Question

СРОЧНО!ПОМОГИТЕ1)Определите результат выполнения следующего фрагмента программы x:=1; while x<=10 do begin y:=x*x+4: x:=x+2; end; write('x=',x,'y=',y); А)x=9;y=53 B)x=11;y=85 C)x=9;y=85 2)Определите вариант правильно составленной программы вычисления произведения нечетных числе на промежутке {1,7} A)x:=1;p:=1;while x>=7 begin p:=p*x; x:=x+2;end;writeln('p=',p) B)x:=1;p:=1;while x<=7 begin p:=p*x; x:=x+2;end;writeln('p=',p) C)x:=1;p:=1;while x<=7 begin p:=p*x; end;writeln('p=',p) D)x:=0;p:=0;while x<=7 p:=p*x;x:=x+2 writeln ('p=',p)

Answer 1

1. Ответ варианта C;
2. Ответ варианта B.

Answer 2

1. С (x=9;y=85)
2. В:
x:=1;
p:=1;
while x<=7 begin
p:=p*x;
x:=x+2;
end;
writeln('p=',p);

Answer 3

Question 1: The given code snippet is as follows: ``` x:=1; while x<=10 do begin y:=x*x+4; x:=x+2; end; write('x=',x,'y=',y); ``` We need to determine the result of executing this code.

Step-by-step execution:

1. Initialize `x` with 1. 2. Enter the `while` loop because `x` is less than or equal to 10. 3. Calculate `y` by squaring the current value of `x` and adding 4. In this case, `y` becomes 5. 4. Increment `x` by 2. Now `x` becomes 3. 5. Go back to the beginning of the `while` loop. 6. Check the condition again. `x` is still less than or equal to 10, so continue the loop. 7. Calculate `y` as before. `y` becomes 13. 8. Increment `x` by 2. Now `x` becomes 5. 9. Repeat steps 5-8 until `x` becomes 11. 10. Exit the `while` loop because `x` is no longer less than or equal to 10. 11. Execute the `write` statement to output the values of `x` and `y`. In this case, it will output "x=11, y=13".

Result:

The correct answer is B) x=11; y=13.

Question 2: There are four options for a correctly written program to calculate the product of odd numbers in the range {1, 7}.

Option A: ``` x:=1; p:=1; while x>=7 do begin p:=p*x; x:=x+2; end; writeln('p=',p); ```

Option B: ``` x:=1; p:=1; while x<=7 do begin p:=p*x; x:=x+2; end; writeln('p=',p); ```

Option C: ``` x:=1; p:=1; while x<=7 do begin p:=p*x; end; writeln('p=',p); ```

Option D: ``` x:=0; p:=0; while x<=7 do begin p:=p*x; x:=x+2; writeln('p=',p); end; ```

We need to determine which option is the correct one.

Option A:

This option is incorrect because the condition `x>=7` in the `while` loop will never be true since `x` is initially set to 1.

Option B:

This option is correct. It correctly calculates the product of odd numbers in the range {1, 7}.

Option C:

This option is incorrect because it is missing the increment statement for `x` inside the `while` loop. As a result, the loop will never terminate and it will result in an infinite loop.

Option D:

This option is incorrect because it initializes `x` and `p` with 0, and the loop condition `x<=7` will not be true for the first iteration. Additionally, the `writeln` statement is inside the loop, so it will print the value of `p` multiple times.

Result:

The correct answer is B) x:=1; p:=1; while x<=7 do begin p:=p*x; x:=x+2; end; writeln('p=',p);.