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Naddika [18.5K]
3 years ago
13

12 times the square root of 8737

Engineering
1 answer:
mixer [17]3 years ago
3 0
12 times the square root of 8737 is 1121.66305101
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Help please thank you​
OleMash [197]

2.4 is the correct answer .

125 ÷ 52

4 0
4 years ago
Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or
n200080 [17]

Answer:

Here is the JAVA program:  

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed that is 4 and assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two lists origList[] and offsetAmount[] are assigned values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // product variable to store the product of 2 lists

for(i = 0; i <= origList.length - 1; i++){

/* loop starts with i at 0th position or index and ends when the end of the origList is reached */

/* multiples each element of origList to corresponding element of offsetAmount and stores result in the form of character string in product*/

   product+= Integer.toString(origList[i] *= offsetAmount[i]) + " ";  }

 System.out.println(product); }}   //displays the product of both lists

Output:

80 180 80 400

Explanation:

If you want to print the product of origList with corresponding value in offsetAmount in vertical form you can do this in the following way:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

 origList[i] *= offsetAmount[i];

System.out.println(origList[i]); } }}

Output:

80                                                                                                                            

180                                                                                                                          

80                                                                                                                            

400

The program along with the output is attached as screenshot with the input given in the example.

8 0
3 years ago
Go give some love to this video, thumbs up needed!!!<br> https://youtu.be/F6w_CqB0bZM
kolezko [41]

Answer:

i liked the vid

Explanation:

can i get brainliest pls

5 0
3 years ago
A metal having a cubic structure has a density of , an atomic weight of , and a lattice parameter of Å One atom is associated wi
Veronika [31]

Answer:

Explanation:

Answer: The crystal structure of the metal is BCC

Explanation:

we first calculate the volume of the unit cell.

Volume of unit cell= (a°)^3.

The lattice parameter here is a°.

Substitute (6.13 * 10^-8)cm for a°.

Volume of unit cell = (6.13 * 10^-8)^3 = 2.3034 * 10^-22 cm^3/cell.

To determine the crystal structure we use

Density (p) = {(Number of atoms per cell) (Atomic mass)} / {(volume of unit cell)(Avogrado constant)}.

Substitute 1.892g/cm^3 for p (6.02*10^23) atoms/mol for Avogrado constant 1.3921g/mol.

For atomic mass and (2.3034 * 10^-22) cm^3/cell for unit cell.

1.892g/cm^3 = {(Number of atoms per cell) (1.3291g/mol)} / {(2.3034 * 10^-22) (6.02 * 10^23 atoms/mol)}.

Changing the subject of formula we have :

Number of atoms per cell = {(2.3034 * 10^-22) * (6.02 * 10^23) * 1.892} / 132.91

Number of atoms per cell = 2.

Since the number of atoms per cell is 2, :. the crystal structure of metal is BCC.

Note: p = density

a° = a subscript o

4 0
3 years ago
tech a says that small vacuum leaks can be found with a vacuum gauge. tech b says that the catalytic converter is typically loca
schepotkina [342]

Both are true. Tech A claims that a vacuum gauge can be used to locate tiny vacuum leaks. According to tech B, the catalytic converter is usually placed behind the muffler.

<h3>Which of the following links the catalytic converter to the exhaust manifold?</h3>

Tailpipe. The catalytic converter has cleaned up the exhaust gases, which are then sent out of the car and into the atmosphere via the tailpipe, the last link in the exhaust system.

<h3>Which of the following exhaust system parts aids in lowering exhaust emissions?</h3>

Hazardous pollutants from engine exhaust are decreased by the catalytic converter. The converter, which is situated between the exhaust manifold and the muffler, works with heat and metals that serve as catalysts.

To know more about vacuum gauge visit:-

brainly.com/question/20989381

#SPJ4

8 0
1 year ago
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