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V125BC [204]
2 years ago
7

Lenny is working as an apprentice to a typesetter. He has been told to sort different font stamps based on their type. Match the

different fonts he picked to their type.

Computers and Technology
1 answer:
DedPeter [7]2 years ago
3 0

Answer:

Serif without kerning is the first one

Italic is the second one

Bold is the third one

Serif with kerning is the fourth one

Explanation:

I had this question and got this correct

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The correct answer is A.True
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7.2 code practice edhesive
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Explanation:

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Read 2 more answers
Consider bears = {"Grizzly":"angry", "Brown":"friendly", "Polar":"friendly"}. Can you replace #blank# so the code will print a g
Goshia [24]

Answer:

bears = {"Grizzly":"angry", "Brown":"friendly", "Polar":"friendly"}

for bear in bears:

   if bears[bear] == "friendly":

       print("Hello, "+bear+" bear!")

   else:

       print("odd")

Explanation:

A dictionary called bears is given. A dictionary consists of key-value pairs.

You need to check each key-value pairs in bears and find the ones that have "friendly" as value using a for loop and if-else structure. In order to access the values of the dictionary, use the dictionary name and the key (inside the loop, the key is represented as bear variable). If a key has a value of "friendly", print the greeting. Otherwise, print "odd".

6 0
3 years ago
Write a program that reads a file name from the keyboard. The file contains integers, each on a separate line. The first line of
Sindrei [870]

Answer:

Explanation:

import java.util.Scanner;

import java.io.*;

public class ArrayProcessing

{

public static void main(String[] args) throws IOException

{

Scanner kb = new Scanner(System.in);

String fileName;

System.out.print("Enter input filename: ");

fileName = kb.nextLine();

File file = new File(fileName);

if(!file.exists())

{

System.out.println("There is no input file called : " + fileName);

return;

}

int[] numbers = inputData(file);

printArray(numbers);

}

public static int[] inputData(File file) throws IOException

{

Scanner inputFile = new Scanner(file);

int index = 1;

int size = inputFile.nextInt();

int[] values = new int[size];

while(inputFile.hasNextInt() && index < values.length)

{

values[index] = inputFile.nextInt();

index++;

}

inputFile.close();

return values;

}

public static void printArray(int[] array)

{

int count = 1;

for (int ndx = 1; ndx < array.length; ndx++)

{

System.out.printf("%5.f\n", array[ndx]);

count++;

}

if(count == 10)

System.out.println("");

}

}

4 0
4 years ago
How are status reports useful
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Answer:

Project managers use status reports to keep stakeholders informed of progress and monitor costs, risks, time and work. Project status reports allow project managers and stakeholders to visualize project data through charts and graphs.

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