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Consider the following expression:8.30 x 10^-5 = x(0.100 + x)^2We can solve for x using a technique called successive approximations.8.30 x 10^-5 =x1 (0.100)^2Step 1: If we assume that x is very small compared to 0.100 (such that 0.100 x ? 0.100) then our first approximation of x (let\'s call it x1) can be calculated asx1 = ?Step 2: Now, take your first approximation of x and plug it into the full equation.8.30 x 10^-5 =x2 (0.100 + x1)^2x2 = ?Step 3: Each successive approximation uses the value from the previous approximation.8.30 x 10^-5 =x3 (0.100 +x2)^2x3 = ?Step 4, etc.: Continue this process until two x values agree within the desired level of precision.x4 = ?x5 = ?Which values are the first to agree to two significant figures?a. x1 and x2b. x2 and x3c. x3 and x4d. x4 and x5Which values are the first to agree to three significant figures?a. x1 and x2b. x2 and x3c. x3 and x4d. x4 and x5

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Answer:

a. x3 and x4 are the first to agree to two significant figures

b. x4 and x5 are the first to agree to three significant figures

Step-by-step explanation:

8.30 x 10⁻⁵ = x1(0.100)²

8.30 x 10⁻⁵ = 0.01x1

x1 = 8.30 x 10⁻⁵/0.01 = 0.0083

8.30 x 10⁻⁵ = x2(0.100 + 0.0083)²

8.30 x 10⁻⁵ = (0.01172889)x2

x2 = 8.30 x 10⁻⁵/0.01172889

x2 = 0.007076543475128

8.30 x 10⁻⁵= x3 (0.100 + 0.007076543475128)²

8.30 x 10⁻⁵ = 0.011465386162581)x3

x3 = 8.30 x 10⁻⁵/0.011465386162581

x3 = 0.007239180505832

8.30 x 10⁻⁵ = x4 (0.100 + 0.007239180505832)²

8.30 x 10⁻⁵ = (0.011500241835562) x4

x4 = 8.30 x 10⁻⁵/0.011500241835562

x4 = 0.007217239531723

8.30 x 10⁻⁵ = x5 (0.1 + 0.007217239531723)²

8.30 x 10⁻⁵ = (0.011495536452803) x5

x5 = 8.30 x 10⁻⁵/0.011495536452803

x5 = 0.007220193710904

From the above results;

a. x3 and x4 are the first to agree to two significant figures

b. x4 and x5 are the first to agree to three significant figures

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