Given the value x=false ,y=5 and z=1 what will be the value of F=(4%2)+2*Y +6/2 +(z&&x)?

F = 13

Given:

x = false

y = 5

z = 1

F = (4%2)+2*Y +6/2 +(z&&x)

We solve this arithmetic using the order of precedence:

i. Solve the brackets first

=> (4 % 2)

This means 4 modulus 2. This is the result of the remainder when 4 is divided by 2. Since there is no remainder when 4 is divided by 2, then

4 % 2 = 0

=> (z && x)

This means (1 && false). This is the result of using the AND operator. Remember that && means AND operator. This will return false (or 0) if one or both operands are false. It will return true (or 1) if both operands are true.

In this case since the right operand is a false, the result will be 0. i.e

(z && x) = (1 && false) = 0

ii. Solve either the multiplication or division next whichever one comes first.

=> 2 * y

This means the product of 2 and y ( = 5). This will give;

2 * y = 2 * 5 = 10

=> 6 / 2

This means the quotient of 6 and 2. This will give;

6 / 2 = 3

iii. Now solve the addition by first substituting the values calculated earlier back into F.

F = (4%2)+2*Y +6/2 +(z&&x)

F = 0 + 10 + 3 + 0

F = 13

Therefore, the value of F is 13