Answer:
coupons = int(input("Enter the number of coupons you win: "))
candy_bars = int(coupons / 10)
gumballs = coupons % 10
print("Candy bars: " + str(candy_bars) + ", Gumballs: " + str(gumballs))
Explanation:
*The code is in Python.
Ask the user to enter the number of coupons
Calculate the number of candy bars, divide the coupons by 10 and typecst the result to int
Calculate the number of gumballs, use the modulo to find the remainder
Print the values
Answer:
Given that:
A= 40n^2
B = 2n^3
By given scenario:
40n^2=2n^3
dividing both sides by 2
20n^2=n^3
dividing both sides by n^2 we get
20 = n
Now putting n=20 in algorithms A and B:
A=40n^2
= 40 (20)^2
= 40 * (400)
A= 16000
B= 2n^3
= 2 (20)^3
= 2(8000)
B= 16000
Now as A and B got same on n = 20, then as given:
n0 <20 for n =20
Let us take n0 = 19, it will prove A is better than B.
We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.
System software are software that can be installed and run separately
Answer:
def rec_dig_sum( num ):
num_list = [ digit for digit in str(num)]
total = 0
for x in num_list:
total += x
return total
def dict_of_rec_dig_sums(low, high):
mydict = dict()
for number in the range(low, high+1):
mydict[rec_dig_sum(number)] = number
return mydict
Explanation:
The python program defines two functions, "rec_dig_sum" and "dict_of_rec_dig_sums". The former accepts a number and returns the sum of the digits of the number while the latter accepts a low and high number range.
The program returns a dictionary with the recursive sum as the keys and the number count as the values.
Answer:
A) Maintaining the shared connections between offices
Explanation:
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