Simplification and Approximation

The square root of 64 is 8 and the square root of 81 is 9. Since 74 is between 64 and 81, the root is between 8 and 9. 8.6^2 = 73.96, which is very close to 74.

sqrt(0.00021025) = sqrt(21025 / 100000000) = 145 / 10000 = 0.0145.

The square root of 0.0009 is 0.03, and the square root of 0.01 is 0.1. Dividing 0.03 by 0.1 gives 0.3, which is the correct value.

2^3 = 8. Since 3.02 is very close to 3, 2^3.02 will be slightly larger than 8. Among the choices, 8 is the closest approximate value.

(8.8)^(1/3) = (8 + 0.8)^(1/3) = 2 * (1 + 0.1)^(1/3). Using binomial expansion (1+x)^n = 1 + nx + n(n-1)x^2/2: 2 * (1 + 1/3 * 0.1 + (1/3 * -2/3 * 0.01) / 2) = 2 * (1 + 0.0333 - 0.0011) = 2 + 0.0666 - 0.0022 = 2 + 0.0644. 58/900 is approx 0.0644.

Using the binomial approximation (1 + x)^n approx 1 + nx for small x, (1 + 0.01)^(-7) approx 1 + (-7)(0.01) = 1 - 0.07 = 0.93.

1 / sqrt(0.8) = 1 / sqrt(4/5) = sqrt(5) / 2. Since sqrt(5) is approximately 2.236, 2.236 / 2 = 1.118, which is approximately 1.1.

The square root of 0.9 is approximately 0.94868. Among the choices, 0.94 is the closest approximation.

0.090909 is 1/11. 0.454545 is 5 times 0.090909, so it is 5 * 1/11 = 5/11.

Using the Taylor expansion for tan^-1(1 + x) near x=0, tan^-1(1 + x) is approximately tan^-1(1) + (1 / (1 + 1^2)) * x = pi/4 + x/2. With x = 0.001, this is 0.785398 + 0.0005 = 0.785898, which is closest to 0.7855.

Number of digits = floor(log10(5^17)) + 1. log10(5^17) = 17 * log10(5) = 17 * log10(10/2) = 17 * (1 - log10(2)) = 17 * (1 - 0.3010) = 17 * 0.699 = 11.883. floor(11.883) + 1 = 11 + 1 = 12.

33^(1/5) is slightly more than 32^(1/5) = 2. Using linear approximation: f(x+dx) = f(x) + f'(x)dx. f(32) = 2, f'(x) = (1/5)x^(-4/5), f'(32) = (1/5)(1/16) = 1/80 = 0.0125. So 33^(1/5) approx 2 + 0.0125 = 2.0125.

Substitute x = 1.1 into the expression: (1.1)^3 + 5(1.1)^2 - 7(1.1) + 10. This equals 1.331 + 5(1.21) - 7.7 + 10 = 1.331 + 6.05 - 7.7 + 10 = 9.681. The closest provided option is 9.69.