Answer:
Following are the answer to this question:
Explanation:
In option 1:
The value of n is= 7, which is (base case)
when n=k for the true condition:
when n=k+1 it tests the value:
since k>6 hence the value is KH>3 hence proved.
In option 2:
when:
for n=1:(base case)
0<=0 \\ condition is true
when the above statement holds value n=1
when n=k
when n=k+1
![[\therefore KH>K \Rightarrow \log(KH>\loK)]](https://tex.z-dn.net/?f=%5B%5Ctherefore%20KH%3EK%20%5CRightarrow%20%20%5Clog%28KH%3E%5CloK%29%5D)
In option 3:
when n=1:
when n=k
![\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\](https://tex.z-dn.net/?f=%5Cto%20%28A_1%5Ccap%20A_2%20%5Ccap.....A_k%29%20%5Ccup%20B%5C%5C%3D%28A_1%5Ccup%20B%29%20%5Ccap%28A_2%5Ccup%20B_2%29....%28A_k%20%5CcapB%29.....%28a%29%5C%5C%5Cto%20n%3D%20k%2B1%5C%5C%20%5Cto%20%28A_1%5Ccap%20A_2%20%5Ccap.....A_%7BkH%7D%29%20%5Ccup%20B%3D%20%28A_1%5Ccup%20B%29%5C%5C%5C%5C%5Cto%20%20%5B%28A_1%5Ccap%20A_2%20%5Ccap.....A_%7Bk%7D%29%20%5Ccup%20B%5D%5Ccap%20%28A_%7BKH%7D%5Ccup%20B%29%5C%5C%5C%5C%5Cto%20%20%5B%28A_1%5Ccup%20B%29%20%5Ccap%20%28A_2%20%5Ccup%20B%29%20%5Ccap%20%28A_3%5Ccup%20B%29.....%28A_k%5Ccup%20B%29%5Ccap%20%28A_%7Bk%2B1%7D%20%5Ccup%20B%29%5C%5C%5C%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20substituting%20%5C%20equation%20%5C%20a%20%5C%5C%5C%5C)
hence n=k+1 is true.
Answer:
i think its C.
if its wrong i'm truly sorry
Explanation:
Opening unfamiler emails and visiting unknown websites
Answer:
def leap_year(y):
if y % 4 == 0:
return 1
else:
return 0
def number_of_days(m,y):
if m == 2:
return 28 + leap_year(y)
elif m == 1 or m == 3 or m == 5 or m == 7 or m == 8 or m ==10 or m == 12:
return 31
elif m == 4 or m == 6 or m == 9 or m == 11:
return 30
def days(m,d):
if m == 1:
return 0 + d
if m == 2:
return 31 + d
if m == 3:
return 59 + d
if m == 4:
return 90 + d
if m == 5:
return 120 + d
if m == 6:
return 151 + d
if m == 7:
return 181 + d
if m == 8:
return 212 + d
if m == 9:
return 243 + d
if m == 10:
return 273 + d
if m == 11:
return 304 + d
if m == 12:
return 334 + d
def days_left(d,m,y):
if days(m,d) <= 60:
return 365 - days(m,d) + leap_year(y)
else:
return 365 - days(m,d)
print("Please enter a date")
day=int(input("Day: "))
month=int(input("Month: "))
year=int(input("Year: "))
choice=int(input("Menu:\n1) Calculate the number of days in the given month.\n2) Calculate the number of days left in the given year.\n"))
if choice == 1:
print(number_of_days(month, year))
if choice == 2:
print(days_left(day,month,year))
Explanation:
Hoped this helped
