Answer:
a = int(input("Enter first number: "))
b = int(input("Enter second number: "))
c = int(input("Enter third number: "))
a,b = a+b,b+c
print(a,b,c)
Explanation:
Assignments like this are easy in python, in the sense that no helper variables are needed!
Answer:
The answer is "Option b".
Explanation:
In the given code, a static method "inCommon" is declared, that accepts two array lists in its parameter, and inside the method two for loop is used, in which a conditional statement used, that checks element of array list a is equal to the element of array list b. If the condition is true it will return the value true, and if the condition is not true, it will return a false value. In this, the second loop is not used because j>0 so will never check the element of the first element.
Answer:
When you use the expression 'At the Drop of a Hat' you mean that something is happening instantly, without any delay. Example of use: “We're all packed and ready to go; we can leave at the drop of a hat.”
Explanation:
Hope this helped!! :)
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Answer:
Reorder terms.
y=52x−1
Cancel the common factor of 22.Factor 22 out of −2-2.
y−4=5x2+52⋅(2(−1))y-4=5x2+52⋅(2(-1))
Cancel the common factor.
y−4=5x2+52⋅(2⋅−1)y-4=5x2+52⋅(2⋅-1)
Rewrite the expression.
y−4=5x2+5⋅−1y-4=5x2+5⋅-1
Multiply 55 by −1-1.
y−4=5x2−5
Answer:
Answer explained
Explanation:
From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.
we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.
