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wariber [46]
3 years ago
7

What is a BUG in terms of computer programming?

Computers and Technology
1 answer:
diamong [38]3 years ago
4 0

Answer:

C

Explanation:

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Please help meeee , you will get 20 points
docker41 [41]

Answer:

true

Explanation:

true trueee 3uehehehdgeyeyehhehehegegegrhrtggrevegrgrhehehru

5 0
3 years ago
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Mirrors on cars exist to____.
TEA [102]

Answer:

c is ur answer

Explanation:

7 0
3 years ago
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The process of learning by making mistakes and by trial and error is called _____.
vodomira [7]
I believe it’s problem solving.
6 0
3 years ago
Given two Strings String s1 = "11223351638791377189728193"; String s2 = "983763978669976862724569282405742578"; and String varia
inn [45]

Answer:

import java.math.BigInteger;  

public class Main {  

public static void main(String[] args)  {  

String s1 = "11223351638791377189728193";   String s2 = "983763978669976862724569282405742578";

BigInteger num1 = new BigInteger(s1);  

BigInteger num2 = new BigInteger(s2);  

BigInteger product = num1.multiply(num2);  

String s3 = product.toString(10);

System.out.println(s3);  

}  

}  

Explanation:

Because s1 and s2 contain large integers, the program need the biginteger module in other to compute the products of s1 and s2 when converted to integers.

This line initializes the two string variables s1 and s2

String s1 = "11223351638791377189728193";   String s2 = "983763978669976862724569282405742578";

The next two lines convert the two strings to BigInteger. This allows the program to multiply large integers

<em> BigInteger a = new BigInteger(s1);   </em>

<em>BigInteger b = new BigInteger(s2);  </em>

<em />

This computes the product of the two big integers

BigInteger product = num1.multiply(num2);  

This converts the product to String s3

String s3 = product.toString(10);

This prints s3

System.out.println(s3);  

}  

}  

7 0
3 years ago
A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2 until it reaches its final angu
Nataly_w [17]
The first thing we are going to do is find the equation of motion:
 ωf = ωi + αt
 θ = ωi*t + 1/2αt^2
 Where:
 ωf = final angular velocity
 ωi = initial angular velocity
 α = Angular acceleration
 θ = Revolutions.
 t = time.
 We have then:
 ωf = (7200) * ((2 * pi) / 60) = 753.60 rad / s
 ωi = 0
 α = 190 rad / s2
 Clearing t:
 753.60 = 0 + 190*t
 t = 753.60 / 190
 t = 3.97 s
 Then, replacing the time:
 θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
 θ1 = 1494.51 rad
 For (10-3.97) s:
 θ2 = ωf * t
 θ2 = (753.60 rad / s) * (10-3.97) s
 θ2 = 4544,208 rad
 Number of final revolutions:
 θ1 + θ2 = (1494.51 rad + 4544.208 rad) * (180 / π)
 θ1 + θ2 = 961.57 rev
 Answer:
 the disk has made 961.57 rev 10.0 s after it starts up
3 0
3 years ago
Read 2 more answers
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