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What is the answer to Evaluate x1 - x-1 plus x0 for x equals 2?
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∙ 11y ago
shut up and do your hw
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∙ 11y ago
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An example of a trinomal with x2?
x2 + x1 + x0
What is the distance between the origin ( 0 0) and a point at ( -15 8)?
Use Pythagoras: For a line joining two points (x0, y0) to (x1, y1) there is a right angle triangle: One side (leg) joins (x0, y0) to (x1, y0) with length (x1 - x0) One side (leg) joins (x1, y0) to (x1, y1) with length (y1 - y0) The hypotenuse joins (x0, y0) to (x1, y1) → length hypotenuse = √((x1 - x0)² + (y1 - y0)²) → length between (0, 0) and (-15, 8) is given by: distance = √((-15 - 0)² + (8 - 0)²) = √((-15)² + (8)²) = √(225 + 64) = √289 = 17 units
Find a c program for Bisection method?
#include#include #include#define ESP 0.001#define F(x) (x)*(x)*(x) + (x)*(x) + (x) + 7void main(){int i=1;float x0,x1,x2;double f1,f2,f0,t;clrscr( );printf( "\nEnter the value of x0: ");scanf( "%f",&x0);printf( "\nEnter the value of x1: ");scanf( "%f",&x1);printf( "\n____________________________________________\n");printf( "\niteration\tx0\tx1\tx2\tf0\tf1\tf2");printf( "\n_____________________________________________\n");do{x2= (x0+x1)/2;f0= F(x0);f1= F(x1);f2= F(x2);printf( "\n%d %f %f %f %lf %lf %lf",i,x0,x1,x2,f0,f1,f2);if (f0*f2ESP);printf( "\n________________________________________\n");printf( "\n\nApp.root = %f",x2);getch( );}
Can you evaluate 4xy for x1 y2?
It is 8.
How do you get the square root of 709?
One way to do this would be with Newton's method: Let: f(x) = 709 - x2 f'(x) = -2x We'll start with 27 as an approximation: x0 = 27 x1 = x0 - f(x0) / f'(x0) = 27 - (709 - 729) / -2(27) = 27 - 10/27 ≈ 26.629629 x2 = x1 - f(x1) / f'(x1) ≈ 26.629629 - (709 - 709.137174) / -2(26.629629) ≈ 26.627053 x3 = x2 - f(x2) / f'(x2) ≈ 26.627053 - (709 - 708.999951) / -2(26.627053) ≈ 26.627053 So we know that the square root is approximately 26.627053. You can check this of course by squaring our answer: 26.6270532 = 708.999951
X plus one over x squared plus two?
it equals x1 it equals x1
Maxz equals 2x1 plus 2x2 stc 5x1 plus 3x2 equals 8 x1 plus 2x2 equals 4 x1 x2 equals 0 and integerssolve by ipp?
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An example of a trinomal with x2?
x2 + x1 + x0
How can you find an equation line between two pair of points?
Assuming you want the equation of the straight line between the two points (x0, y0) and (x1, y1), the equation is: y - y0 = m(x - x0) where m is the gradient between the two points: m = (y1 - y0) ÷ (x1 - x0) Note: if the two x coordinates are equal, that is x0 = x1, then the equation of the line is x = x0.
What hooks up to X0?
On a transformer connection H1 and H2 are the primary connections. X1 and X2 are the secondary connections. If your transformer has a split secondary that is grounded, that terminal is X0. The sequence is X1 - X0 - X2. The X0 usually indicates that there is a connection to a neutral wire along with the ground wire.
What is the distance between the origin ( 0 0) and a point at ( -15 8)?
Use Pythagoras: For a line joining two points (x0, y0) to (x1, y1) there is a right angle triangle: One side (leg) joins (x0, y0) to (x1, y0) with length (x1 - x0) One side (leg) joins (x1, y0) to (x1, y1) with length (y1 - y0) The hypotenuse joins (x0, y0) to (x1, y1) → length hypotenuse = √((x1 - x0)² + (y1 - y0)²) → length between (0, 0) and (-15, 8) is given by: distance = √((-15 - 0)² + (8 - 0)²) = √((-15)² + (8)²) = √(225 + 64) = √289 = 17 units
Explain how to find the distance between the points (28 -17) and (-15-17) on a coordinate plane?
To find the distance between any two points on the Cartesian plane use Pythagoras: The distance between (x0, y0) and (x1, y1) is given by: distance = √((x1 - x0)² + (y1 - y0)²) → distance between (28, -17) and (-15, -17) is: distance = √((x1 - x0)² + (y1 - y0)²) = √((-15 - 28)² + (-17 - -17)²) = √((-43)² + (0)) = √1849 = 43 ------------------------ In this case, the y-coordinates are the same (y0 = y1 = -17), so this becomes: distance = √((x1 - x0)² + (y0 - y0)²) = √((x1 - x0)² + 0²) = √((x1 - x0)²) = |x1 - x0| The vertical bars around the expression mean the absolute value of the expression, which is the numerical value of the expression ignoring the sign. distance = |x1 - x0| = |-15 - 28| = |-43| = 43.
How do you construct the Lagrange interpolation for a polynomial cosx plus sinx given the four values of x?
Construct the Lagrange interpolating polynomial P1(x) for f(x) = cos(x)+sin(x) when x0 = 0; x1 = 0:3. Find the absolute error on the interval [x0; x1].
What is Newton raphson's method in r programing?
It's a method used in Numerical Analysis to find increasingly more accurate solutions to the roots of an equation. x1 = x0 - f(x0)/f'(x0) where f'(x0) is the derivative of f(x0)
Find a c program for Bisection method?
#include#include #include#define ESP 0.001#define F(x) (x)*(x)*(x) + (x)*(x) + (x) + 7void main(){int i=1;float x0,x1,x2;double f1,f2,f0,t;clrscr( );printf( "\nEnter the value of x0: ");scanf( "%f",&x0);printf( "\nEnter the value of x1: ");scanf( "%f",&x1);printf( "\n____________________________________________\n");printf( "\niteration\tx0\tx1\tx2\tf0\tf1\tf2");printf( "\n_____________________________________________\n");do{x2= (x0+x1)/2;f0= F(x0);f1= F(x1);f2= F(x2);printf( "\n%d %f %f %f %lf %lf %lf",i,x0,x1,x2,f0,f1,f2);if (f0*f2ESP);printf( "\n________________________________________\n");printf( "\n\nApp.root = %f",x2);getch( );}
What is the difference between Regula-falsi method and secant method?
Regula-Falsi Method evaluates using assumed variables like "a", "b", f(a), f(b) Secant Method Directly works with x1, x2, f(x1), f(x2) Difference is in the Assignment pattern only, otherwise both are used to find root of Non-Linear equations using the same procedure which is: x1= [a * f(b) - b * f(a)]/[f(b)-f(a)] x1= [x0 * f(x1) - x1 * f(x0)]/[f(x1)-f(x0)] Thank You :-)
What is the formula to calculate 5474115504 to the 9th root?
The only "formula" is 54741155041/9 However, you can calculate the value iteratively, using the Newton-Raphson method as follows: Define f(x) = x9 - 547411504 and solve for f(x) = 0 The first derivative is f'(x) = 9*x8 So take a guess at x, say x0. Calculate x1 = x0 - f(x0)/f'(x0) Continue: calculate x2 = x1 - f(x1)/f'(x1). If you started with a reasonably good estimate you will find that the the estimates converge to the answer. In this case, the answer is 12.079 (approx).
