answer: D. 5

The function foo takes two arguments, x and y. x is just a regular variable, while y is a function that takes one argument. In the body of the function, y(x) is the first function call. This calls the function y with x as its argument. In the example code, y is a lambda function that squares its argument, so y(x) computes the square of x. The second function call is y(x+1). This calls the same y function, but with x+1 as its argument. In the example code, this computes the square of x+1. Finally, the two results from the function calls are added together, and the sum is returned by foo. In the example code, foo(1, lambda x: x*x) is called. This passes 1 as the value of x and a lambda function that squares its argument as the value of y. Therefore, y(x) evaluates to 1 squared, which is 1, and y(x+1) evaluates to 2 squared, which is 4. The sum of these two values is 5, which is the value returned by foo.

Let's analyze the function `foo` with the provided arguments: 1. `x` is `1`, and `y` is a lambda function defined as `lambda x: x*x`, which squares its input. Here's how `foo` works in this case: - `y(x)` with `x` being `1`, evaluates to `1*1`, which is `1`. - `y(x+1)` with `x+1` being `2` (since `x` is `1`), evaluates to `2*2`, which is `4`. Adding these results together gives `1 + 4`, which equals `5`. Thus, `print(foo(1, lambda x: x*x))` will output `5`.

Can someone please walk me though how we get the answer 5 from this? def foo(x,y): return (y(x) + y(x+1)) print(foo(1, lambda x: x*x))