more print devices. Print devices so equipped have their own IP addresses and typically have an embedded web-based configuration interface. With network-attached print devices, the primary deployment decision the administrator must make is to decide which computer will function as th.

ANSWERFundamental configurations of Windows printer Deployments.pdf

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Since HCl is strong acid , it dissociates fully t.pdf

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/** * @author * */ public class Person { String sname, iname; int fid, sid; } /** * @author * */ public class Student extends Person { String sname; int sid; int no_of_credits, tot_grade_pts; /** * @param sname * @param sid */ public Student(String sname, int sid) { this.sname = sname; this.sid = sid; } /** * @return */ public String get_name() { return sname; } public int get_sid() { return sid; } public boolean two_equal(int sid) { if ((this.sid) == sid) return true; else return false; } public void set_credits(int credits) { no_of_credits = credits; } public int get_credits() { return no_of_credits; } public void set_tot_gpts(int gpts) { tot_grade_pts = gpts; } public int get_tgpts() { return tot_grade_pts; } public float get_GPA() { return ((tot_grade_pts) / (no_of_credits)); } } /** * @author * */ public class Instructor extends Person { String iname; int fid; String dept; /** * @param iname * @param fid */ public Instructor(String iname, int fid) { this.iname = iname; this.fid = fid; } /** * @param dept */ void set_dept(String dept) { this.dept = dept; } /** * @return */ public String get_dept() { return dept; } } import java.util.ArrayList; /** * @author * */ public class Course { String cname, instructor_name; int creg_code, max_students, no_of_students; ArrayList reg_students = new ArrayList(); /** * @param cname * @param creg_code * @param max_students */ Course(String cname, int creg_code, int max_students) { this.cname = cname; this.creg_code = creg_code; this.max_students = max_students; } /** * @param instructor_name */ void set_instructor(String instructor_name) { this.instructor_name = instructor_name; } /** * method to search student id * * @param sid * @return */ public boolean search_student(int sid) { if (reg_students.contains(sid)) return true; else return false; } /** * method to add student id * * @param sid */ public void add_student(int sid) { try { if ((reg_students.size()) == max_students) { throw new MyException(); } else reg_students.add(sid); } catch (Exception e) { System.out.println(\"Course has maximum students\"); } } /** * method to remove student id * * @param sid */ public void rem_student(int sid) { try { if (reg_students.contains(sid)) reg_students.remove(new Integer(sid)); else throw new MyException(); } catch (Exception e) { System.out.println(\"NO STUDENT FOUND WITH THE GIVEN id\"); } } } /** * @author * */ public class MyException extends Exception { public MyException() { // TODO Auto-generated constructor stub super(\" Invalid Data\"); } } import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Registrar { /** * @param args * @throws NumberFormatException * @throws IOException */ public static void main(String args[]) throws NumberFormatException, IOException { Course c1 = new Course(\"default_course\", 123, 10); int ch, regcode, max_stu, sid; String cname, sname; BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); do { System..

@author public class Person{ String sname, .pdf

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no even covalent coupunds also relese heat thos.pdf

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organic layer note this ammonium salt (itself is.pdf

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Glucose is more sluble in water because it has moreno . of -OH groups on it that can hydrogen bond withwater molecules. C-OH-----------H-O-H But NaCl is an ionic compound it can dissociates to Na+ & Cl-ions only, it will not form hydrogen bonding like glucose. Solution Glucose is more sluble in water because it has moreno . of -OH groups on it that can hydrogen bond withwater molecules. C-OH-----------H-O-H But NaCl is an ionic compound it can dissociates to Na+ & Cl-ions only, it will not form hydrogen bonding like glucose..

Glucose is more sluble in water because it has mo.pdf

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compounds 1 and 3 are the same be cause they are .pdf

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C) Its a double displacement reaction .pdf

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B)light interacting with a metal and producing a .pdf

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