Area increases by a factor of (3r)²/r² = 9.

The focal length (f) of a mirror is given by f = R/2. For a convex mirror, R is positive, so f = 20 cm / 2 = 10 cm.

Using the Arrhenius equation, if the rate doubles for a 10°C increase, Ea is approximately 40 kJ/mol.

The order of the reaction is third because the rate is proportional to the concentration raised to the power of 3.

The reaction is second-order because the rate depends on the square of the concentration of A.

According to Ohm's Law, if the resistance (R) is doubled and the voltage (V) remains constant, the current (I) will be halved (I = V/R).

According to Ohm's law, I = V/R. If R is doubled, I will be halved.

Max revenue occurs at x = -b/(2a) = 100/(2*2) = 25.

Using the sum and product of roots: 1 + (-3) = -2 and 1 * (-3) = -3, thus k = 3.

The discriminant must be greater than zero: 2² - 4*1*k > 0, which simplifies to k < 1.

Using the product of roots: k = (-2)(-3) = 6.

Using the sum and product of roots: k = 1*4 = 4, and sum = 1 + 4 = 5, thus k = 7.

Using the sum of roots (-3 + -4 = -7) and product of roots (-3*-4 = 12), we find p = 12.

Using the sum of the roots: p = -(3 + 4) = -7.

The speed of light in a medium is given by v = c/n. Therefore, v = (3 x 10^8 m/s) / 1.5 = 2 x 10^8 m/s.

Impedance (Z) = Voltage (V) / Current (I) = 120V / 5A = 24Ω.

Using Raoult's Law: P_solution = X_solvent * P_solvent. X_solvent = P_solution / P_solvent = 80/100 = 0.8, thus X_solute = 1 - X_solvent = 0.2.

Using Ohm's law, I = V/Xc = 100V/50Ω = 2A.

Using Ohm's law, I = V/Xc = 100V/50Ω = 2A.

Current (I) = Voltage (V) / Capacitive Reactance (XC) = 100V / 50Ω = 2A.

I = V/Xc = 100/50 = 2 A.

I = V / R = 240V / 60Ω = 4A.

According to Ohm's law, if voltage is halved and resistance remains constant, the current also halves.

According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. Therefore, if the volume is doubled, the pressure is halved.

According to Boyle's Law, if the volume is halved, the pressure doubles.

The position of the first minimum in single-slit diffraction is given by a sin(θ) = λ. Increasing λ will cause θ to increase, moving the minimum away from the center.

At the midpoint, the electric fields due to +Q and -Q cancel each other out, resulting in a net electric field of 0.

Possible combinations for sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 combinations. Total outcomes = 36. Probability = 6/36 = 1/6.

C_eq = 1 / (1/C1 + 1/C2) = 1 / (1/4 + 1/4) = 2 µF.

Using the formula for parallel resistors, 1/R_eq = 1/R1 + 1/R2 = 1/4 + 1/6 = 5/12, R_eq = 12/5 = 2.4Ω.