MATHEMATICS HIGH SCHOOL

What is the result when the following code is run? double x = 1; double y = 1; int i = 0; do { y = x / 2; x = x + y; i = i + 1; } while (x < 2.5); System.out.print(i + " ");

Answers

Answer 1
Answer:

Answer:

The answer is: 3

Step-by-step explanation:

1. At the begining of the program we start by declairing the variables:

 double x=1, double y=1 and int i=0.

2. The structure do...while is used to defined the loop. x<2.5 is the finalization condition of the loop. i is the counter of the loop.

y=x/2 is the first calculation

x=x+y is the second one. Here is where the values of the variable x changes.

a) for the first iteration, the values of y and x are shown below:

\n\nx=1\ny=1\ny=1/2=0.5\nx=1+0.5=1.5\ni=1

The variable x is minor to 2.5 so the loop will continue computing.

b)   the second iteration, the values of y and x are shown below::

 y=0.5\nx=1.5\ny=(1.5)/(2)=0.75\n x=1.5+0.75=2.25\ni=2

The variable x is still minor to 2.5 so the loop will continue computing.

c) third iteration:

y=0.75\nx=2.25\ny=(2.25)/(2) =1.1125\nx=2.25+1.125=3.375\ni=3

The condition x<2.5 is not true so the loop ends.

3. System.out.print(i + " "); displays the value of the variable i wich value is 3.

Therefore the number 3 is display.

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How do you solve this equation?

Answers

p = -1
8=8p+13-3p
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Thats 5p
Now
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Answers

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Answers

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Answers

Answer:

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Step-by-step explanation:
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