answer:Dan originally had 62.5 seashells. After giving some to Jessica, he has 30.75 seashells left. To find out how many seashells he gave to Jessica, we subtract the number of seashells he has left from the original number he had: 62.5 (original number of seashells) - 30.75 (seashells left) = 31.75 (seashells given to Jessica). So, Dan gave boxed{31.75} seashells to Jessica.

answer:Solve (9-k)x = 17. When k=9, there is no solution; When k > 9, x = frac{17}{9-k} < 0, which does not meet the requirement; When k < 9, x = frac{17}{9-k} > 0, Therefore, when k=8 or k=-8, x=17 or x=1. Thus, the final answer is boxed{k=8 text{ or } k=-8}.

answer:The reaction between sodium hydride (NaH) and water (H2O) to form sodium hydroxide (NaOH) and hydrogen gas (H2) can be represented by the following balanced chemical equation: NaH + H2O → NaOH + H2 From the balanced equation, we can see that 1 mole of NaH reacts with 1 mole of H2O to produce 1 mole of NaOH. Therefore, the reaction is in a 1:1:1 molar ratio. If we have 3 moles of NaH and 3 moles of H2O, the reaction will proceed with both reactants in stoichiometric amounts, meaning they will react completely without any excess of either reactant. Thus, 3 moles of NaH will react with 3 moles of H2O to produce 3 moles of NaOH (since the ratio is 1:1:1). Therefore, the number of moles of NaOH formed is boxed{3} moles.

answer:1. **Construct the quadrilateral and draw diagonals**: (triangle EGH) is a right triangle because (angle H = 90^circ) and (HE = GH = 20). (triangle EFG) is isosceles with (EF = FG = 15). 2. **Compute diagonals**: (triangle EGH) is right-angled, thus (EG = sqrt{HE^2 + HG^2} = sqrt{20^2 + 20^2} = 20sqrt{2}). 3. **Find (MN) using midpoint properties**: - (M) and (N) are midpoints, so they split (FG) and (HE) into equal parts (FG/2 = 15/2 = 7.5) and (HE/2 = 20/2 = 10). - Both triangles (FGH) and (HEG) are isosceles, thus the perpendiculars from (H) to (EF) (let this be point (P)) and from (E) to (GH) (let this be point (Q)) split the bases in half, giving (EP = PF = 7.5) and (HQ = QG = 10). Since (EFGH) is a rectilinear shape consiting of an isosceles triangle and a right-angle triangle, (overline{MN}) intersects (overline{PQ}) at right angles as it also does in a kite (though (EFGH) is not a kite). 4. **Apply the Pythagorean theorem**: (overline{MN}) is parallel and congruent to (overline{PQ}) due to the properties of midpoints, and each sections off the respective diagonal halves: [ MN^2 = MP^2 + PN^2 = left(frac{20sqrt{2}}{2}right)^2 = 200. ] Hence, (MN^2 = boxed{200}).