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slava [35]
4 years ago
11

. On Juan's twenty-sixth birthday, he deposited $7,500 in a retirement account. Each year thereafter, he deposited $1,000 more t

han the previous year. Using a gradient series factor, determine how much was in the account immediately after his thirty-fifth deposit if
Engineering
1 answer:
never [62]4 years ago
7 0

Answer:

$1,783,805

Explanation:

Okay, if we are going to answer thus question efficiently, we need to take time to understand each and every sentence in the question. Hence, from the question we have the following data or information and they are;

1. The initial deposit in which Juan deposited on his 26th birthday = $7500.

2. After the statement in [1] above, Juan decided to be depositing $1000 more than the one he deposited on his 26th birthday.

Therefore, we are given that there is 5% in the interest rate. Thus, after his thirty-fifth deposit he will have;

$7500[90.3203] + [5.516] × $1000 × 200.581] = $1,783,805

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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

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3 years ago
What does the plasma membrane do
aniked [119]

<u>Answer:</u>

The plasma membrane encloses specific structures.

<u>Explanation: </u>

plasma membrane is also called as cell membrane. Plasma membrane only  allows the particle to get in and pass out of the cell by osmosis and diffusion method from the outside environment. It is responsible for the molecular traffic inside the cell.

It helps in maintaining the shape of the cell. It has many proteins in it. Therefore cell membrane are responsible for having specific structures.

7 0
3 years ago
The speed of sound in a fluid can be calculated using the following equation:
wlad13 [49]

Answer:

Jesus is always the answer

4 0
3 years ago
Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and
jonny [76]

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:    

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ =  0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

p = 8 bar = 800 kPa

Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ +  17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

3 0
3 years ago
How do heat and our use of electricity affect our daily activities and the environment
Andrews [41]

Answer:

there hope it can help.......

3 0
2 years ago
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