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Naddika [18.5K]
4 years ago
6

What is the value of vals[4][1]? double[][] vals = {{1.1, 1.3, 1.5}, {3.1, 3.3, 3.5}, {5.1, 5.3, 5.5}, {7.1, 7.3, 7.5}};

Computers and Technology
1 answer:
neonofarm [45]4 years ago
4 0

Answer:

When the user concludes the value of "vals[4][1]", then it will give an exception of "ArrayIndexOutOfBoundsException".

Explanation:

  • It is because the size of the above array is [4*3] which takes the starting index at [0][0] and ending index at [3][2]. It is because the array index value starts from 0 and ends in (s-1).
  • When the double dimension array size is [5][5], then it will conclude the value of [4][1].
  • The above array have following index which value can be calculated :-- [0][0],[0][1],[0][2],[1][0], [1][1],[1][2], [2][0], [2][1], [2][2],[3][0],[3][1] and [3][2].
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convert the following c code to mips. assume the address of base array is associated with $s0, n is associated with $s1, positio
Brums [2.3K]

Answer:

Explanation:

hello we will follow a step by step process for this code, i hope you find it easy.

Mips Equivalent code:

  sw      $0,0($fp)

.L7:

       lw      $2,12($fp)

       addiu   $2,$2,-1

       lw      $3,0($fp)

       slt     $2,$3,$2

       beq     $2,$0,.L2

       nop

       lw      $2,0($fp)

       sw      $2,8($fp)

       lw      $2,0($fp)

       addiu   $2,$2,1

       sw      $2,4($fp)

.L5:

       lw      $3,4($fp)

       lw      $2,12($fp)

       slt     $2,$3,$2

       beq     $2,$0,.L3

       nop

       lw      $2,8($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $3,24($2)

       lw      $2,4($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $2,24($2)

       slt     $2,$2,$3

       beq     $2,$0,.L4

       nop

       lw      $2,4($fp)

       sw      $2,8($fp)

.L4:

       lw      $2,4($fp)

       addiu   $2,$2,1

       sw      $2,4($fp)

       b       .L5

       nop

.L3:

       lw      $3,8($fp)

       lw      $2,0($fp)

       beq     $3,$2,.L6

       nop

       lw      $2,0($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $2,24($2)

       sw      $2,16($fp)

       lw      $2,8($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $3,24($2)

       lw      $2,0($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       sw      $3,24($2)

       lw      $2,8($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $3,16($fp)

       sw      $3,24($2)

.L6:

    lw      $2,0($fp)

       addiu   $2,$2,1

       sw      $2,0($fp)

       b       .L7

        nop

cheers  i hope this helps

7 0
3 years ago
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omeli [17]

Answer:

1) 10cos ( 2πfct + 10sin 2πfmt )

2) 50 watts

3) 1000 Hz

Explanation:

m(t) = 10cos(1000πt)

c(t) = 10cos(2πfct )

modulation index (  = 10

1) expression for modulated signal ( u(t) )

u(t) = Ac(cos 2πfct ) + A (cos 2πfct m(t) ) Am cos2πf mt

      = 10cos 2πfct + 100 cos 2πfct  cos2π500t

      = 10cos ( 2πfct + 10sin 2πfmt )

2) power of modulated signal

power of modulated signal  μ^2 / 2 ]

           = 10^2/2 ]

           = 50 watts

3) Bandwidth

B = 2fm  = 2 * 500 = 1000 Hz

4 0
3 years ago
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