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.

T is in the middle as S is to the left and U is to the right.

S is the sister of U as they share the same parents, T.

T is the son of V, making V the grandfather of S.

S is the sister of T, and U is the brother of T, so S is also the sister of U.

Since today is Tuesday, her next meeting will be the following Tuesday.

Robert is Sarah's grandfather.

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.

The difference A - B includes elements in A that are not in B, which are {1, 2}.

The intersection of sets A and B includes the elements that are common to both sets. Therefore, the intersection is {3, 4}.

A - B represents the elements in A that are not in B. Thus, A - B = {1}.

The intersection of sets A and B includes only the elements that are present in both sets, which are {b, c}.

The intersection A ∩ B includes elements that are both even and prime, which is {2}.

The number of elements in the union of sets A and B is given by |A ∪ B| = |A| + |B| - |A ∩ B| = 10 + 15 - 5 = 20.

The number of elements in the union of sets A and B is given by |A ∪ B| = |A| + |B| - |A ∩ B| = 10 + 15 - 5 = 20.

The number of elements in the union of sets A and B is given by |A ∪ B| = |A| + |B| - |A ∩ B| = 15 + 10 - 5 = 20.

The number of elements in the union of sets A and B is given by |A ∪ B| = |A| + |B| - |A ∩ B| = 15 + 10 - 5 = 20.

The intersection of sets A and B is the set of elements that are common to both sets. Therefore, the intersection is {3, 4}.

The intersection of sets A and B is the set of elements that are common to both sets, which is {4, 5}.

The multiples of 3 less than 30 are {3, 6, 9, 12, 15, 18, 21, 24, 27}, totaling 9 elements.

Hybrid sets can include elements that are exclusive to either of the original sets, making them versatile in composition.

By definition, a hybrid set formed from set A and set B must include all elements from set A.

Both 12 and 9 are multiples of 3 and less than 15, but 12 is the first option listed.

The integer 10 is not divisible by 3, hence it is not a member of set C.