Live practice

For a 3+ ion, e=Z−3. Use N=1.304e and A=N+Z.

Let \(x\in C-B\). Then \(x\in C\) and \(x\notin B\). Since \(A\subset B\), membership of \(x\) in \(A\) would force \(x\in B\), a contradiction. Hence \(x\notin A\). Thus \(x\in C-A\), proving \(C-B\subset C-A\).

Name the inner functions, differentiate each factor, and then simplify.

Substitution into the original expression verifies the stated domain and final value.

Translate every condition into an equation and retain exact values throughout the ellipse calculation.

The last line satisfies the original conditions and supplies every requested part.

Translate every condition into an equation and retain exact values throughout the straight-line calculation.

The last line satisfies the original conditions and supplies every requested part.

Use the sequence definition or geometric-progression formula, keeping the first-term offset explicit.

The displayed substitutions cover every requested part and verify the result against the original data.

Proceed from the given expression and keep every transformation explicit.

The last line satisfies the required domain and establishes the result.

Treat the first dissociation of H₂S as a weak-acid equilibrium. Added HCl supplies a common H⁺ ion. The second dissociation then gives [S²⁻]=Ka₂[HS⁻]/[H⁺].

The electromagnetic spectrum increases in frequency from radio through microwave and visible light to X-rays and the very high-frequency cosmic radiation.

Every element of \(A\) either lies in \(B\), and hence in \(A\cap B\), or does not lie in \(B\), and hence lies in \(A-B\). The two cases exhaust \(A\), proving the first equality.

For the second, an element of \(A\cup B\) either belongs to \(A\), or belongs to \(B\) but not \(A\); the latter elements form \(B-A\). Hence both sides contain exactly the same elements.