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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
