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Alex
3 years ago
10

Roses are red, violets are blue, I'm going to ki.ll myself, to make life better for you.

Computers and Technology
2 answers:
Shtirlitz [24]3 years ago
6 0

Answer:

i like this poem helps out the world

Explanation:

p,s. deserves brainiest

kvasek [131]3 years ago
6 0
No don’t kill urself lol
You might be interested in
An additional factor in how an element is rendered is that properties are passed from a parent element to its children in a proc
belka [17]

Answer:

style inheritance

Explanation:

What happens when a property on an element has no value supplied is governed by style inheritance. A property should inherit its value from its parent element, according to this specification.

6 0
1 year ago
As the team leader, John ensures that all his teammates are clear in the team goals they need to achieve. He demonstrates the qu
zimovet [89]

Delegation ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

7 0
3 years ago
Read 2 more answers
Write a filter that reads in a sequence of integers and prints the integers, removing repeated values that appear consecutively.
lara31 [8.8K]

Answer:

Following is the code for filter:

public class filter

{ public static void main(String[] args)

{ int x = StdIn.readInt();

System.out.print(" " + x + " ");

while(!StdIn.isEmpty())

{ int y = StdIn.readInt();

    if(y != x)

          System.out.print(" " + y + " ");

    x = y;

    }

    }

    }

Explanation:

  • A public class filter is used.
  • The main function will accept a single argument as string[], it is also known as java command line argument.
  • Now the Stdln.readInt is used to read the integers in the sequence and store it in integer x.
  • The value stored in variable x will be printed using System.out.print
  • Now unless the Stdln.readInt gets an empty value, check each value of sequence and store in variable y.
  • If y is not equal to previous value x, print it and shift the value of y into x.
  • Repeat the loop again.

i hope it will help you!

7 0
3 years ago
Read 2 more answers
A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the atte
11Alexandr11 [23.1K]

Answer:

Attenuation = 0.458\ db

Explanation:

Given

Power at point A = 100W

Power at point B = 90W

Required

Determine the attenuation in decibels

Attenuation is calculated using the following formula

Attenuation = 10Log_{10}\frac{P_s}{P_d}

Where P_s = Power\ Input and P_d  = Power\ output

P_s = 100W

P_d = 90W

Substitute these values in the given formula

Attenuation = 10Log_{10}\frac{P_s}{P_d}

Attenuation = 10Log_{10}\frac{100}{90}

Attenuation = 10 * 0.04575749056

Attenuation = 0.4575749056

Attenuation = 0.458\ db <em>(Approximated)</em>

7 0
3 years ago
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
2 years ago
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