The answer will be google docs since it depends on a web and its an application. hope it helps
<h2>
Answer:</h2>
F = 13
<h2>
Explanation:</h2>
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:
<em>i. Solve the brackets first</em>
=> (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
<em>ii. Solve either the multiplication or division next whichever one comes first.</em>
=> 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
<em>iii. Now solve the addition by first substituting the values calculated earlier back into F.</em>
F = (4%2)+2*Y +6/2 +(z&&x)
F = 0 + 10 + 3 + 0
F = 13
Therefore, the value of F is 13
1. The current is the same everywhere in the circuit. This means that wherever I try to measure
the current, I will obtain the same reading.
2. Each component has an individual Ohm's law Voltage Drop. This means that I can calculate
the voltage using Ohm's Law if I know the current through the component and the resistance.
3. Kirchoff's Voltage Law Applies. This means that the sum of all the voltage sources is equal to
the sum of all the voltage drops or
VS = V1 + V2 + V3 + . . . + VN
4. The total resistance in the circuit is equal to the sum of the individual resistances.
RT = R1 + R2 + R3 + . . . + RN
5. The sum of the power supplied by the source is equal to the sum of the power dissipated in
the components.
<span>PT = P1 + P2 + P3 + . . . + PN</span>
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Answer:
def isdivisible():
maxint=input("Enter the Max Int")
int1=0
int2=0
int1=input("Enter the first Integer")
int2=input("Enter the second Integer")
tup1=(int1, int2)
print(tup1)
i = 1
for i in range(1, int(maxint)-1):
if int(tup1[0])%i==0 & int(tup1[1])%i==0:
print(i)
else:
continue
isdivisible()
1.2 Outputs
First test case:
Enter the Max Int6
Enter the first Integer2
Enter the second Integer8
('2', '8')
1
2
Second test case: returning empty list
Enter the Max Int2
Enter the first Integer13
Enter the second Integer27
('13', '27')
Test case 3:
Enter the Max Int4
Enter the first Integer8
Enter the second Integer10
('8', '10')
1
2
Explanation:
The program is as above, and the three test cases are also mentioned. We have created a tuple out of two input integer, and performed the output as required.
