log_4(64) = log_4(4^3) = 3.

log_5(125) = log_5(5^3) = 3.

log_5(25) = 2 and log_5(5) = 1. Therefore, 2 - 1 = 1.

In the equation x^2 = 4py, we have 4p = 16, thus p = 4.

In the equation x^2 = 4py, we have 4p = 20, thus p = 20/4 = 5.

Setting the two pieces equal at x = 2: 3(2) + p = 2^2 - 4. Solving gives p = -2.

Planck's constant is approximately 6.626 x 10^-34 Joule seconds.

Setting the two pieces equal at x = 1: 5 - q = 3(1) + 2. Solving gives q = 0.

The ideal gas constant R is 0.0821 L·atm/(K·mol).

The value of R in L·atm/(K·mol) is 0.0821.

R can have different values depending on the units used, including 0.0821 L·atm/(K·mol), 8.314 J/(K·mol), and 62.36 L·mmHg/(K·mol).

The value of the ideal gas constant R in L·atm/(K·mol) is 0.0821.

The value of R in L·kPa/(K·mol) is 8.314.

sec(60°) = 1/cos(60°) = 1/(1/2) = 2.

sec(sin^(-1)(1/2)) = 1/cos(π/6) = 2.

sec(sin^(-1)(3/5)) = √(34)/3

Using the triangle with opposite = 1 and adjacent = √3, hypotenuse = 2. Thus, sec(tan^(-1)(1/√3)) = 2.

sec(θ) = 1/cos(θ) = 1/(1/3) = 3.

sec(θ) = 1/cos(θ) = 1/(3/5) = 5/3.

sec^(-1)(2) = π/3, since sec(π/3) = 2.

The double angle formula states that sin 2θ = 2sin θ cos θ.

Using the double angle formula sin(2x) = 2sin x cos x. Since sin x = 1/2, cos x = √(1 - (1/2)^2) = √(3/4) = √3/2. Thus, sin(2x) = 2 * (1/2) * (√3/2) = √3/2.

Using the double angle formula sin(2θ) = 2sin θ cos θ. First, find cos θ using sin^2 θ + cos^2 θ = 1. cos θ = sqrt(1 - (1/3)^2) = sqrt(8/9) = 2sqrt(2)/3. Thus, sin(2θ) = 2 * (1/3) * (2sqrt(2)/3) = 4sqrt(2)/9.

Using the double angle formula, sin(2θ) = 2sin(θ)cos(θ) = 2(1/√2)(1/√2) = 1.

Using the double angle formula, sin(2θ) = 2sin θcos θ.

Using the double angle formula, sin(2θ) = 2sin(θ)cos(θ).

sin(30°) = 1/2 and cos(60°) = 1/2. Therefore, sin(30°) + cos(60°) = 1/2 + 1/2 = 1.

sin(30°) = 1/2.

sin(45°) = cos(45°) = √2/2. Therefore, sin(45°) + cos(45°) = √2/2 + √2/2 = √2.

Using the co-function identity, sin(90° - A) = cos A.