Given that a horizontal platform 6m in diameter revolves so that a point on its rim moves 6.50 m/s. We need to find the angular speed in rpm.
Step 1:We know that Angular speed of a body is given byω = v/r Whereω is the angular velocity of the body v is the velocity of the body r is the radius of the circular path of the body
Step 2:Diameter of the platform, D = 6m Radius of the platform,
r = D/2
= 6/2
= 3 m Velocity of the point on the rim of the platform, v = 6.50 m/s
Step 3: Let's plug in the values in the formula of angular velocity
ω = v/rω
= 6.50/3ω
= 2.1667 rad/s
Step 4:We need to convert radians per second to rpm (revolutions per minute)We know that 2π radians = 1 revolution or 1 rotation 1 revolution = 2π radians Angular velocity in revolutions per minute (rpm),ω = (2.1667)/(2π) * (60) = 20.71 rpm Therefore, the angular speed of the platform in rpm is 20.71 rpm.
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Without using malloc or free, and using as little stack space as possible, * write a standard C program to insert an unsigned long into a linked list * ordered from smallest to largest. * * You may include stddef.h and stdio.h but do not use any other libraries. * * You may take as much time as you like, but you can expect a full * implementation to take you about an hour. */
Here's an example of a C program for inserting an O long into a linked list ordered from smallest to largest.
It is written without using malloc or free and uses as little stack space as possible:```#include #include struct node { unsigned long value; struct node *next; }; void insert(struct node **head, unsigned long value) { struct node *new = (struct node *)alloca (sizeof(struct node)); new->value = value; while (*head && (*head)->value < value) { head = &(*head)->next; } new->next = *head; *head = new; } void print(struct node *head) { while (head) { printf("%lu ", head->value); head = head->next; } print f("\n"); } int main() { struct node *head = NULL; insert(&head, 3); insert(&head, 2); insert(&head, 1); print(head); return 0; }```The insert function takes a pointer to the head of the list and the value to insert. It creates a new node on the stack using alloca, and then iterates through the list until it finds the correct position to insert the new node. It then updates the head pointer and the new node's next pointer.
The print function simply iterates through the list and prints each value. The main function demonstrates how to use the insert and print functions. It creates an empty list, inserts three values in reverse order, and then prints the list.
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Can you write a shell script to compute the multiplication of the first n integer numbers? The n is passed as the first argument. Hint: use some loop structure, and keep increasing the i value to n.
A shell script to calculate the multiplication of the first n integer numbers can be written using a loop structure and by increasing the i value to n.
To compute the multiplication of the first n integer numbers using a shell script, we can use a loop structure and increment the value of i in each iteration until it reaches n. Below is the shell script to accomplish this:
#!/bin/bash
result=1
for ((i=1;i<= $1;i++))
do
result=$((result*i))
done
echo "
The multiplication of the first $1 integer numbers is: $result".
The script first initializes the result variable to 1. Then, it uses a for loop to iterate from 1 to n (which is passed as the first argument to the script). In each iteration, the value of i is multiplied to the current value of the result variable and the result is stored back in the result variable. After the loop completes, the script prints the final result using the echo command.
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How can I return a rounded amount of 3 decimal places in C# if I have
return(currentGallons/Capacity()) * 100;?
In order to return a rounded amount of 3 decimal places in C#, you can make use of the Math.
Round() method.
Here's how you can modify your code to achieve this:
double percentage = (current Gallons / Capacity()) * 100;
double rounded Percentage = Math.Round (percentage, 3);
return rounded Percentage;
Here, we are first calculating the percentage by dividing current Gallons by Capacity() and then multiplying by 100. We then use the Math.Round() method to round this value to 3 decimal places and store it in the variable rounded Percentage. Finally, we return the rounded percentage as the result of the method.
The explanation of the code snippet above is given as below:
You can modify your code as shown below to achieve this:
double percentage = (current Gallons / Capacity()) * 100;
double rounded Percentage = Math.Round(percentage, 3);
return rounded Percentage;
Explanation: Here, we are first calculating the percentage by dividing current Gallons by Capacity() and then multiplying by 100. We then use the Math.Round() method to round this value to 3 decimal places and store it in the variable rounded Percentage. Finally, we return the rounded percentage as the result of the method.
Conclusion: In this way, you can easily return a rounded amount of 3 decimal places in C# using the Math.Round() method.
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Write a function called main which takes three parameters: a string to be encrypted, an integer shift value, and a code group size. Your method should return a string which is its cyphertext equivalent. The main method should start by asking the user to input: • Message to work with • Distance to shift the alphabet • Number of letters to group on • Whether they want to encrypt or decrypt the message
The function called main which takes three parameters are as follows:
def main (message, shift, group_size):
ciphertext = ""
for char in message:
if char.isalpha():
if char.isupper():
shifted_char = chr ((ord (char) - 65 + shift) % 26 + 65)
else:
shifted_char = chr ((ord (char) - 97 + shift) % 26 + 97)
ciphertext + = shifted_char
else:
ciphertext += char
grouped_ciphertext = ""
for i in range(0, len(ciphertext), group_size):
group = ciphertext[i:i+group_size]
grouped_ciphertext += group + " "
return grouped_ciphertext.strip()
How can I encrypt or decrypt a message using a shift cipher and code groups?The provided main function takes three parameters: the message to be encrypted or decrypted, the shift value which determines the distance to shift the alphabet, and the group_size which specifies the number of letters to group together.
The function returns the ciphertext equivalent of the input message. To encrypt or decrypt the message, the function applies the shift cipher algorithm. For each character in the message, the function checks if it is an alphabetic character.
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For each student in the GRADE table, list the student's first name and last name, the section number, the corresponding course number and course name, the grade, and the section instructor's name. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BI V S Paragraph Arial 10pt % 三三三 v I. Q For each student in the GRADE table, list the student's first name and last name, the section number, the corresponding course number and course name, the grade, and the section instructor's name. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BI V S Paragraph Arial 10pt % 三三三 v I. Q
For listing student's first name, last name, section number, corresponding course number and course name, grade, and section instructor's name, we have to use INNER JOIN, ON clause, and SELECT statement.
Inner join, also known as the join clause, is used to combine rows from two or more tables based on a related column between them. In this case, we have to list down the first name, last name, section number, corresponding course number and course name, grade, and section instructor's name for every student in the GRADE table.
We can achieve this using the INNER JOIN clause, ON clause, and SELECT statement to select the desired columns from the required tables. For instance, we can use the INNER JOIN clause on the GRADE, STUDENT, COURSE, and SECTION tables based on the related columns such as the student id, section id, and course id. The ON clause can be used to link the tables by the related columns. Finally, the SELECT statement can be used to choose the desired columns for outputting.
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How many types of Memory Operand Addressing Modes are there in the 8088/8086 Microprocessor and what are they called?
There are 5 types of Memory Operand Addressing Modes in the 8088/8086 Microprocessor. The memory operand addressing modes are named as follows: Direct Addressing ModeThe register points to the memory address in the direct addressing mode. The direct addressing mode is a simple addressing mode. In a direct addressing mode, the operand is stored at the memory location that is specified by an 8-bit or 16-bit address.
Displacement Addressing Mode A register contains a pointer to the memory location, and a displacement specifies an offset from the location indicated by the register. In the displacement addressing mode, the operand is located at the memory location that is equal to the sum of the contents of the specified register and an 8-bit or 16-bit displacement.
Indexed Addressing Mode The indexed addressing mode is similar to the displacement addressing mode. In this mode, an index register is added to a displacement to form a memory address. The indexed addressing mode is used to access the memory array elements.
Base-indexed Addressing Mode The address of a memory operand in this mode is computed by adding a displacement to the sum of the contents of two registers, one of which is called the index register and the other the base register. The base-indexed addressing mode is also called the base pointer addressing mode.
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** Using Scala ** Must use Scala - no other languages are acceptable.
a. Create a class Employee with fname, lname, and daily_rate (define daily_rate as float). Within this class create named function that calculates the wage by multiplying the daily rate to the hours worked.
b. Create a class SalesRep with fname, lname, daily_rate and commission_pct. This SalesRep class extends Employee and has a named function that calculates the commission that is the sales multiplied by the commission_pct.
c. Instantiate the Employee class – call this office_manager.
d. Instantiate the SalesRep class – call this sales_rep
e. Print a statement like: Office Manager, John Doe earns $60000.00 per year
f. Print a statement like: Sales Rep, Ricky Johnson earns $50000 per year
I have instantiated the two classes `Employee` and `SalesRep` and printed the required statements as per the problem statement.
Here is the solution to the given problem statement using Scala:Code:```scala
class Employee(val fname:String, val lname:String, val daily_rate:Float){
def calculateWage(hours:Float): Float = daily_rate * hours
}
class SalesRep(override val fname:String, override val lname:String, override val daily_rate:Float, val commission_pct:Float) extends Employee(fname, lname, daily_rate){
def calculateCommission(sales:Float): Float = sales * commission_pct
}
val office_manager = new Employee("John", "Doe", 100.0f)
val sales_rep = new SalesRep("Ricky", "Johnson", 50.0f, 0.2f)
println(s"Office Manager, ${office_manager.fname} ${office_manager.lname} earns $${office_manager.calculateWage(8.0f) * 365} per year")
println(s"Sales Rep, ${sales_rep.fname} ${sales_rep.lname} earns
$${(sales_rep.calculateWage(8.0f) * 365) + (sales_rep.calculateCommission(1000000))} per year")```
In the above code, I have created two classes `Employee` and `SalesRep`. `Employee` class contains the `daily_rate` and `calculateWage()` function that takes hours as a parameter and returns the wage of an employee.
`SalesRep` class extends the `Employee` class and contains an additional parameter `commission_pct` along with the `calculateCommission()` function that returns the commission of an employee based on the sales made.Then I have instantiated the two classes `Employee` and `SalesRep` and printed the required statements as per the problem statement.
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Given is the following NFA AN (QN, E, SN.AN, FN). 90 € b 91 a 93 a, b 92 a Following the construction presented in class, to prove that for every NFA there exists an equivalent DFA, give Qp op qp. and Fp for equivalent DFA Ap=(QD,E,Sp.9D. Fp) with L(Ap)=L(AN).
Qp = {0,1,2,3,4,5}, Sp = {a, b}, qp=0 and Fp = {4, 5}. These are the required values of Qp, Sp, qp, Fp for the equivalent DFA Ap=(QD,E,Sp.9D. Fp) with L(Ap)=L(AN).
Given is the following NFA AN (QN, E, SN.AN, FN). 90 € b 91 a 93 a, b 92 a Following the construction presented in class, to prove that for every NFA there exists an equivalent DFA, give Qp op qp and Fp for equivalent DFA Ap=(QD,E,Sp.9D. Fp) with L(Ap)=L(AN).
Firstly, we construct the table for ε-closure which is as follows: NFA state ε-closure0 0, 13 34 35 6, 57 8Now, we construct DFA which is as follows:
State State in AN Final0 {0, 1, 3} no1 {2, 4} no2 {5} no3 {6, 7} no4 {8} yes5 φ yes State table for DFA is as follows:States b a 0 {0, 1, 3} {2, 4}1 {2, 4} {2, 4}2 {5} {6, 7}3 {6, 7} {2, 4}4 {8} {5}5 φ φ
Now, we can conclude that Qp = {0,1,2,3,4,5} which are the states of equivalent DFA and qp=0 is the start state of equivalent DFA.So, Fp = {4,5} are the final states of the equivalent DFA. The equivalent DFA is as follows:
Therefore, we have Qp = {0,1,2,3,4,5}, Sp = {a, b}, qp=0 and Fp = {4, 5}. Hence, these are the required values of Qp, Sp, qp, Fp for the equivalent DFA Ap=(QD,E,Sp.9D. Fp) with L(Ap)=L(AN).
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Overhead service conductors shall have a vertical clearance of not less than Select one: Oa. 900 mm Ob. 2.5 m Oc. 3.0 m Od. 3.7 m above the roof surface.
The vertical clearance for overhead service cables must be at least 2.5 meters. Therefore, choice (B) is right.
Radio frequency (RF) signals are transferred over coaxial cables in the cable television system, or in more contemporary versions, light pulses are transmitted through fiber-optic cables to provide television content to viewers. Unlike broadcast television (also referred to as terrestrial television),
where the television signal is transmitted over the air by radio waves and received by an antenna attached to the television, or satellite television, when a communications satellite in orbit of the Earth broadcasts radio waves that are picked up by a satellite dish antenna on the roof and sent as the television feed. the FM radio spectrum,
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An overhang beam is shown below. q=430 N/m, and F=610 N. Do the following: A. Draw the shear and moment diagrams. Make sure you show the shape properly. B. Label the maximum positive and maximum negative values on the shear diagram, C. Label the maximum positive and maximum negative values on the moment diagram. Maximum Shear:
The maximum positive and negative values on the shear and moment diagrams can be labeled based on the magnitudes of the shear force and bending moment at different locations along the beam.
To draw the shear and moment diagrams for the given overhang beam, one need to determine the reactions at the supports and analyze the distribution of forces along the beam.
Here given that:
q = 430 N/m (distributed load)
F = 610 N (applied load)
First, the reactions at the supports is determined.
Taking moments about the fixed support:
Sum of moments = 0
-610(4) + R2(7) = 0
R2 = 1740 N
Taking vertical forces equilibrium:
Sum of vertical forces = 0
R1 + R2 - (430)(7) - 610 = 0
R1 = 2360 N
So, R1 = 2360 N and R2 = 1740 N.
Step 2:
At the left end (support R1), the shear force is equal to R1 = 2360 N.
Moving along the beam, there is a downward distributed load of 430 N/m, which causes a linear decrease in shear force.
At the overhang support (support R2), the shear force suddenly drops by 610 N due to the applied load F.
Step 3:
At the left end (support R1), the bending moment is zero.
Moving along the beam, there is an increasing positive moment due to the distributed load q.
At the overhang support (support R2), there is a sudden positive moment caused by the applied load F.
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A pump with an efficiency of 88.8% lifts 3.2 m³ of water per minute to a height of 26 metres. An electrical motor having an efficiency of 87.7% drives the pump. The motor is connected to a 220-V dc supply. Assume that 1 m³ of water has a mass of 1000 kg. 2.1 Calculate the input power of the motor. [8] 2.2 Calculate the current drawn from the source. [2] 2.3 Calculate the electrical energy consumed in kWh when the motor runs at this load for eight hours. [2] For the toolhar i
Answer:2.1: Calculation of input power of the motor:First, we need to find out the volume of water lifted per second.Volume of water lifted per second = (3.2 / 60) m³/s = 0.0533 m³/sNow, the mass of 1 m³ of water is 1000 kg.Power required to lift water = weight x speed of liftingPower required to lift water = mg x h/t= (1000 x 9.8) x 26/60= 4266.67 watts or 4.2667 kWSo, the output power of the pump will be = Power required to lift water / Efficiency of the pump= 4.2667 / 0.888= 4.8 kW
Therefore, the input power of the motor is:Input power of the motor = output power of the pump / Efficiency of the motor= 4.8 / 0.877= 5.469 kW2.2: Calculation of the current drawn from the source:Formula to calculate the current drawn from the source is:I = P / VWhere I is the currentP is the powerV is the voltageHere, P = 5.469 kWV = 220 VSo, I = 5.469 / 220= 0.02486 A2.3: Calculation of the electrical energy consumed in kWh when the motor runs at this load for eight hours:The total time for which the motor will run = 8 hours = 8 x 60 x 60 s= 28,800 secondsSo, the energy consumed = Power x Time= 5.469 x 28,800= 157,699.2 Joules= 43.8 kWhTherefore, the electrical energy consumed in kWh when the motor runs at this load for eight hours is 43.8 kWh.Explanation:In the first step, we calculated the volume of water lifted per second, which was 0.0533 m³/s. Then, we calculated the power required to lift the water, which was 4.2667 kW.We found out the output power of the pump using the formula "output power of the pump = Power required to lift water / Efficiency of the pump". It was found to be 4.8 kW.
To calculate the input power of the motor, we used the formula "Input power of the motor = output power of the pump / Efficiency of the motor". We got 5.469 kW as the input power of the motor.Then, we calculated the current drawn from the source using the formula "I = P / V". It was 0.02486 A.In the final step, we calculated the electrical energy consumed in kWh when the motor runs at this load for eight hours. The total time for which the motor will run was 28,800 seconds. Therefore, the energy consumed was 43.8 kWh.
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Pipes are a mechanism for Inter Process Communication. Pipes have a limitation that they cannot be accessed from outside the process that created it. Which IPC mechanism overcomes this limitation while offering similar functionality? Explain briefly.
There are several IPC mechanisms available in the computer architecture that has different advantages and disadvantages. Signals, pipes, shared memory, and sockets are the most commonly used IPC mechanisms. Among these IPC mechanisms, socket provides the solution to overcome the limitation of the pipe that it cannot be accessed from outside the process that created it.
IPC, or inter-process communication, is the term used to describe a collection of communication techniques between multiple threads in one or more processes. The primary purpose of IPC is to allow data to be shared between multiple processes.Pipes are one of the oldest and most commonly used IPC mechanisms. However, a pipe has a significant limitation: it cannot be accessed from outside the process that created it.Socket-based IPC is another widely used mechanism for interprocess communication that overcomes this limitation while providing similar functionality.
Sockets are endpoints of a two-way communication link that can be accessed by multiple processes concurrently.Socket-based IPC has some advantages over pipe-based IPC. One of the main benefits of socket-based IPC is that sockets can be accessed by multiple processes concurrently. Additionally, sockets provide a more robust and flexible mechanism for communication than pipes because they can be used to communicate between processes on different hosts.In summary, while pipes are an effective IPC mechanism, they have limitations, including the inability to access them from outside the process that created them.
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Simulation: Write and simulate a MIPS assembly-language routine that uses a procedure called LISTADD to add the elements of an integer array. The array, along with its length, is stored in memory words. Use the array: [23, -2, 45, 67, 89, 12, -100, 0, 120, 6]. The arguments passed to LISTADD are the base address of the array (passed through register $a0) and the length of the array (passed through register $al). The procedure LISTADD returns the result in register $v0. The main procedure saves the result in memory and prints it on console preceded by an appropriate text message. Procedure call should look like the following: jal LISTADD LISTADD: jr Sra
Here is a sample program in MIPS assembly language that populates an array with Fibonacci numbers based on the given requirements:
# Initialize array size and address
li $s0, 8 # size of array
li $s1, 5000 # address of first element
# Populate array with Fibonacci numbers
li $t0, 0 # first Fibonacci number
sw $t0, 0($s1) # store first number in array
addi $s1, $s1, 4 # move to next array element
li $t1, 1 # second Fibonacci number
sw $t1, 0($s1) # store second number in array
addi $s1, $s1, 4 # move to next array element
# Calculate and store remaining Fibonacci numbers in array
addi $t2, $t0, $t1 # calculate next Fibonacci number
addi $s0, $s0, -2 # decrement remaining elements counter
loop:
sw $t2, 0($s1) # store Fibonacci number in array
addi $s1, $s1, 4 # move to next array element
addi $t0, $t1, 0 # shift numbers to calculate next Fibonacci number
addi $t1, $t2, 0
addi $t2, $t0, $t1
addi $s0, $s0, -1 # decrement remaining elements counter
bne $s0, $zero, loop # loop until all elements are filled
# End program
li $v0, 10 # exit syscall
syscall
Thus, this program initializes the size of the array to 8 and the starting address of the array to 5000.
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Write a comprehension explanation of what is transport and application layer systems, their role and their importance in the network system you have designed.
Transport layer is the fourth layer in the seven-layer Open Systems Interconnection (OSI) reference model. This layer is responsible for providing a reliable end-to-end communication channel by dividing data into smaller segments and ensuring that each segment is sent and received correctly.
The transport layer is responsible for connection-oriented or connectionless communication, message segmentation, and reassembly. This layer is essential because it ensures that data is transferred from one end of the network to the other end in a timely, reliable, and efficient manner. The transport layer is responsible for protocols like TCP and UDP.
On the other hand, the application layer is the seventh and highest layer of the OSI model. This layer interacts with the user and provides services that support end-to-end communication between network applications. The application layer includes protocols like HTTP, FTP, SMTP, and DNS. The application layer is responsible for data exchange between the user's device and the server and is used to interact with the system.
The application layer protocols are responsible for data storage and retrieval, directory services, email exchange, web browsing, and other network-related tasks. The application layer protocols are very important because they provide users with a way to access the network.
They also provide security, data exchange, and data storage. These protocols are responsible for the data exchange and communication between the user's device and the network system.In summary, the transport and application layer systems play a significant role in the network system.
They ensure that data is transmitted correctly, reliably, and securely from the user's device to the network system. The transport layer divides the data into smaller segments and ensures that each segment is sent and received correctly. The application layer provides a way for users to access the network and communicate with it. These layers work together to create a network system that is efficient, reliable, and secure.
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Problem 1 (10 Points) What is the height of the shortest binary tree that contains 21 nodes? Is this tree full? Is it balanced?
The height of the shortest binary tree that contains 21 nodes is four, with a full tree. A binary tree is a tree structure where each node can have at most two children, known as the left and right children. The root node of the tree is the node at the top of the tree, while the leaf nodes are the nodes at the bottom of the tree that have no children.
A full binary tree is a tree in which every node has either zero or two children, but never only one child. A balanced binary tree is a binary tree where the left and right subtrees' heights differ by no more than one node. If the height of the tree is h, the number of nodes in the tree is 2^(h+1)-1.
If we have 21 nodes in a binary tree, then 2^(h+1)-1 = 21. By solving the equation, h = 4.
So, the height of the shortest binary tree that contains 21 nodes is four.
Since 2^(4+1)-1 = 31, the tree is not full because it has only 21 nodes.
In addition, since the left and right subtrees' heights differ by more than one, the tree is not balanced.
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1.1 Describe Boosting, Bagging and Ensemble Learning in the Machine Learning context.
1.2 What is the difference between K-Means, Support Vectors Machine, and KNN algorithms?
1.3 Briefly explain why the dimensionality reduction is important in Machine Learning.
1.1 Boosting, Bagging and Ensemble Learning are machine learning techniques used to improve the accuracy of a model. Boosting is a method of creating a strong classifier from a set of weak classifiers.
1.2 K-Means, Support Vector Machines (SVM), and K-Nearest Neighbors (KNN) algorithms are used in machine learning for clustering and classification.K-Means is an unsupervised learning algorithm used for clustering. The goal of K-Means is to divide a set of observations into K clusters based on their similarity. The algorithm assigns each observation to the nearest cluster based on the distance between them.SVM is a supervised learning algorithm used for classification and regression analysis. The goal of SVM is to find the optimal hyperplane that separates the different classes.
1.3 Dimensionality reduction is important in machine learning because it can improve the accuracy and efficiency of a model. The more features a data set has, the more complex the model becomes.
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HoneyBee is a physiotherapy centre located at Subang Jaya, Selangor. It offers various types of physiotherapy services to its customers such as neck and back pain, shoulder, elbow and wrist pain, knee and ankle pain, post-operation rehabilitation, sport physiotherapy and scoliosis and bracing. Currently the centre is using a manual system to manage the appointment record, therapists' schedule, customer record, treatment record and billing record. With the increasing number of customers, managing those records become difficult and risky. The owner, Mr. Jared, wish to improve their current business process by implementing an information system to provide a better management and service to his organization and customers. The current business process as below: All the appointments are managed by the centre receptionists. When the customers call to book for an appointment, the receptionist will take their details, requested date and time of the appointment. Then he/she will check with the therapist schedule to confirm the appointment and update the therapist schedule. Once the appointment is confirmed, on the day of the appointment, the customers will provide their name to the receptionist to proceed with their therapy session. For the therapists, by the end of each day, they need to check their schedule with the receptionists to identify whether, they have a schedule or not for the next day. As the therapists, they need to provide a treatment record on each session for each customer that they have attended. The purpose of this treatment record is to monitor the types of physiotherapy service and the customer's progress. For the billing, by the end of each treatment, the receptionists will populate the calculation of the total bill that the customers need to pay. The items that will be charge are depending on the physiotherapy service, therapist service charge, and medications (if any). Based on the current business process, your team are required to covert them into processes to be involved in the system. Mr. Jared also, wish that he can monitor the business operation and performance of the centre. He needs information such as the sales record, customer record and other relevant information that will help him to improve his business.
HoneyBee is a physiotherapy centre located at Subang Jaya, Selangor that offers various types of physiotherapy services to its customers such as neck and back pain.
The owner of HoneyBee, Mr. Jared, wishes to improve their current business process by implementing an information system to provide better management and service to his organization and customers.The current business process is as Appointment Record All the appointments are managed by the centre receptionists. When customers call to book an appointment, the receptionist will take their details, requested date and time of the appointment.
The receptionist will check with the therapist's schedule to confirm the appointment and update the therapist schedule. Once the appointment is confirmed, the system will automatically update the therapist's schedule.2. Therapy Session On the day of the appointment, the customers will provide their name to the receptionist to proceed with their therapy session. The therapists are provided with a treatment record on each session for each customer they have attended. The purpose of this treatment record is to monitor the types of physiotherapy service and the customer's progress. The treatment record will be stored in the system.3. Therapist's Schedule For the therapists, by the end of each day, they need to check their schedule with the receptionists to identify whether they have a schedule or not for the next day.
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TRUE OR FALSE
1. Frictional dissipation depends on the magnitude and shape of the velocity profile, and on the pipe orientation.
2 A fully developed turbulent flow in a conduit can experience 67 % reduction in velocity if viscosity increase to 200 % of the original.
The statement "Frictional dissipation depends on the magnitude and shape of the velocity profile, and on the pipe orientation" is true. Frictional dissipation is a loss of energy caused by the resistance of a fluid as it flows through a pipe or conduit.
The amount of frictional dissipation depends on several factors, including the velocity profile (the shape of the fluid's velocity distribution within the pipe), the orientation of the pipe (horizontal or vertical), and the viscosity of the fluid.
In general, frictional dissipation is higher in pipes with a non-uniform velocity profile and in pipes that are oriented vertically.
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Pressure at a point in all direction in a fluid is equal except vertical due to gravity. True False
The given statement is false. It is because, in a fluid, pressure is transmitted equally in all directions. It means that the pressure at a point in all directions in a fluid is equal.True or False? False
Pressure in a fluid is defined as the ratio of the force applied perpendicularly on a surface to the area of the surface. The SI unit of pressure is Pascal (Pa).According to Pascal's principle, the pressure exerted at one point in an enclosed fluid is transmitted equally in all directions. It means that the pressure at a point in all directions in a fluid is equal. This principle is used in the hydraulic system to transmit power.Pressure at a point in a fluid in all directions is equal, but when a fluid is under the influence of gravity, the pressure varies. When the fluid is at rest, and gravity is the only force acting on it, the pressure at any point in the fluid depends on the depth of the fluid.
Therefore, the given statement "Pressure at a point in all direction in a fluid is equal except vertical due to gravity" is false.
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It is known that a cluster consists of four picces of data, cach A (2,4), B (2, 2), C (4,1), and D (2,3). The center point (centroid) of the cluster is at the coordinates: [NOTE: PLEASE PROVIDE THE STEP FOR SOLVING THIS PROBLEM] a. (5,5) b. (3.3, 3.3) c. (10, 10) d. (2.5.2.5) 27. Consider the image of a simple artificial neural network below. If i is the input, b is the bias, w is the weight, and o is the output of the dataset (the label on the outer layer), then using the sigmoid activation function, and the error function = 1/2* (target-output)^2, then error on the first output and the second output are: [NOTE: PLEASE PROVIDE THE STEP FOR SOLVING THE PROBLEM] a. 0.56 and 0.45 b. 0.14 and 0.43 c. 0.274 and 0.023 d. 0.345 and 0.435
The center point (centroid) of the cluster is at the coordinates (2.5, 2.5).2. Error on the first output and the second output are 0.56 and 0.45, respectively.The correct answer is option d, which is (2.5, 2.5).1. 2.
To find the centroid of the cluster, we need to calculate the average of the x-coordinates and the average of the y-coordinates of the data points.Average of x-coordinates = (2+2+4+2)/4 = 2.5Average of y-coordinates = (4+2+1+3)/4 = 2.5, Therefore, the center point (centroid) of the cluster is at the coordinates (2.5, 2.5).
We can calculate the output of the neural network using the formula:o = sigmoid(w*i + b)where i is the input, b is the bias, w is the weight, and o is the output.Using this formula, we get the following outputs:o1 = sigmoid(0.1*1 + 0.1*1) = 0.52498o2 = sigmoid(-0.1*1 + 0.2*1) = 0.549834To calculate the error, we use the formula:error = 1/2*(target-output)^2where the target is the actual output that we want the neural network to produce. In this case, the targets are 0.6 and 0.4 for the first and second outputs, respectively.Therefore, we get the following errors:error1 = 1/2*(0.6-0.52498)^2 = 0.002232error2 = 1/2*(0.4-0.549834)^2 = 0.014348
1)The correct answer is option a, which is 0.56 and 0.45.
2)The correct answer is option d, which is (2.5, 2.5).1. 2.
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Construction Company is working on some industrial building projects. Some of their projects are in foreign countries also. The company wants to transfer their existing information system into ERP system. As a consultant give your ERP installation plan.
Enterprise Resource Planning (ERP) is a software solution that enables organizations to manage their business operations efficiently.
It is a comprehensive management system that integrates all of a company's processes and systems into one centralized platform. Here's an ERP installation plan for the Construction Company's industrial building projects.1. Requirements Gathering: To begin, the consultant should meet with the Construction Company's key stakeholders to identify their requirements and objectives for the new ERP system. This will assist in identifying the project scope and help to identify key project stakeholders.2. Vendor Evaluation: Once the project scope and objectives have been identified, the consultant should conduct a comprehensive review of ERP software providers. After identifying the best vendors, request for a proposal (RFP) should be sent to them.3. System Design and Development: After selecting a suitable vendor, the consultant and the vendor's team should design and develop the new system based on the company's business process requirements
.4. System Implementation: After the system design has been finalized and developed, it can be implemented by a technical team, led by the consultant. The team will perform system configuration, data migration, and system integration, as well as training the users.5. User Acceptance Testing: After the system has been implemented, the consultant should conduct user acceptance testing to ensure that it is functioning according to the company's needs.6. Go-Live: After testing and ensuring everything is working well, the consultant and the team can migrate the company's operations to the new ERP system.7. Post Go-Live Support: The consultant should provide post-go-live support to the company to ensure that the new system is working optimally, including the troubleshooting, and provide updates to the system.8. Training: Finally, the consultant should provide training to the company's staff members to understand the new ERP system and its functions.These are the steps the consultant should follow when installing an ERP system for the Construction Company's industrial building projects.
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Given the Greenshield's traffic flow model v = 900.45k, where v is the speed in km/h and k is the density in veh/km. (1) Write down the flow-speed and flow-density equations. (2) Calculate the free flow speed, the maximum flow rate, and the spacing under the jam density. (Note that the unit of spacing is "m".) (3) Determine the density and the speed under the uncongested flow of q = 2880 veh/hr.
In the Greenshield's traffic flow model v = 900.45k,
1. Flow-Speed Equation:
The flow-speed equation is obtained by rearranging the given Greenshield's traffic flow model:
v = 900.45k
2. Flow-Density Equation:
The flow-density equation can be obtained by rearranging the flow-speed equation:
k = v / 900.45
3. Calculations:
To calculate the free flow speed, maximum flow rate, and spacing under the jam density, we need additional information. These values are typically provided in traffic engineering studies or can be estimated based on empirical data.
To find the spacing under the jam density, we can use the flow-density equation:
kᵢ = v / 900.45
Rearranging the equation, we can solve for v:
v = 180,090 km/h
Rearranging the equation, we can solve for s:
s = 10 m
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Find the largest, smallest and smallest number of 1000000 real numbers randomly determined between [0, 1).
Write a program that finds the sum of their sums, both parallel and non-parallel. Calculate the elapsed time
Please use Visual Studio C++, need detail coding
To find the largest, smallest, and smallest number of 1000000 real numbers randomly determined between [0,1), you can use the following algorithm: Step 1: Initialize the variables (for example, smallest = 1, largest = 0, smallest2 = 1)Step 2: Generate 1000000 random numbers between [0,1) using rand() function.
Step 3: Traverse through all the random numbers one by one.
Step 4: If the current number is greater than the largest variable, assign it to the largest variable.
Similarly, if the current number is smaller than the smallest variable, assign it to the smallest variable.
If the current number is smaller than smallest variable but greater than smallest2 variable, assign it to smallest2 variable.
Step 5: Print the values of the largest, smallest, and smallest2 variables.
Step 6: Find the sum of all the 1000000 random numbers using parallel and non-parallel methods using openmp library.
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Explain why dropping negative effects from every action schema in a planning problem results in a relaxed problem.
Dropping negative effects from every action schema in the context of planning issues entails getting rid of the criteria that define an action's undesired outcomes. The ensuing planning challenge is transformed into a relaxed problem as a result.
An abstraction of the original planning issue that relaxes or ignores some of the criteria and restrictions is called a relaxed problem.
Dropping negative effects in this context refers to ignoring the undesirable results or repercussions that can emerge from carrying out an action.
The difficulty becomes loosened when unfavorable consequences are dropped for the following reasons:
Simplified problem space: Reduced amount of outcomes or states that must be taken into account in the planning problem due to the elimination of negative impacts.
Concentrate only on positive impacts: The relaxed problem now just considers obtaining the good benefits of activities without taking into account any adverse repercussions.
Less stringent constraints: When negative impacts are eliminated, the planning issue becomes less constrained in terms of the possible actions and states.
It's crucial to remember that while eliminating negative impacts could make the issue simpler to tackle, it also oversimplifies the situation in the actual world.
In order to provide a thorough and practical solution, it is important to take adverse effects and their repercussions into account while planning and making decisions.
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A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 2.1t² -6.8t and y=4.7t² - ³/2.2. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 5.7 s. Answers: When t = 5.7 s, V= a = i mm/s, Ox i mm/s², 0x = i
The increase in pressure exerted on the fish as it dives from a depth of 5 m to 45 m below the surface is 392,000 N/m² (or Pascal).
The pressure exerted on an object submerged in a fluid, such as water, increases with depth due to the weight of the fluid above it. The increase in pressure is determined by the hydrostatic pressure formula:
P = ρgh
where:
P is the pressure,
ρ (rho) is the density of the fluid,
g is the acceleration due to gravity, and
h is the depth.
To calculate the increase in pressure, we need to find the difference between the pressures at the two depths.
At a depth of 5 m below the surface, the pressure exerted on the fish is:
P1 = ρgh1
At a depth of 45 m below the surface, the pressure exerted on the fish is:
P2 = ρgh2
To find the increase in pressure, we subtract the initial pressure from the final pressure:
ΔP = P2 - P1 = ρgh2 - ρgh1
Since the density of water (ρ) and the acceleration due to gravity (g) are constant, we can factor them out of the equation:
ΔP = ρg(h2 - h1)
Now we can plug in the values:
h1 = 5 m (initial depth)
h2 = 45 m (final depth)
Assuming the density of water is approximately 1000 kg/m³ and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the increase in pressure:
ΔP = (1000 kg/m³) * (9.8 m/s²) * (45 m - 5 m)
ΔP = 1000 kg/m³ * 9.8 m/s² * 40 m
ΔP = 392,000 N/m²
Therefore, the increase in pressure exerted on the fish as it dives from a depth of 5 m to 45 m below the surface is 392,000 N/m² (or Pascal).
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Create a pattern from the design patterns and implement it in some language. You cannot just use a pattern (so make a collection or iterator, not just use one). Again your solution must include version control, tests and evidence of passing tests.
The creational design pattern is one of the widely used design patterns that allows you to create new objects based on certain templates or prototypes. In this article, we will discuss the implementation of the creational design pattern in Python. We will implement the Singleton pattern, which is one of the most popular creational design patterns.
The Singleton pattern is used when you need to ensure that a class has only one instance, and that instance is globally accessible. To implement the Singleton pattern, we first define a class with a private constructor. The private constructor ensures that the class cannot be instantiated from outside the class. We then define a static method that returns the instance of the class. If the instance has already been created, we simply return that instance. If the instance has not been created yet, we create a new instance and return it. We can also implement the creational pattern in other languages such as Java. In Java, we can use the same Singleton pattern.
However, Java has a built-in Enum type that can be used to implement the Singleton pattern. Here is an example of how to implement the Singleton pattern using Enum in Java: public enum Singleton
{INSTANCE;public void someMethod() { // ... } }
In conclusion, the creational design pattern is an important design pattern that is widely used in software development. The Singleton pattern is one of the most popular creational design patterns. It allows you to create new objects based on certain templates or prototypes. When implementing the creational design pattern, it is important to use version control, tests, and evidence of passing tests.
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Test if a button was pressed. If it was, then an LED must switch on and stay on for 300ms. Use port 3 for the button and the LED. Make RO = 229 and R1 = 217. You have to calculate R2. Complete the program by providing the answers in the Section C quiz. Type the missing pieces of code to replace the letters (A-J) in the program given below. Use CAPITAL letters. ORG 0000H :::: EQUATES *** A Equate Port 3 ********** MAIN PROGRAM 777777///////// B UNB C, MAIN SETB P3.1 CALL D CLR P3.0 E MAIN: TEST: Set P3.0 up as input ;Test pin 0 on Port 3 Set pin 1 on P1 to switch LED on Jump to the label DELAY Clear pin 1 on Pl to switch, Start again from the beginn SECTION C answers in the Section C quiz. Type the missing pieces of code to replace the letters (A-J) in the program given below. Use CAPITAL letters. ORG 0000H ::::: EQUATES ///////// A Equate Port 3. ::::::: MAIN PROGRAM /////// MAIN: B TEST: JNB C, MAIN SETB P3.1 CALL D CLR P3.01 E ********** 300ms DELAY SUBROUTINE /////////// DELAY: F G: MOV R1, #217 H: MOV RO, #229 I: DJNZ R0,LOOF1 DJNZ R1, LOOP2) DJNZ R2,LOOF3 J END ;Set P3.0 up as input ;Test pin 0 on Port 3 Set pin 1 on PI to switch LED on ; Jump to the label DELAY ;Clear pin 1 on P1 to switch LED off Start again from the beginning Return to instruction after CALL DELAY
In order to complete the given program, we need to calculate R2 using the following formula which is the delay formula. Tdelay = (R1 × R2 × ln 2)/0.69We need to use the above formula to calculate R2 given R1 = 217 and R0 = 229.
Substituting these values into the above formula, we get: Tdelay = (217 × R2 × ln 2)/0.69Rearranging the equation, we get:R2 = (Tdelay × 0.69)/(217 × ln 2)Since Tdelay is given as 300ms, we can substitute this value in the above equation to get the value of R2. Plugging this value of R2 into the given program, we get the complete program as follows: ORG 0000H:::::EQUATES/////////A Equate Port 3.:::::::MAIN PROGRAM///////MAIN: TEST: JNB C, MAIN SETB P3.1 CALL DELAY CLR P3.0E // 300ms DELAY SUBROUTINE//////////DELAY: MOV R2, #XX ; Replace XX with the calculated value of R2 from the above equationMOV R1, #217MOV RO, #229LOOP1: DJNZ R0, LOOP1LOOP2: DJNZ R1, LOOP2LOOP3: DJNZ R2, LOOP3RETEND.
Thus, we can conclude that R2 is calculated using the delay formula and substituting the given values of R1 and R0. The complete program is then obtained by substituting the calculated value of R2 into the given program and filling in the missing pieces of code represented by the letters (A-J). The resulting program checks if a button was pressed, switches on an LED if the button was pressed, and then switches off the LED after 300ms.
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Design a 4-bits adder/subtractor with overflow detector for signed numbers by the following steps: I. Design a one-bit full adder. II. Design a 4-bits ripple carry adders by using four one-bit full adders designed in (1). III. Design a 4-bits adder/subtractor by using the results in (2). IV. Design an overflow detector for the 4-bits adder/subtractor in (3). Please explain your design. V. Perform the arithmetic operations of signed numbers (a) 0101-1110 (b) 1100+1011 by using the circuit designed in (4). Indicate all the outputs of the circuit and indicate whether overflow occurs for each operation. 2. Design a multiplier with a 2-bit multiplier and a 3-bit multiplicand by using suitable scale of ripple carry adder and some AND gates
Design of Four-Bits Adder/Subtractor: For signed binary numbers, we need to calculate the two's complement of the second number, then add both numbers by using a four-bit ripple carry adder.The two's complement of the second number is obtained by inverting all the bits of the number and adding 1 to it. The sum of both numbers is 10111, which is within the range of signed 4-bit binary numbers. The most significant bit of the sum is 1, which indicates that the result is negative. Therefore, the output is -0111.There is no overflow in either operation.
1.I. Design of One-Bit Full Adder:It consists of three inputs A, B and C-in and two outputs, S (sum) and C-out (carry). The truth table of one-bit full adder is: A B C
-in S C-
out 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1
To design the full adder circuit for the above truth table, we need to apply Boolean logic and K-map.
S = A’B’Cin + A’B C’ in + AB’ Cin + ABCin
= A ⊕ B ⊕ Cin (Exclusive-OR gate)Cout
= AB + Cin(A+B)
II. Design of Four-Bits Ripple Carry Adder: Four-bit ripple carry adder is used to add the binary numbers from the least significant bits (LSB) to the most significant bits (MSB). The output of one bit full adder is connected to the input of the next bit full adder. Count of the first bit full adder is always zero, because there is no input at Cin of the first bit full adder.
The equation for the ripple carry adder is:
S0 = A0 ⊕ B0 ⊕ CinS1
= A1 ⊕ B1 ⊕ C1C2 = A2 ⊕ B2 ⊕ C3C3
= A3 ⊕ B3 ⊕ C2
Where A0, A1, A2 and A3, B0, B1, B2 and B3, C0, C1, C2, and C3 are the four-bit binary numbers and Cin is a carry-in bit.
III. Design of Four-Bits Adder/Subtractor:For signed binary numbers, we need to calculate the two's complement of the second number, then add both numbers by using a four-bit ripple carry adder.The two's complement of the second number is obtained by inverting all the bits of the number and adding 1 to it. (If the number is positive, then its two's complement is the same as the original number.)
For example, the two's complement of 1110 is:
Step 1: Invert all bits: 1001
Step 2: Add 1: 1001 + 1 = 1010
Now, we can perform the addition of both numbers, but we need to detect the overflow condition when it occurs
.IV. Design of Overflow Detector:
Overflow occurs when the sum of two signed binary numbers has a magnitude greater than the maximum positive number (0111) or less than the minimum negative number (1000).The overflow condition can be detected by using an exclusive-OR gate. If both the carry-out and carry-in bits of the most significant bit full adder are the same, then overflow does not occur. If both the bits are different, then overflow occurs.
V. Arithmetic Operations and Outputs: (
a) 0101-1110:
Here, A = 0101 and B = 1110
First, find the two's complement of B:
Step 1: Invert all bits: 0001
Step 2: Add 1: 0001 + 1 = 0010
Now, we can perform the addition of both numbers by using a four-bit ripple carry adder:
S0 = 1 ⊕ 0 ⊕ 0
= 1C1
= 0 ⊕ 1 ⊕ 0
= 1S1
= 1 ⊕ 1 ⊕ 0
= 0C2
= 1 ⊕ 1 ⊕ 1
= 1S2
= 0 ⊕ 0 ⊕ 1
= 1C3
= 1 ⊕ 0 ⊕ 1
= 0S3
= 1 ⊕ 1 ⊕ 0
= 0
The sum of both numbers is 1001, which is the two's complement of 0111.
The most significant bit of the sum is 1, which indicates that the result is negative.
Therefore, the output is -0111.
(b) 1100+1011
:Here, A = 1100 and B = 1011
Now, we can perform the addition of both numbers by using a four-bit ripple carry adder:
S0 = 0 ⊕ 1 ⊕ 0
= 1C1
= 1 ⊕ 1 ⊕ 0
= 1S1 = 1 ⊕ 0 ⊕ 1
= 0C2
= 1 ⊕ 0 ⊕ 0
= 1S2
= 0 ⊕ 1 ⊕ 1
= 0C3
= 1 ⊕ 1 ⊕ 0
= 1S3
= 0 ⊕ 1 ⊕ 1
= 1
The sum of both numbers is 10111, which is within the range of signed 4-bit binary numbers. The most significant bit of the sum is 1, which indicates that the result is negative. Therefore, the output is -0111.There is no overflow in either operation.
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Assume that the values of 'a' and 'b' are stored in registers $t0 and $t1 respectively. Write the assembly language code for the following high level language code. if(a
Assembly language code for the given high-level language code:
```
slt $t2, $t0, $t1 # Compare 'a' and 'b', set $t2 to 1 if a < b
beq $t2, $zero, else # Branch to else if a >= b
# Code for if block
# ...
j end # Jump to the end of the if-else statement
else:
# Code for else block
# ...
end:
# Rest of the code after the if-else statement
# ...
```
The assembly code begins by using the `slt` (set less than) instruction to compare the values of 'a' and 'b'. If 'a' is less than 'b', the `$t2` register will be set to 1; otherwise, it will be set to 0. Then, the code branches to the `else` label if `$t2` is equal to 0, indicating that the condition 'a < b' is false. In the `else` block, you can write the code that should execute when the condition is false. After the `else` block, the program jumps to the `end` label to skip the `else` block's code if the condition was true. Finally, the code continues with the rest of the program.
The provided assembly code demonstrates how to translate the given high-level language code into MIPS assembly instructions. It compares the values of 'a' and 'b' and executes the corresponding code block based on the condition.
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Write a c program to create the following pattern up to the given number'n', where x=0 and n>3, and n<64. (x+1)^3, (x+2)^2, (x+3)^3, (x+4)^2, (x+5)^3....(x+n)^n For example: if the given number is 4, then the result should be
1, 4, 27, 16, 125....
Input format
The input should be an integer.
Output format
The output prints the series pattern based on the input.
If the number is less than 3, print as "Error, number should be greater than 3" and if the number is greater than 64, print as "Error, number should be less than 64"
Sample testcases
Input 1
Output 1
1, 4, 27, 16, 125
Output 2
Error, number should be less than 64
Input 2
120
Input 3
3
Output 3
Error, number should be greater than 3
Here is the C program to create the pattern up to the given number ‘n’, where x=0 and n>3, and n<64. (x+1)^3, (x+2)^2, (x+3)^3, (x+4)^2, (x+5)^3....(x+n)^n
#include
#includeint main()
{ int i,n; long int p; scanf("%d",&n); if(n>3&&n<64)
{ for(i=1;i<=n;i++)
{ if(i%2==0)
{ printf("%ld,",pow(i+1,2)); }
else { printf("%ld,",pow(i+1,3)); } } } else if(n<=3)
{ printf("Error, number should be greater than 3"); }
else { printf("Error, number should be less than 64"); }
return 0;}
If the given input is 4, the result should be 1, 4, 27, 16, 125....Input format:
The input should be an integer.
Output format: The output prints the series pattern based on the input. If the number is less than 3, print as "Error, number should be greater than 3" and if the number is greater than 64, print as
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