07.12.2021

What is the area of the tabletop?

. 5

Faq

Computers and Technology
Step-by-step answer
P Answered by Specialist

-

CODE:

#include <iostream>

#include<cmath>

using namespace std;

float calcBaseArea(float a);

float calcSideArea(float s,float l);

void prntSprfArea(float base_area,float side_area);

int main()

{

float h;

float a;

float base_area

float side_area;

cout<<"Enter the side length of the base of the square pyramid in feet : ";

cin>>a;

cout<<"Enter the height of the square pyramid in feet : ";

cin>>h;

base_area=calcBaseArea(a);

side_area=calcSideArea(a,h);

cout<<"Base surface area of the square pyramid is "<<base_area<<" square feet. "<<endl;

cout<<"Side area of the square pyramid is "<<side_area<<" square feet."<<endl;

prntSprfArea(base_area,side_area);

return 0;

}

float calcBaseArea(float a)

{

return pow(a,2);

}

float calcSideArea(float s,float l)

{

float area=(s*l)/2;

return 4*area;

}

void prntSprfArea(float base_area,float side_area)

{

cout<<"Total surface area of the pyramid is "<<base_area+side_area<<" square feet.";

OUTPUT:


This means that the surface area is composed of the base area (i.e., the area of bottom square) plus
Computers and Technology
Step-by-step answer
P Answered by Specialist

-

CODE:

#include <iostream>

#include<cmath>

using namespace std;

float calcBaseArea(float a);

float calcSideArea(float s,float l);

void prntSprfArea(float base_area,float side_area);

int main()

{

float h;

float a;

float base_area

float side_area;

cout<<"Enter the side length of the base of the square pyramid in feet : ";

cin>>a;

cout<<"Enter the height of the square pyramid in feet : ";

cin>>h;

base_area=calcBaseArea(a);

side_area=calcSideArea(a,h);

cout<<"Base surface area of the square pyramid is "<<base_area<<" square feet. "<<endl;

cout<<"Side area of the square pyramid is "<<side_area<<" square feet."<<endl;

prntSprfArea(base_area,side_area);

return 0;

}

float calcBaseArea(float a)

{

return pow(a,2);

}

float calcSideArea(float s,float l)

{

float area=(s*l)/2;

return 4*area;

}

void prntSprfArea(float base_area,float side_area)

{

cout<<"Total surface area of the pyramid is "<<base_area+side_area<<" square feet.";

OUTPUT:


This means that the surface area is composed of the base area (i.e., the area of bottom square) plus
Biology
Step-by-step answer
P Answered by Master

Question 1 & 2. Image 1. All the high or low pressure areas are marked. Standard atmospheric pressure is 1,013.25 millibars. Everything above is high pressure and everything below is low pressure.  

Question 3. Low pressure areas. These areas are basically storms. As hurricanes and blizzard are classified as storms, one expects snow or rain in areas with low atmospheric pressure.  

Question 4. High pressure areas. These areas are associated with light winds and the phenomenon known as subscidence. Through adiabatic or compressional heating the air mass is ”dried out”. This is known as subscidence.  

Question 5 & 6. Image 2. Northern Hemisphere: high pressure – clockwise direction, low pressure – counterclockwise

Question 7. As winds move in an anticlockwise manner around a low-pressure area, currently, the wind will blow from a southwestern to northeastern direction. However, as soon as the low-pressure area moves in, the wind will blow from a southeastern to a northwestern direction.  

Question 8. As winds move in a clockwise manner around a low-pressure area, currently, the wind will blow from a southeastern to a northwestern direction. However, as soon as the high-pressure area moves in, the wind will blow from a southwestern to northeastern direction.  

Question 9. Image 3. Strong winds are indicated by closely packed isobars as they indicate a steep horizontal pressure gradient.  


***kinda need this in under 24hours**  add the data to the isobar map and use that data to answer th
***kinda need this in under 24hours**  add the data to the isobar map and use that data to answer th
***kinda need this in under 24hours**  add the data to the isobar map and use that data to answer th
Mathematics
Step-by-step answer
P Answered by Specialist
1. Photo description: A picture of the Eiffel tower, to be stuck on a mat.
Dimensions (including units): 4 in x 6 in
2. Since 2x would be added to each dimension:
Length: 6 + 2x (inches)
Width: 4 + 2x (inches)
3. Area: A = LW = (6+2x)(4+2x) square inches
4. F: (6)(4) = 24, O: (6)(2x) = 12x, I: (2x)(4) = 8x, L: (2x)(2x) = 4x^2
Polynomial expression: Adding the FOIL terms up: 4x^2 + 20x + 24
5. The area should be in square inches, since we multiplied length (in inches) by width (in inches).
6. Multiply factors using the distribution method:
(6+2x)(4+2x) = 6(4+2x) + 2x(4+2x) = 24 + 12x + 8x + 4x^2 = 24 + 20x + 4x^2
This is identical to the expression in Part 4.
7. x: 24 + 20x + 4x^2
If x = 1.0 in: Area = 24 + 20(1) + 4(1)^2 = 48 in^2
If x = 2.0 in: Area = 24 + 20(2) + 4(2)^2 = 80 in^2
8. If a white mat costs $0.03 per square inch and a black mat costs $0.05 per square inch, determine the cost of each size of black and white mat.
x Total area of matCost of white matCost of black mat
1.0 in, A = 48 in^2, (0.03)(48) = $1.44, (0.05)(48) = $2.40
2.0 in, A = 80 in^2, (0.03)(80) = $2.40, (0.05)(80) = $4.00
9. The cheapest option would be the white mat with 1-in margins on all sides, which would cost $1.44. Without any further criteria on aesthetics or size limitations, this is the most viable option.
Physics
Step-by-step answer
P Answered by PhD
Answers to the questions:
Question 1 and 2. (Image 1.) All the high or low pressure areas are marked. Standard atmospheric pressure is 1,013.25 millibars. Everything above is high pressure and everything below is low pressure.

Question 3. Low pressure areas. These areas are basically storms. As hurricanes and blizzard are classified as storms, one expects snow or rain in areas with low atmospheric pressure.

Question 4. High pressure areas. These areas are associated with light winds and the phenomenon known as subscidence. Through adiabatic or compressional heating the air mass is ”dried out”. This is known as subscidence.

Question 5 and 6. (Image 2.) Northern Hemisphere: high pressure – clockwise direction, low pressure – counterclockwise

Question 7. As winds move in an anticlockwise manner around a low-pressure area, currently, the wind will blow from a southwestern to northeastern direction. However, as soon as the low-pressure area moves in, the wind will blow from a southeastern to a northwestern direction.

Question 8. As winds move in a clockwise manner around a low-pressure area, currently, the wind will blow from a southeastern to a northwestern direction. However, as soon as the high-pressure area moves in, the wind will blow from a southwestern to northeastern direction.

Question 9. (Image 3). Strong winds are indicated by closely packed isobars as they indicate a steep horizontal pressure gradient.
Answers to the questions:
Question 1 and 2. (Image 1.) All the high or low pressure areas are marke
Answers to the questions:
Question 1 and 2. (Image 1.) All the high or low pressure areas are marke
Biology
Step-by-step answer
P Answered by Specialist

Question 1 & 2. Image 1. All the high or low pressure areas are marked. Standard atmospheric pressure is 1,013.25 millibars. Everything above is high pressure and everything below is low pressure.  

Question 3. Low pressure areas. These areas are basically storms. As hurricanes and blizzard are classified as storms, one expects snow or rain in areas with low atmospheric pressure.  

Question 4. High pressure areas. These areas are associated with light winds and the phenomenon known as subscidence. Through adiabatic or compressional heating the air mass is ”dried out”. This is known as subscidence.  

Question 5 & 6. Image 2. Northern Hemisphere: high pressure – clockwise direction, low pressure – counterclockwise

Question 7. As winds move in an anticlockwise manner around a low-pressure area, currently, the wind will blow from a southwestern to northeastern direction. However, as soon as the low-pressure area moves in, the wind will blow from a southeastern to a northwestern direction.  

Question 8. As winds move in a clockwise manner around a low-pressure area, currently, the wind will blow from a southeastern to a northwestern direction. However, as soon as the high-pressure area moves in, the wind will blow from a southwestern to northeastern direction.  

Question 9. Image 3. Strong winds are indicated by closely packed isobars as they indicate a steep horizontal pressure gradient.  


***kinda need this in under 24hours**  add the data to the isobar map and use that data to answer th
***kinda need this in under 24hours**  add the data to the isobar map and use that data to answer th
***kinda need this in under 24hours**  add the data to the isobar map and use that data to answer th
Mathematics
Step-by-step answer
P Answered by Master
1. Photo description: A picture of the Eiffel tower, to be stuck on a mat.
Dimensions (including units): 4 in x 6 in
2. Since 2x would be added to each dimension:
Length: 6 + 2x (inches)
Width: 4 + 2x (inches)
3. Area: A = LW = (6+2x)(4+2x) square inches
4. F: (6)(4) = 24, O: (6)(2x) = 12x, I: (2x)(4) = 8x, L: (2x)(2x) = 4x^2
Polynomial expression: Adding the FOIL terms up: 4x^2 + 20x + 24
5. The area should be in square inches, since we multiplied length (in inches) by width (in inches).
6. Multiply factors using the distribution method:
(6+2x)(4+2x) = 6(4+2x) + 2x(4+2x) = 24 + 12x + 8x + 4x^2 = 24 + 20x + 4x^2
This is identical to the expression in Part 4.
7. x: 24 + 20x + 4x^2
If x = 1.0 in: Area = 24 + 20(1) + 4(1)^2 = 48 in^2
If x = 2.0 in: Area = 24 + 20(2) + 4(2)^2 = 80 in^2
8. If a white mat costs $0.03 per square inch and a black mat costs $0.05 per square inch, determine the cost of each size of black and white mat.
x Total area of matCost of white matCost of black mat
1.0 in, A = 48 in^2, (0.03)(48) = $1.44, (0.05)(48) = $2.40
2.0 in, A = 80 in^2, (0.03)(80) = $2.40, (0.05)(80) = $4.00
9. The cheapest option would be the white mat with 1-in margins on all sides, which would cost $1.44. Without any further criteria on aesthetics or size limitations, this is the most viable option.
Computers and Technology
Step-by-step answer
P Answered by PhD

Following are the code to this question:

Explanation:

//For Solid:

public class Solid//defining a class Solid

{

private String myName;//defining a String variable

public Solid(String name)//defining parameterized constructor

{

myName = name;//holding String value in myName variable

}

public String getName()//defining getName method  

{

return myName;//return value of myName

}

public double volume()//defining method volume that overriden its subclass

{

return 0;//return value

}

public double surfaceArea()//defining method surfaceArea that overriden its subclass

{

return 0;//return value

}

}

//For RectangularPrism:

public class RectangularPrism extends Solid//defining a class RectangularPrism that inherits Solid

{

  protected double length,width,height;//defining double variable

RectangularPrism(String str,double l,double w,double h)//defining parameterized constructor

{

  super(str);//use super method

  this.length=l;//use this to hold value

  this.width=w;//use this to hold value

  this.height=h;//use this to hold value

}

public double volume()//defining volume method  

{

return length*width*height;//calculate and return volume value

}

public double surfaceArea()//defining a  method surfaceArea  

{

return 2*(length*width+width*height+height*length);//calculate and return the surfaceArea

}

}

// For  Cube:

public class Cube extends RectangularPrism//defining Cube class that inherits RectangularPrism  

{

  Cube(String name, double side) //defining parameterized constructor

   {

       super(name, side, side, side);//use super method

   }

  public double volume()//defining volume method  

  {

   return length * length * length;//calculate and return volume

  }

public double surfaceArea()//defining method surfaceArea  

  {

  return 6 * length*length;//calculate and return surfaceArea

   }

}

//for Cylinder:

import java.lang.*;//import lang package  

public class Cylinder extends Solid//defining Cylinder class that inherits Solid

{

  private double radius, height;//defining double variable

  public Cylinder(String str, double r, double h)//defining parameterized constructor

  {

      super(str);//use super Method

      this.radius = r;//use this keyword to hold value

      this.height = h;//use this keyword to hold value

  }

  public double volume()//defining volume method

  {

      return Math.PI * radius * radius * height;//calculate and return volume

  }

  public double surfaceArea()//defining surfaceArea method

  {

     return 2 * Math.PI * radius * (height + radius);//calculate and return surfaceArea

  }

}

//For Pyramid

import java.lang.*;//import package

public class Pyramid extends Solid//defining a Pyramid class that inherits Solid

{  

  private double length, width,height;//defining double variable

  public Pyramid(String str, double l, double w, double h)//defining parameterized constructor  

  {

      super(str);//use super method

      this.length = l;//use this keyword to hold value

      this.width = w;//use this keyword to hold value

      this.height = h;//use this keyword to hold value

  }

public double volume()//defining volume method

{

return (length*width*height)/3.0;//calculate and return volume

}

public double surfaceArea()//defining a method surfaceArea

{

return (length*width)+length*Math.sqrt((Math.pow((width/2),2)+(height*height)))+(width*Math.sqrt((Math.pow((length/2),2)+(height*height))));//calculate and return surfaceArea

}

}

//For Sphere

import java.lang.*;//import package

public class Sphere extends Solid //defining Sphere class that inherits Solid

{

  private double radius;//defining double variable

  public Sphere(String str, double r)//defining parameterized constructor  

  {

      super(str);//use super method

      this.radius = r;//use this to hold value

  }

  public double volume()//defining volume method

  {

      return (4.0/3.0)*Math.PI*radius*radius*radius;//calculate and return volume

  }

  public double surfaceArea()//defining method surfaceArea  

  {

    return 4 * Math.PI * radius * radius;//calculate and return surfaceArea

  }

}

please find attachment.


Given the Solid class, extend it with:

Pyramid 
Cylinder 
RectangularPrism 
Sphere 
Make sure to cr
Given the Solid class, extend it with:

Pyramid 
Cylinder 
RectangularPrism 
Sphere 
Make sure to cr
Given the Solid class, extend it with:

Pyramid 
Cylinder 
RectangularPrism 
Sphere 
Make sure to cr
Given the Solid class, extend it with:

Pyramid 
Cylinder 
RectangularPrism 
Sphere 
Make sure to cr

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