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jeka57 [31]
3 years ago
12

Question 13 (1 point)

Computers and Technology
1 answer:
Mrac [35]3 years ago
8 0
The second option “1 inch”

Is correct for default word documents
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Given parameters b and h which stand for the base and the height of an isosceles triangle (i.e., a triangle that has two equal s
schepotkina [342]

Answer:

The area of the triangle is calculated as thus:

Area = 0.5 * b * h

To calculate the perimeter of the triangle, the measurement of the slant height has to be derived;

Let s represent the slant height;

Dividing the triangle into 2 gives a right angled triangle;

The slant height, s is calculated using Pythagoras theorem as thus

s = \sqrt{b^2 + h^2}

The perimeter of the triangle is then calculated as thus;

Perimeter = s + s + b

Perimeter = \sqrt{b^2 + h^2} + \sqrt{b^2 + h^2} +b

Perimeter = 2\sqrt{b^2 + h^2} + b

For the volume of the cone,

when the triangle is spin, the base of the triangle forms the diameter of the cone;

Volume = \frac{1}{3} \pi * r^2 * h

Where r = \frac{1}{2} * diameter

So, r = \frac{1}{2}b

So, Volume = \frac{1}{3} \pi * (\frac{b}{2})^2 * h

Base on the above illustrations, the program is as follows;

#include<iostream>

#include<cmath>

using namespace std;

void CalcArea(double b, double h)

{

//Calculate Area

double Area = 0.5 * b * h;

//Print Area

cout<<"Area = "<<Area<<endl;

}

void CalcPerimeter(double b, double h)

{

//Calculate Perimeter

double Perimeter = 2 * sqrt(pow(h,2)+pow((0.5 * b),2)) + b;

//Print Perimeter

cout<<"Perimeter = "<<Perimeter<<endl;

}

void CalcVolume(double b, double h)

{

//Calculate Volume

double Volume = (1.0/3.0) * (22.0/7.0) * pow((0.5 * b),2) * h;

//Print Volume

cout<<"Volume = "<<Volume<<endl;

}

int main()

{

double b, h;

//Prompt User for input

cout<<"Base: ";

cin>>b;

cout<<"Height: ";

cin>>h;

//Call CalcVolume function

CalcVolume(b,h);

//Call CalcArea function

CalcArea(b,h);

//Call CalcPerimeter function

CalcPerimeter(b,h);

 

return 0;

}

3 0
3 years ago
Which of these is not one of the main parts of an email?
pav-90 [236]
The header is not a main part of the email. 
3 0
3 years ago
Read 2 more answers
What does the hard disk drive do?
Airida [17]
It stores data and retrieving digital information using one or more rigid rapidly rotating disks (platters) coated in magnetic material
4 0
3 years ago
Read 2 more answers
This rights protected document cannot be opened because the rights management feature has been disabled on your machine by Polic
erastovalidia [21]

This rights protected document cannot be opened because the rights management feature has been disabled on your machine by Policy is known to be an error message.

<h3>What is the error message about?</h3>

The Error message above is known to be one that shows that  your IT department has made a group policy in one's company's Active Directory. It is known to be one that tends to disables the use of ADRMS (Rights Management Feature) in all to all users.

Note that in the case above, one need to ask your IT department to be able to disable or make changes to the group policy so that it will not apply to the users who are said to require access secured Microsoft Office documents in any of Ansarada Rooms.

Another option is to look at documents off the network with the use of a personal computer or a mobile phone.

Hence, This rights protected document cannot be opened because the rights management feature has been disabled on your machine by Policy is known to be an error message.

Learn more about  error message from

brainly.com/question/25671653

#SPJ1

4 0
2 years ago
Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first
irinina [24]

Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires is True.

True

<u>Explanation:</u>

In network packet loss is considered as connectivity loss. In this scenario host A send two segment to host B and acknowledgement from host B Is awaiting at host A.

Since first acknowledgement is lost it is marked as packet lost. Since in network packet waiting for acknowledgement is keep continues process and waiting or trying to accept acknowledgement for certain period of time, once period limits cross then it is declared as packet loss.

Meanwhile second comes acknowledged is success. For end user assumes second segments comes first before first segment. But any how first segment expires.

3 0
3 years ago
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