Solve the inequality. Graph the solution set. \[ |x| \leq 14 \]

If a solution containing 53.75 g of mercury(II) nitrate is allowed to react completely with a solution containing 14.334 g of sodium sulfate according to the equation below. \[ \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{HgSO}_{4}(\mathrm{~s}) \] How many grams of solid precipitate will be formed? mass: 30.00 How many grams of the reactant in excess will remain after the reaction? mass:

4) Charlie och Kim löser ekvationen \( \frac{16-x^{2}}{4-x}=0 \). Charlie får lösningen \( x=-4 \), Kim säger att även \( x=4 \) är en lösning. Vem har rätt? Visa hur ekvationen ska lösas och förklara vem av dem som har rätt.

Simplify: \(5 - (3 - 1)\)

\( \left. \begin{array} { | l | l | l | l } { f ^ { \prime } ( 3 ) } & { } & { f ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } } \\ \hline & { } & { } & { } \\ \hline & { } & { } & { } \\ \hline \end{array} \right. \)

Un cycliste quitte une ville \( A \) à 4 h 55 mn vers une ville B. Un deuxième cycliste quitte ville \( B \) vers la ville \( A \) à 6 h 10 mn . La somme de leurs vitesses est \( 50 \mathrm{~km} / \mathrm{h} \), la distance A et \( B \) est 125 km et ils se rencontrent a 7 h 52 mn . Calcule la vitesse de chacun.