Define a function calc_pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base. Sample output with inputs: 4.5 2.1 3.0Volume for 4.5, 2.1, 3.0 is: 9.45Relevant geometry equations: Volume = base area x height x 1/3 Base area = base length x base width.
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Define a function calc_pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base.
Sample output with inputs: 4.5 2.1 3.0Volume for 4.5, 2.1, 3.0 is: 9.45Relevant geometry equations:
Volume = base area x height x 1/3
Base area = base length x base width.
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- Define a function calc_pyramid_volume() with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base. calc_pyramid_volume() calls the given calc_base_area() function in the calculation.Relevant geometry equations:Volume = base area x height x 1/3(Watch out for integer division).Sample output with inputs: 4.5 2.1 3.0 Volume for 4.5, 2.1, 3.0 is: 9.45 python def calc_base_area(base_length, base_width): return base_length * base_width ''' Your solution goes here ''' length = float(input())width = float(input())height = float(input())print('Volume for', length, width, height, "is:", calc_pyramid_volume(length, width, height))Create a function that scales, for example, an analog input value (counts) to measured value. Mao. the function can, for example, have 5 inputs and 1 output. The 5 inputs are then: In_value, Inn_Min, Inn_Max, Ut_Min and Ut_Max. The function value is then a scaled value.Example: If the input value is from an analog input (counts) which should correspond to a temperature between -20 and + 300 degrees then Inn_Min = 0, Inn_Max = 32767 (WAGO analog in), Ut_Min = -20 and Ut_Max = 300. Test the function and reflect on why we can realize this as a function and not have to create a function block. use e!CockPITImplement a C function, named triangle, that outputs Pascal’s triangle. In general, the Pascal’s triangle can be represented as: 0C0 1C1 1C0 2C2 2C1 2C0 3C3 3C2 3C1 3C0 4C4 4C3 4C2 4C1 4C0. . .where nCr represents how many ways there are to choose r from n, not counting duplicates. The formula used to calculate nCr can be written as:nCr = n!/r!(n −r)!where n! is the factorial of n. The function triangle:• is called by value where exactly one parameter (the number of rows of Pascal’s triangle)is passed to it.• returns void (i.e., no value).• prints out Pascal’s triangle to the standard output.• Employs two other functions, namely:– int choose(int n, int r);That chooses r from n– int factorial(int n);That calculates the factorial of n.Note: You are allowed — in fact, encouraged — to design and use any other functions as needed. Note:• The number of the rows is limited to integers between 0 and 13 inclusive.• If 0 is entered…
- Define a function calc_pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base.Sample output with inputs: 4.5 2.1 3.0Volume for 4.5, 2.1, 3.0 is: 9.45 Relevant geometry equations:Volume = base area x height x 1/3Base area = base length x base width. ''' Your solution goes here ''' length = float(input())width = float(input())height = float(input())print('Volume for', length, width, height, "is:", calc_pyramid_volume(length, width, height))Define a function calc_pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base.Sample output with inputs: 4.5 2.1 3.0Volume for 4.5, 2.1, 3.0 is: 9.45 Relevant geometry equations:Volume = base area x height x 1/3Base area = base length x base width. ''' Your solution goes here ''' length = float(input())width = float(input())height = float(input())print('Volume for', length, width, height, "is:", calc_pyramid_volume(length, width, height)) pythonImplement the quadratic_formula() function. The function takes 3 arguments, a, b, and c, and computes the two results of the quadratic formula: x1=−b+b2−4ac2a x2=−b−b2−4ac2a The quadratic_formula() function returns the tuple (x1, x2). Ex: When a = 1, b = -5, and c = 6, quadratic_formula() returns (3, 2). Code provided in main.py reads a single input line containing values for a, b, and c, separated by spaces. Each input is converted to a float and passed to the quadratic_formula() function. Ex: If the input is: 2 -3 -77 the output is: Solutions to 2x^2 + -3x + -77 = 0 x1 = 7 x2 = -5.50 code: # TODO: Import math module def quadratic_formula(a, b, c):# TODO: Compute the quadratic formula results in variables x1 and x2return (x1, x2) def print_number(number, prefix_str):if float(int(number)) == number:print("{}{:.0f}".format(prefix_str, number))else:print("{}{:.2f}".format(prefix_str, number)) if __name__ == "__main__":input_line = input()split_line = input_line.split(" ")a =…
- Define a function which takes one parameter N, and use loop(s) to control and output a pyramid pattern as follows: print a total of N rows, the number of asterisks in each row equals the row number (e.g. the 1st row has 1 asteroid, the Nth row has N asteroids). Note the pattern should be center aligned. E.g. when N is 5, you will print the following pattern * * * * * * * * * * * * * * *(Python) Define a function calc_pyramid_volume() with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base. calc_pyramid_volume() calls the given calc_base_area() function in the calculation.Relevant geometry equations:Volume = base area x height x 1/3(Watch out for integer division).Sample output with inputs: 4.5 2.1 3.0Volume for 4.5, 2.1, 3.0 is: 9.45In Python Define a function pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base.Sample output with inputs: 4.5 2.1 3.0 Volume for 4.5, 2.1, 3.0 is: 9.45 Relevant geometry equations:Volume = base area x height x 1/3Base area = base length x base width. ''' Your solution goes here ''' length = float(input())width = float(input())height = float(input())print('Volume for 4.5, 2.1, 3.0 is:', pyramid_volume(length, width, height))
- Write a function that makes the base conversion. It takes two integers n and b as input (n >= 0 andb > 1, both in base 10) and expresses n in base b. For example, the decimal expansion of 345 inbase 10 is 345 = 3 × 102 + 4 × 101 + 5 × 100.Sample input: n = 12, b =2. Sample output: 1100Explanation: (12)10 = (1100)2Write a program using Scilab that includes the following lines x = linspace(0,1,N); y = sin(%pi*x/2); xi = linspace(0,1,100); The program should use the interp1 function to perform spline interpolation of the (x,y) data at the 100 xi points. These interpolated values should be compared to the exact values of the function sin(π x /2) and your program should use this to find the smallest number of samples N that gives an interpolation with an error of no more than ±0.00001Define a function CalcPyramidVolume with double data type parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volume of a pyramid with a rectangular base. Relevant geometry equations:Volume = base area x height x 1/3Base area = base length x base width.(Watch out for integer division). 419894.2696446.qx3zqy7 #include <iostream>using namespace std; /* Your solution goes here */ int main() { double userLength; double userWidth; double userHeight; cin >> userLength; cin >> userWidth; cin >> userHeight; cout << "Volume: " << CalcPyramidVolume(userLength, userWidth, userHeight) << endl; return 0;}