For l=2, the orbital type is d.

The variable resistor allows for adjustment to achieve balance in the bridge.

Substituting x = 2 into the equation gives 2^2 - 3*2 + p = 0 => 4 - 6 + p = 0 => p = 2.

Using the root, we substitute: 2^2 - 6*2 + k = 0 => 4 - 12 + k = 0 => k = 8.

Using Vieta's formulas, if one root is 3, the other root is 7 - 3 = 4. Thus, k = 3 * 4 = 12.

A relation is a function if every element in the domain (set A) maps to exactly one element in the codomain. Here, 1 maps to 2, and 2 maps to 3, but 3 has no mapping, so R is not a function.

A relation R is reflexive if every element in set A is related to itself. Since R contains (1, 1), (2, 2), and (3, 3), R is reflexive.

A relation is symmetric if for every (a, b) in R, (b, a) is also in R. Since R only contains pairs of the form (a, a), it is symmetric.

A relation R is transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also be in R. Here, (1, 2) and (2, 3) are in R, but (1, 3) is not, so R is not transitive.

R is not reflexive as (1,1), (2,2), (3,3) are not in R. It is symmetric as (2,3) implies (3,2) is not in R. It is transitive as (1,2) and (2,3) implies (1,3) is not in R. Thus, R is not all of the above.

R is reflexive because it contains all pairs (a, a) and symmetric because (1,2) implies (2,1).

R is not a partial order because it is not transitive; (1,2) and (2,2) do not imply (1,2).

R is reflexive and symmetric, but not transitive. Thus, both 1 and 2 are true.

R is neither reflexive, symmetric, nor transitive as it does not satisfy any of the properties.

R is not symmetric as (b,c) does not imply (c,b) is in R. It is reflexive and transitive.

R is neither reflexive, symmetric, nor transitive.

Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (15 * 5) / 10 = 7.5Ω.

For balance, R1/R2 = R3/R4, thus R4 = (R2 * R3) / R1 = (20 * 30) / 10 = 60Ω.

The bridge is not balanced because 10/20 ≠ 30/60.

Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.

Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.

Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (10 * 15) / 5 = 30Ω.

For balance, R4 must be 30Ω since 5/15 = 10/30.

Since sec A = 1/cos A, we have cos A = 1/2.

Since sec θ = 1/cos θ, if sec θ = 2, then cos θ = 1/2.

sec(x) = 1/cos(x), so cos(x) = 1/2.

Using the identity sin^2 A + cos^2 A = 1, we have cos A = sqrt(1 - (0.6)^2) = sqrt(1 - 0.36) = sqrt(0.64) = 0.8.

Using the identity sin^2 A + cos^2 A = 1, we have cos A = sqrt(1 - (0.6)^2) = sqrt(1 - 0.36) = sqrt(0.64) = 0.8.

Using the identity tan A = sin A / cos A, we find cos A = sqrt(1 - (0.6)^2) = 0.8, thus tan A = 0.6 / 0.8 = 0.75.

sin A = 1/2 at A = 30° and A = 150°.