The first iteration of the Fixed point method to solve the following system near (0.8,0.4) x- x2 - y2 = 0 y - x2 + y2 = 0 a) (0.6, 0.5) b) (0.8 , 0.5) c) (0.7, 0.04) d) ( 0.06, 0.05) If x=(1.25, 0.02, - 5.15, 0), ||x|l = a)5.15 b) 5.8 c) 6.4 d) 28 By using Bisection method to find a root of the equation x3 – 4x – 8.95 = 0, a root lies between

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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3- The first iteration of the Fixed point method to solve the following
system near (0.8,0.4) x- x2 - y2 = 0
y - x? + y? = 0
a) (0.6 , 0.5)
b) (0.8 , 0.5)
c) ( 0.7 , 0.04) d) ( 0.06 , 0.05)
4- If x=(1.25, 0.02, – 5.15, 0), ||x||, =
a) 5.15
b) 5.8
c) 6.4
d) 28
5- By using Bisection method to find a root of the equation
x3 – 4x – 8.95 = 0, a root lies between
a) a=0 and b=1 b) a=1 and b=2 c)a=2 and b=3
d) a=3 and b=4
Transcribed Image Text:3- The first iteration of the Fixed point method to solve the following system near (0.8,0.4) x- x2 - y2 = 0 y - x? + y? = 0 a) (0.6 , 0.5) b) (0.8 , 0.5) c) ( 0.7 , 0.04) d) ( 0.06 , 0.05) 4- If x=(1.25, 0.02, – 5.15, 0), ||x||, = a) 5.15 b) 5.8 c) 6.4 d) 28 5- By using Bisection method to find a root of the equation x3 – 4x – 8.95 = 0, a root lies between a) a=0 and b=1 b) a=1 and b=2 c)a=2 and b=3 d) a=3 and b=4
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