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Lostsunrise [7]
3 years ago
6

How to walk a dog dududududududududesssss

Engineering
2 answers:
bagirrra123 [75]3 years ago
6 0
Pick up a leash put the dog on the leash open the door/cage take the dog out and walk it
rusak2 [61]3 years ago
4 0
Putting them on a leash and picking a route and walking it
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Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from the person if the exposed su
Ghella [55]

Answer:

dfggf

Explanation:

3 0
3 years ago
A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind
Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

N_F=n_F\times \frac{N_S}{n_S} here,

n_F = the number of unseeded faults = 6

N_S = number of seeded faults = 30

n_s = number of seeded faults found = 15

So NF will be calculated as,

N_F=6\times \frac{30}{15}=12

And the estimate of faults remaining is  N_F-n_F = 12 - 6 = 6

8 0
3 years ago
Implement
kolbaska11 [484]

Answer:

#include <iostream>

using namespace std;

// Pixel structure

struct Pixel

{

unsigned int red;

unsigned int green;

unsigned int blue;

Pixel() {

red = 0;

green = 0;

blue = 0;

}

};

// function prototype

int energy(Pixel** image, int x, int y, int width, int height);

// main function

int main() {

// create array of pixel 3 by 4

Pixel** image = new Pixel*[3];

for (int i = 0; i < 3; i++) {

image[i] = new Pixel[4];

}

// initialize array

image[0][0].red = 255;

image[0][0].green = 101;

image[0][0].blue = 51;

image[1][0].red = 255;

image[1][0].green = 101;

image[1][0].blue = 153;

image[2][0].red = 255;

image[2][0].green = 101;

image[2][0].blue = 255;

image[0][1].red = 255;

image[0][1].green = 153;

image[0][1].blue = 51;

image[1][1].red = 255;

image[1][1].green = 153;

image[1][1].blue = 153;

image[2][1].red = 255;

image[2][1].green = 153;

image[2][1].blue = 255;

image[0][2].red = 255;

image[0][2].green = 203;

image[0][2].blue = 51;

image[1][2].red = 255;

image[1][2].green = 204;

image[1][2].blue = 153;

image[2][2].red = 255;

image[2][2].green = 205;

image[2][2].blue = 255;

image[0][3].red = 255;

image[0][3].green = 255;

image[0][3].blue = 51;

image[1][3].red = 255;

image[1][3].green = 255;

image[1][3].blue = 153;

image[2][3].red = 255;

image[2][3].green = 255;

image[2][3].blue = 255;

// create 3by4 array to store energy of each pixel

int energies[3][4];

// calculate energy for each pixel

for (int i = 0; i < 3; i++) {

for (int j = 0; j < 4; j++) {

energies[i][j] = energy(image, i, j, 3, 4);

}

}

// print energies of each pixel

for (int i = 0; i < 4; i++) {

for (int j = 0; j < 3; j++) {

// print by column

cout << energies[j][i] << " ";

}

cout << endl;

}

}

// function prototype

int energy(Pixel** image, int x, int y, int width, int height) {

// get adjacent pixels

Pixel left, right, up, down;

if (x > 0) {

left = image[x - 1][y];

if (x < width - 1) {

right = image[x + 1][y];

}

else {

right = image[0][y];

}

}

else {

left = image[width - 1][y];

if (x < width - 1) {

right = image[x + 1][y];

}

else {

right = image[0][y];

}

}

if (y > 0) {

up = image[x][y - 1];

if (y < height - 1) {

down = image[x][y + 1];

}

else {

down = image[x][0];

}

}

else {

up = image[x][height - 1];

if (y < height - 1) {

down = image[x][y + 1];

}

else {

down = image[x][0];

}

}

// calculate x-gradient and y-gradient

Pixel x_gradient;

Pixel y_gradient;

x_gradient.blue = right.blue - left.blue;

x_gradient.green = right.green - left.green;

x_gradient.red = right.red - left.red;

y_gradient.blue = down.blue - up.blue;

y_gradient.green = down.green - up.green;

y_gradient.red = down.red - up.red;

int x_value = x_gradient.blue * x_gradient.blue + x_gradient.green * x_gradient.green + x_gradient.red * x_gradient.red;

int y_value = y_gradient.blue * y_gradient.blue + y_gradient.green * y_gradient.green + y_gradient.red * y_gradient.red;

// return energy of pixel

return x_value + y_value;

}

Explanation:

Please see attachment for ouput

6 0
3 years ago
How do engineering and technology impact the natural world and environment
ddd [48]

Answer:

Answer to the following question is as follow.

Explanation:

Civil engineering industrial development projects have a significant influence on the environment, particularly in terms of excessive noise, environmental pollution, and land shrinkage.

Engineers have totally transformed our environment, from contemporary dwellings to bridges, space flight, automobiles, and cutting-edge mobile technologies. Engineers utilise their expertise to build new and exciting prospects and address any difficulties that may occur, and they employ their innovative ideas to accomplish so.

8 0
3 years ago
BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of
Tanya [424]

(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ

<u>Explanation:</u>

<u />

Given-

Lattice parameter of Li  = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200

(a)

Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b)

Density is represented by ρ

ρ = n X (AM) / v X Nₐ

where,

Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

4 0
4 years ago
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