Consider a function: f (n) = n - 1. What function is it. Select one: a. onto b. one to one c. not a valid function d. Bijective
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A: Answer: -------- None of the other statements is true.
Q: The function f(x)=x2 from set of positive real numbers to positive real numbers is __________
A: Bijective
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A: given: explicit formula for a function from the set of all integers tothe set of positive integers…
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A: The solution for the above given question is given below:
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A: Answer: F is a onto function. So , the correct answers are d and e .
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A: Given f(x) = 3x+1 g(x) = x2 h(x) = 2x
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A: Here I explained Step by step answer for given Function S. I hope you like it.
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A: I have given an answer in step 2.
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Q: Determine whether or not the function f : Z × Z ! Z is onto, if f((m, n)) =m-n.
A: Determine whether or not the function f : Z × Z ! Z is onto, if f((m, n)) =m-n.
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A: the code is an given below :
Q: Define a function S : Z+ → Z+ as follows. For each positive integer n, S(n) = the sum of the…
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A: Answer: I have given answer in handwritten format and in brief.
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A:
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A: The properties for the given function is as follows.
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- Write a function that, given a list of non-overlapping intervals of items, accepts an item as an input and decides which, if any, interval that item falls in. If the elements are integers and the intervals are 1643-2033, 5532-7643, 8999-10332, 5666653-5669321, then the question point 9122 is in the third interval and 8122 is not.Write a function double lagrange(double x_int[], double y_int[], int n, double x) which calculates the value of the n-th order Lagrange’s polynomial Pn(x) defined for the set of nodes {x(0), y(0)}, ..., {x(n), y(n)}, for a given value of xWhich functions are one-to-one? Which functions are onto? Describe the inversefunction for any bijective function.(a) f : Z → N where f is defined by f (x) = x4 + 1(b) f : N → N where f is defined by f (x) = { x/2 if x is even, x + 1 if x is odd}(c) f : N → N where f is defined by f (x) = { x + 1 if x is even, x − 1 if x is odd}
- Consider the function, f(n), defined by the following code int f(int n) { int r = 0; if ( n % 2 == 0 ) return 1; // when n is even while ( n > 1 ) { n = n – 3; r++; } return r; } From the definitions of big-Theta and little-oh: Prove or disprove that f(n) is Θ ( n ) 2. Prove or disprove that f(n) is o ( n^2 ) ( note: function f(n) as defined by the code, that is, the return value, and is not asking about the runtime of the code.)a) Implement a function start( ) that takes as input a positive number a, rounds its logarithm base 4 to the nearest integer n, and returns the value of 1.0 x 2^n.Write a program in Python with a function squareRoot(x, epsilon) that uses bisection search to return a number y that is close enough to the square root of x, so that abs(y**2 - x) < epsilon.
- A triple (x, y, z) of positive integers is pythagorean if x2 + y2 = z2. Using the functions studied in class, define a function pyth which returns the list of all pythagorean triples whose components are at most a given limit. For example, function call pyth(10) should return [(3, 4, 5), (4, 3, 5), (6, 8, 10), (8, 6, 10)]. [Hint: One way to do this is to construct a list of all triples (use unfold to create a list of integers, and then a for-comprehension to create a list of all triples), and then select the pythagorean ones. def unfold[A, S](z: S)(f: S => Option[(A, S)]): LazyList[A] = f(z) match { case Some((h, s)) => h #:: unfold(s)(f) case None => LazyList() } code in scalaInteresting, intersecting def squares_intersect(s1, s2): A square on the two-dimensional plane can be defined as a tuple (x, y, r) where (x, y) are the coordinates of its bottom left corner and r is the length of the side of the square. Given two squares as tuples (x1, y1, r1) and (x2, y2, r2), this function should determine whether these two squares intersect, that is, their areas have at least one point in common, even if that one point is merely the shared corner point when these two squares are placed kitty corner. This function should not contain any loops or list comprehensions of any kind, but should compute the result using only integer comparisons and conditional statements. This problem showcases an idea that comes up with some problems of this nature; it is actually far easier to determine that the two squares do not intersect, and negate that answer. Two squares do not intersect if one of them ends in the horizontal direction before the other one begins, or if the same…Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] """ def generate_parenthesis_v1(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) return ifright>0: add_pair(res, s+")", left, right-1) ifleft>0: add_pair(res, s+"(", left-1, right+1) res= [] add_pair(res, "", n, 0) returnres def generate_parenthesis_v2(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) ifleft>0: add_pair(res, s+"(", left-1, right) ifright>0andleft<right: add_pair(res, s+")", left, right-1).
- Write a Matlab function that will: read in the spreadsheet file, accept two textual/string arguments: first metric abbreviation/acronym (for example GP) from the list above, Second metric abbreviation/acronym (for example 3P%) from the list above, finds the player (A) who has the highest/maximum value of the FIRST METRIC (if there is more than one with that value, pick the first one - lowest row number) and stores his name and corresponding FIRST METRIC value, finds the player (B) who has the lowest/minimum value of the FIRST METRIC (if there is more than one with that value, pick the first one - lowest row number) and stores his name and corresponding FIRST METRIC value, computes FIRST METRIC average for all players, finds SECOND METRIC value for both players found above, computes SECOND METRIC average for all players,Prepare the SPIM program for the function: int fun(int n) { int i,f=1; for (i=n; i>0; i--) f=f+1; return f; }Write the code in python to define a function has_repeat(mystr), which takes a string parameter and returns a boolean result. - If mystr has a character that is repeated (count is more than 1 for that character), return True - Otherwise, return False Hint: You can solve this problem by iterating over the characters in mystr and comparing each character's count with 1. For example: Test Result print(has_repeat("Happy")) True print(has_repeat("sad")) False print(has_repeat("salmons")) True print(has_repeat("Trouts")) False