Is there a way of how to delete this if so pleas tell me

3. Which property is used to select full Row of data grid view

Tju [1.3M]

Answer:

I think a is correct answer.

4 0

3 years ago

A linear representation of a hierarchical file directory is known as what?

MissTica

It is known as a directory path

5 0

3 years ago

Recursively computing the sum of the first n positive odd integers, cont. About (a) Use induction to prove that your algorithm t

julia-pushkina [17]

The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.

int recursiveOddSum(int n) {

if(2n-1==1) return 1;

return (2n-1) + recursiveOddSum(n-1);

}

To prove the correctness of this algorithm by induction, we start from the base case as usual:

$f(1)=1$

by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).

Now we can assume that $f(n-1)$ returns indeed the sum of the first n-1 odd numbers, and we have to proof that $f(n)$ returns the sum of the first n odd numbers. By the recursive logic, we have

$f(n)=f(n-1)+2n-1$

and by induction, $f(n-1)$ is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So, $f(n)$ is the sum of the first n odd numbers, as required:

$f(n)=\underbrace{\underbrace{f(n-1)}_{\text{sum of the first n-1 odds}}+\underbrace{2n-1}_{\text{n-th odd}}}_{\text{sum of the first n odds.}}$

6 0

3 years ago

Half of the integers stored in the array data are positive, and half are negative. Determine the

jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3). cmp ecx, ARRAY_LENGTH

4). jge loop_end

5). mv temp_a, array[ecx]

6). cmp temp_a, 0

7). branch on nge

8). mv array[ecx], temp_a*2

9). add ecx, 1

10). jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0

3 years ago

Homework 8 Matlab Write a function called fibonacciMatrix. It should have three inputs, col1, col2, and n. col1 and col2 are ver

Amiraneli [1.4K]

In this exercise we have to use the knowledge in computational language in python to write the following code:

We have the code can be found in the attached image.

So in an easier way we have that the code is:

function v = myfib(n,v)

if nargin==1

v = myfib(n-1,[0,1]);

elseif n>1

v = myfib(n-1,[v,v(end-1)+v(end)]);

end

function v = myfib(n,v)

if nargin==1

v = myfib(n-1,[0,1]);

elseif n>1

v = myfib(n-1,[v,v(end-1)+v(end)]);

elseif n<1

v = 0;

end

function [n] = abcd(x)

if (x == 1 || x==0)

n = x;

return

else

n = abcd(x-1) + abcd(x-2);

end

fibonacci = [0 1];

for i = 1:n-2

fibonacci = [fibonacci fibonacci(end)+fibonacci(end-1)];

end

>> myfib(8)

ans =

0 1 1 2 3 5 8 13

>> myfib(10)

ans =

0 1 1 2 3 5 8 13 21 34

See more about python at brainly.com/question/18502436

8 0

2 years ago