0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
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
Find the multiplicity root of the equation x3-x2-x plus 1 equals 0 near x0 equals 0.9?
The end part of the question does not seem to make sense. The equation has three real roots, a single root at x = -1 and a double root at x = 1
When x0 how many solutions is this?
x0 = 1 because any number raised to the power of 0 is always equal to 1
How do you integrate of e -2x for x0?
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
What is the net displacement of the particle between 0 seconds and 80 seconds?
That would besqrt[ (x80 - x0)2 + (y80 - y0)2 ) at an angle of tan-1 (y80 - y0) / (x80 - x0)or(x80 - x0) i + (y80 - y0) j
The square root of -x if x0?
If x is 0, the square root is 0 also.
Equation for linear approximation?
The general equation for a linear approximation is f(x) ≈ f(x0) + f'(x0)(x-x0) where f(x0) is the value of the function at x0 and f'(x0) is the derivative at x0. This describes a tangent line used to approximate the function. In higher order functions, the same concept can be applied. f(x,y) ≈ f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) where f(x0,y0) is the value of the function at (x0,y0), fx(x0,y0) is the partial derivative with respect to x at (x0,y0), and fy(x0,y0) is the partial derivative with respect to y at (x0,y0). This describes a tangent plane used to approximate a surface.
How many x intercepts can a quadratic function have?
For a quadratic function, there is one minimum/maximum (the proof requires calculus) and also it is either always convex or concave (prove is also calculus) it is continuous every where, hence, it can have a maximum of 2 roots. Graph it. If there is more than 2 roots, by Intermediate Value Theorem, it cannot be convex/concave everywhere. It will HAVE to have two intervals of increasing or decreasing. It can be easily proven that given any quadratic function f(x), if x = x0 is a minimum/maximum, and x=a != x0 is a root, then 2x0-a is also a root. It is still true that a = x0 as 2x0-x0=x0 implying it is the only root. But the concept of min/max requires Calculus to prove existence. So, this is Calculus, not algebra.
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)
How do you find the square root of 350?
Use the Newton-Raphson method. Let f(x) = x2 - 350 and let f'(x) = 2*x Then start with any guess, x0. The next estimate is x1 = x0 - f(x0)/f'(x0). Continue iterations with xn+1 = xn - f(xn)/f'(xn). If you start with x0 = 15 (quite a long way off, given that 152 is only 225), x2 is accurate to 6 decimal places (error = 8 in 10 million) and x3 to 13 dp (2 in 100 trillion).
The measure of the supplement of an angle exceeds twice the measure of the supplement of the complemant of the angle by 40?
The answer is -13 1/3ohere is the detailed calculation for the problem:Let x0 be the angle, then;(180 - x0) - 2[180 - (90 - x0)] =40(180 -x0) - 2[90+x0]=40180 -x0 - 180 - 2x0=40-3x0=40hencex0= -13 1/3oAny comments are welcome
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
what is 4E8374832E374684237eX72372+399x0+273646728-1888?
0! You said x0! anything x0=0!
Find the multiplicity root of the equation x3-x2-x plus 1 equals 0 near x0 equals 0.9?
The end part of the question does not seem to make sense. The equation has three real roots, a single root at x = -1 and a double root at x = 1
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 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).
When x0 how many solutions is this?
x0 = 1 because any number raised to the power of 0 is always equal to 1
