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graywombat44Lv1

6 Nov 2019

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-133.33 points SCalcET8 4.8.006. Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your ans places.) 2x3 3x2+2 -0, x1 x3 = -133.33 points SCalcET8 4.8.502.XPMI. Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your ans places.) x3 = Â·1-53.34 ports SCalcET8 4.8 505.XP. Use Newton's method to approximate the indicated root of the equation correct to six decimal places The root of x4-2x3 + 4x2-5 = O in the interval [1, 2] Show transcribed image text

-133.33 points SCalcET8 4.8.006. Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your ans places.) 2x3 3x2+2 -0, x1 x3 = -133.33 points SCalcET8 4.8.502.XPMI. Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your ans places.) x3 = Â·1-53.34 ports SCalcET8 4.8 505.XP. Use Newton's method to approximate the indicated root of the equation correct to six decimal places The root of x4-2x3 + 4x2-5 = O in the interval [1, 2]

Show transcribed image text Deanna HettingerLv2

22 Jan 2019