4^2 * 4^3 = 4^(2+3) = 4^5 = 1024

Using the power of a power property, (x^2)^3 = x^(2*3) = x^6.

1 astronomical unit (AU) is approximately equal to 1.496 × 10^11 meters, which is the average distance from the Earth to the Sun.

1 Joule is defined as 1 kg·m²/s², which is the product of mass, distance squared, and time squared.

1 light year is the distance light travels in one year, which is approximately 9.46 × 10^15 meters.

By the double angle identity, 2sin(θ)cos(θ) = sin(2θ).

Using the property of exponents, 2^(3x) * 2^(2x) = 2^(3x + 2x) = 2^(5x).

Using the property of exponents, a^m × a^n = a^(m+n), we get 2^(5+3) = 2^8.

2^5 = 32

3^(2*1) = 3^2 = 9.

Any non-zero number raised to the power of 0 is 1.

4^3 = (2^2)^3 = 2^6 = 64.

Any non-zero number raised to the power of 0 is 1.

5C2 = 5! / (2!(5-2)!) = 10.

Using the property of exponents, 5^2 * 5^3 = 5^(2+3) = 5^5.

Any non-zero number raised to the power of 0 is 1.

9^(1/2) = √9 = 3.

The standard value of acceleration due to gravity at the surface of the Earth is approximately 9.8 m/s².

cos(0°) = 1.

cos(180°) = -1.

Using the double angle formula cos(2x) = 1 - 2sin^2 x. Since sin x = 1/2, we have cos(2x) = 1 - 2*(1/2)^2 = 1 - 2*(1/4) = 1 - 1/2 = 1/2.

Using the double angle formula cos(2θ) = 1 - 2sin^2 θ. Here, sin θ = 1/2, so cos(2θ) = 1 - 2(1/2)^2 = 1 - 2(1/4) = 1 - 1/2 = 1/2.

cos(60°) = 1/2.

cos(90°) = 0.

cos(π/3) = 1/2.

cos^(-1)(-1) = π, since cos(π) = -1.

cos^(-1)(0) corresponds to the angle where the cosine value is 0, which is π.

cot(cos^(-1)(1/2)) = √3

f(2) = 2^2 - 2(2) + 1 = 4 - 4 + 1 = 1.

The standard value of gravitational acceleration at the Earth's surface is approximately 9.81 m/s².