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Scilla [17]
3 years ago
5

Side mirror using convex mirror or concave mirror?​

Computers and Technology
2 answers:
yarga [219]3 years ago
6 0

Answer:

A convex mirror bends light as it reflects the light, and the farther away a point is from the center, the more the light is bent. ... The other kind of mirror you ask about is a concave mirror. This is the opposite side of the glass bowl from earlier; a concave mirror bends light inwards, towards you.

s2008m [1.1K]3 years ago
3 0
A vehicles side mirror is convex, a concave mirror would enlarge everything but convex gives a bigger field of view
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1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

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3 years ago
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What frequency band is used by bluetooth, 802.11b, and 802.11g?
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Nookie1986 [14]

Answer:

1111000

Explanation:

perform the following. write from right to left:

- is the number even? then write down a 0

- is the number odd? then write down a 1 and subtract 1

- divide by 2

- repeat until you reach 0.

So for 120:

120 is even, so write down a 0 and continue with 120/2=60

60 is even,  so write down a 0 and continue with 60/2=30

30 is even,  so write down a 0 and continue with 30/2=15

15 is odd,  so write down a 1 and continue with 14/2=7

7 is odd, so write down a 1 and continue with 6/2=3

3 is odd, so write down a 1 and continue with 2/2=1

1 is odd, so write down a 1 and finish with 0

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3 years ago
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