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(Geometry: intersecting point) Two points on line 1 are given as (x1, y1) and (x2, y2) and on line 2 as (x3, y3) and (x4, y4), as shown in Figure 3.8a and b.
The intersecting point of the two lines can be found by solving the following linear equations:
Write a
FIGURE 3.8 Two lines intersect in (a and b) and two lines are parallel in (c).
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Introduction to Java Programming and Data Structures, Comprehensive Version (11th Edition)
- (Mechanics) The deflection at any point along the centerline of a cantilevered beam, such as the one used for a balcony (see Figure 5.15), when a load is distributed evenly along the beam is given by this formula: d=wx224EI(x2+6l24lx) d is the deflection at location x (ft). xisthedistancefromthesecuredend( ft).wistheweightplacedattheendofthebeam( lbs/ft).listhebeamlength( ft). Eisthemodulesofelasticity( lbs/f t 2 ).Iisthesecondmomentofinertia( f t 4 ). For the beam shown in Figure 5.15, the second moment of inertia is determined as follows: l=bh312 b is the beam’s base. h is the beam’s height. Using these formulas, write, compile, and run a C++ program that determines and displays a table of the deflection for a cantilevered pine beam at half-foot increments along its length, using the following data: w=200lbs/ftl=3ftE=187.2106lb/ft2b=.2fth=.3ftarrow_forward(Geometry: great circle distance) The great circle distance is the distance between two points on the surface of a sphere. Let (x1, y1) and (x2, y2) be the geographi- cal latitude and longitude of two points. The great circle distance between the two points can be computed using the following formula: d = radius * arccos(sin(x1) * sin(x2) + cos(x1) * cos(x2) * cos(y1 - y2)) Write a program that prompts the user to enter the latitude and longitude of two points on the earth in degrees and displays its great circle distance. The average earth radius is 6,378.1 km. The latitude and longitude degrees in the formula are for north and west. Use negative to indicate south and east degrees.arrow_forward(True or False) Seven different positive integers are randomly chosen between 1 and 2022 (including 1 and 2022).There must be a pair of these integers has a difference that is a multiple of 6.arrow_forward
- (proof by contraposition) If the product of two integers is not divisible by an integer n, then neither integer is divisible by narrow_forward(Random Walk Robot) A robot is initially located at position (0, 0) in a grid [−5, 5] × [−5, 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move in and the current position of the robot. Use formatted output to print the direction (Down, Up, Left or Right) in the left. The direction takes 10 characters in total and fill in the field with empty spaces. The statement to print results in such format is given below: cout << setw(10) << left << ‘Down’ << ... ; cout << setw(10) << left << ‘Up’ << ...; If the robot moves back to the original place (0,0), print “Back to the origin!” to the console and stop the program. If it reaches the boundary of the grid, print “Hit the boundary!” to the console and stop the program. A successful run of your code may look like: Due to randomness, your results may have a different trajectory…arrow_forward(Random Walk Robot) A robot is initially located at position (0, 0) in a grid [−5, 5] × [−5, 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move in and the current position of the robot. Use formatted output to print the direction (Down, Up, Left or Right) in the left. The direction takes 10 characters in total and fill in the field with empty spaces. The statement to print results in such format is given below: cout << setw(10) << left << ‘Down’ << ... ; cout << setw(10) << left << ‘Up’ << ...; If the robot moves back to the original place (0,0), print “Back to the origin!” to the console and stop the program. If it reaches the boundary of the grid, print “Hit the boundary!” to the console and stop the program. A successful run of your code may look like: Due to randomness, your results may have a different…arrow_forward
- (Algebra: solve 2 X 2 linear equations) You can use Cramer's rule to solve the following 2 x 2 system of linear equations: ed – bf af – ec y bc ax + by = e X = cx + dy = f ad ad – bc Write a function with the following header: void solveEquation(double a, double b, double c, double d, double e, double f, double& x, double& y, bool& isSolvable) If ad – bc is 0, the equation has no solution and isSolvable should be false. Write a program that prompts the user to enter a, b, c, d, e, and f and displays the result. If ad – bc is 0, report that "The equation has no solution." See Program- ming Exercise 3.3 for sample runs.arrow_forward(x² If h(x) X 2 , then 2arrow_forwardPlease, solve the following problem and explain how the loop works .arrow_forward
- x4 + 2x3 – 7x2 + 3 = 0 a) One of the root of the equation lies in the range (1.0, 2.0). Find this root in 100 iterations using the bisection method. b) Draw the graph of the function between points (0, 2). Your code should include the following steps: • Write the steps of the bisection function (if, else...) and explain each step. (Explain each step in English or Turkish.) • Your code should calculate the root. • Graphic; Variables of x and y axes should be written, x and y axis names should be written, Series should be written to calculate x axis. Use the linspace() for the x series of the graph and section the range 0-2 into 100 pięces.arrow_forward(Algebra: solve quadratic equations) The two roots of a quadratic equation ax? + bx + c = 0 can be obtained using the following formula: -b + VB - 4ac and -b - VB - 4ac 2a 2a b - 4ac is called the discriminant of the quadratic equation. If it is positive, the equation has two real roots. If it is zero, the equation has one root. If it is negative, the equation has no real roots. Write a program that prompts the user to enter values for a, b, and c and displays the result based on the discriminant. If the discriminant is positive, display two roots. If the discriminant is 0, display one root. Otherwise, display "The equation has no real roots."arrow_forward(strongly connected digraph) A digraph is said to be strongly connected if there exists a path in both directions for every u-v pair, from u to v and v to u. python code for given statementarrow_forward
C++ for Engineers and ScientistsComputer ScienceISBN:9781133187844Author:Bronson, Gary J.Publisher:Course Technology Ptr
C++ Programming: From Problem Analysis to Program...Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage Learning