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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_forwardQ1) Write a computer program that uses Newton's method to find the root of a given function, and apply this program to find the root of the following functions, using co as given. Stop the iteration when the error as estimated by n+1 - Enl is less than 10-6. Compare to your results for bisection. (a) f(x) = 1-2xe-/2, xo = 0; (b) f(x)=5-x-¹, x = ¹; (c) f(x)= x³ - 2x - 5, xo = 2; (d) f(x)=e-2, xo = 1; (e) f(x)=x-e, xo = 1; (f) f(x)=x-x-1, xo = 1; (g) f(x)=x²-sinx, xo =/; (h) f(x)= x³-2,0 = 1; (i) f(x) = x + tan x, zo = 3; (j) f(x)=2x-¹ In x, xo = 3.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(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(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(x² If h(x) X 2 , then 2arrow_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
- Please, solve the following problem and explain how the loop works .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(Python) Numerous engineering and scientific applications require finding solutions to a set of equations. Ex: 8x + 7y = 38 and 3x - 5y = -1 have a solution x = 3, y = 2. Given integer coefficients of two linear equations with variables x and y, use brute force to find an integer solution for x and y in the range -10 to 10. Ex: If the input is: 8 7 38 3 -5 -1 Then the output is: x = 3 , y = 2 Use this brute force approach: For every value of x from -10 to 10 For every value of y from -10 to 10 Check if the current x and y satisfy both equations. If so, output the solution, and finish. Ex: If no solution is found, output: "There is no solution" You can assume the two equations have no more than one solution. ''' Read in first equation, ax + by = c '''a = int(input())b = int(input())c = int(input()) ''' Read in second equation, dx + ey = f '''d = int(input())e = int(input())f = int(input())arrow_forward
- C++ for Engineers and ScientistsComputer ScienceISBN:9781133187844Author:Bronson, Gary J.Publisher:Course Technology PtrC++ Programming: From Problem Analysis to Program...Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage Learning