Define hexadecimal number system

As a student, why do you need to know and follow the steps printing a document?

Kazeer [188]

So I am able to print my math notes and keep up in geometry.

7 0

2 years ago

In Java.Use a single for loop to output odd numbers, even numbers, and an arithmetic function of an odd and even number. Use the

rosijanka [135]

Answer:

This program is written using java programming language

Difficult lines are explained using comments (See Attachment for Source file)

Program starts here

import java.util.*;

import java.lang.Math;

public class oddeven{

public static void main(String [] args)

{

double even,odd;//Declare variables even and odd as double

//The next iteration prints odd numbers

for(int i = 1;i<=173;i+=2)

{

odd = i;

System.out.format("%.4f",odd);//Round to 4 decimal places and print

System.out.print("\t");

}

System.out.print('\n');//Start printing on a new line

//The next iteration prints even numbers

for(int i = 0;i<=173;i+=2)

{

even=i;

System.out.format("%.4f",even);//Round to 4 decimal places and print

System.out.print("\t");

}

System.out.print('\n');//Start printing on a new line

double ssqrt;//Declare ssqrt to calculate the square root of sum of even and odd numbers

for(int i = 0;i<=173;i+=2)

{

ssqrt = Math.sqrt(2*i+1);//Calculate square root here

System.out.format("%.4f",ssqrt); //Round to 4 decimal places and print

System.out.print("\t");

}

Explanation:

Libraries are imported into the program

import java.util.*;

import java.lang.Math;

The following line declares variables even and odd as double

double even,odd;

The following iteration is used to print odd numbers with in the range of 0 to 173

for(int i = 1;i<=173;i+=2) {

odd = i;

System.out.format("%.4f",odd); This particular line rounds up each odd numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

The following code is used to start printing on a new line

System.out.print('\n');

The following iteration is used to print even numbers with in the range of 0 to 173

for(int i = 0;i<=173;i+=2)

{

even=i;

System.out.format("%.4f",even); This particular line rounds up each even numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

The following code is used to start printing on a new line

System.out.print('\n');

The next statement declares ssqrt as double to calculate the square root of the sum of even and odd numbers

double ssqrt;

for(int i = 0;i<=173;i+=2)

{

ssqrt = Math.sqrt(2*i+1);This line calculates the square root of sum of even and odd numbers

System.out.format("%.4f",ssqrt); This particular line rounds up each even numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

Download java

7 0

3 years ago

Write a function (named n_pointed_star) to make the turtle draw an n-pointed star. The function should return nothing, and accep

Andreyy89

Answer:

import sys

import turtle

import random

def n_pointed_star(total_points):

if total_points <= 4:

raise ValueError('Not enough total_points')

area = 150

for coprime in range(total_points//2, 1, -1):

if greatest_common_divisor(total_points, coprime) == 1:

start = turtle.position()

for _ in range(total_points):

turtle.forward(area)

turtle.left(360.0 / total_points * coprime)

turtle.setposition(start)

return

def greatest_common_divisor(a, b):

while b != 0:

a, b = b, a % b

return a

turtle.reset()

n_pointed_star(5)

Explanation:

Inside the n_pointed_star function, check whether the total no. of points are less than or equal to 4 and then throw an exception.
Loop through the total_points variable and check whether the result from greatest_common_divisor is equal to 1 or not and then set the starting position of turtle and move it.
Create the greatest_common_divisor which takes two parameters a and b to find the GCD.
Finally reset the turtle and call the n_pointed_star function.

8 0

3 years ago

Here it is crews my profile pic

Murrr4er [49]

It looks niceee
Also can you mark me brainlyest for extra points

7 0

3 years ago

Read 2 more answers

A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an

yulyashka [42]

The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

= 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x 10⁹ x 0.75 / 3 Ghz

= 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

= 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore, numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x 10⁹ / 0.75 = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall that:

MIPS = Clock rate / CPI X 10⁶

So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

= 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

MFLOPS₁ = 1777.6

MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

3 0

2 years ago

Define hexadecimal number system ​

Define hexadecimal number system