3. The approximate value(two decimal places) of a root of x3−x−1=0x3−x−1=0 then 1.51.5 as the initial value, after tw

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answerhappygod
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3. The approximate value(two decimal places) of a root of x3−x−1=0x3−x−1=0 then 1.51.5 as the initial value, after tw

Post by answerhappygod »

3. The approximate value(two decimal places) of a root
of x3−x−1=0x3−x−1=0 then 1.51.5 as
the initial value, after two iterations using the Newton-Raphson
method, is
Select one: A. 1.47 B. 1.32 C.
1.91 D.1.82
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