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Murljashka [212]

3 years ago

8

You can access cloud files from any computer smart phone or tablet as long as it is connected to the Internet is it true or fals

e

Computers and Technology

1 answer:

Zepler [3.9K]3 years ago

8 0

Answer:yes

Explanation:

because it is the internet

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Consider the following modification to the Merge Sort algorithm: divide the input array into thirds (rather than halves), recurs

ehidna [41]

Answer:

The correct answer to the following question will be "n logn".

Explanation:

Time Complexity should be the same as the whole last particular instance, just the log base keeps changing with three.
Throughout the circumstance of separating 2 arrays, we want one correlation. however, for splitting between 3-way arrays, will require 2 comparisons to sort.
However, after breaking the array into 3, we can lower the number of transfers by growing the comparison. So the time complexity would then persist the same, but the log should get base 3 as we've classified into different parts.

So the time complexity will be:

⇒ $T(n)=3T(\frac{n}{3})+ O(n)$

⇒ $=O(n \ logn) \ with \ base \ 3$

7 0

3 years ago

The section of the Telecommunications Act that attracted the most attention was the one calling for the creation of ______, whic

jekas [21]

Answer:

the V-chip

Explanation:

8 0

3 years ago

How can you create a safe and secure password? Choose a password unassociated with your personal history. Choose the same passwo

Simora [160]

Answer:

sudoors787389

Explanation:

in other to get more secured password a combination of password like this will be more secured (sudoors787389)

8 0

3 years ago

Which of the following form a set of angles of a triangle ?

Vsevolod [243]

Answer:

I think it is the first set

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3 years ago

Assume that k corresponds to register $s0, n corresponds to register $s2 and the base of the array v is in $s1. What is the MIPS

BlackZzzverrR [31]

Answer:

hello your question lacks the C segment so here is the C segment

while ( k<n )

{v[k] = v[k+1];

k = k+1; }

Answer : while:

bge $s0, $s2, end # while (k < n)

addi $t0, $s0, 1 # $t0 = k+1

sll $t0, $t0, 2 # making k+1 indexable

add $t0, $t0, $s1 # $t0 = &v[k+1]

lw $t0, 0($t0) # $t0 = v[k+1]

sll $t1, $s0, 2 # making k indexable

add $t1, $t1, $s1 # $t1 = &v[k]

sw $t0, 0($t1) # v[k] = v[k+1]

addi $s0, $s0, 1

j while

end:

Explanation:

The MIPS assembly code corresponding to the C segment is

while:

bge $s0, $s2, end # while (k < n)

addi $t0, $s0, 1 # $t0 = k+1

sll $t0, $t0, 2 # making k+1 indexable

add $t0, $t0, $s1 # $t0 = &v[k+1]

lw $t0, 0($t0) # $t0 = v[k+1]

sll $t1, $s0, 2 # making k indexable

add $t1, $t1, $s1 # $t1 = &v[k]

sw $t0, 0($t1) # v[k] = v[k+1]

addi $s0, $s0, 1

j while

end:

4 0

3 years ago