answer:1. From e=frac{c}{a}=frac{sqrt{3}}{2}, we get frac{c^{2}}{a^{2}}=frac{3}{4}. Since c^{2}=a^{2}-b^{2}, we have a^{2}=4b^{2}. The line passing through F_{1} and perpendicular to the x-axis intersects the ellipse (C) with a segment length of 1. Substituting x=-c,y=frac{1}{2} into the ellipse equation frac{x^{2}}{a^{2}}+ frac{y^{2}}{b^{2}}=1(a > b > 0), we get frac{c^{2}}{4b^{2}}+ frac{(frac{1}{2})^{2}}{b^{2}}=1. This simplifies to frac{3b^{2}}{4b^{2}}+ frac{(frac{1}{2})^{2}}{b^{2}}=1, and solving for b^{2} gives b^{2}=1. Thus, a^{2}=4. Therefore, the equation of the ellipse (C) is boxed{frac{x^{2}}{4}+y^{2}=1}. 2. Let A(x_{1},y_{1}) and B(x_{2},y_{2}) be the points on the ellipse, and the equation of line AB is y=kx+b. Eliminating y from the system of equations begin{cases} frac{x^{2}}{4}+y^{2}=1 y=kx+b end{cases}, we get (1+4k^{2})x^{2}+8kbx+4b^{2}-4=0. Then, x_{1}+x_{2}=-frac{8kb}{4k^{2}+1},x_{1}x_{2}=frac{4b^{2}-4}{4k^{2}+1}. From k_{1}+k_{2}=k_{1}k_{2}-1, we have frac{y_{1}}{x_{1}}+ frac{y_{2}}{x_{2}}=frac{y_{1}y_{2}}{x_{1}x_{2}}-1, which simplifies to x_{2}y_{1}+x_{1}y_{2}=y_{1}y_{2}-x_{1}x_{2}. Substituting y_{1}=kx_{1}+b and y_{2}=kx_{2}+b, we get (k^{2}-2k-1)x_{1}x_{2}+b(k-1)(x_{1}+x_{2})+b^{2}=0. Thus, b^{2}=-frac{4}{3}k^{2}+frac{8}{3}k+frac{4}{3}. Solving the system of inequalities begin{cases} 16k^{2}-8k-1 > 0 -k^{2}+2k+1geqslant 0 end{cases}, we get 1-sqrt{2}leqslant k < frac{1-sqrt{2}}{4} or frac{1+sqrt{2}}{4} < kleqslant 1+sqrt{2}. Therefore, the range of values for k is boxed{[1-sqrt{2},frac{1-sqrt{2}}{4})cup(frac{1+sqrt{2}}{4},1+sqrt{2}]}.

answer:Let's denote the distance to the school as (d) miles. When Johnny jogs to school, he travels at a speed of 5 miles per hour. The time it takes him to jog to school is therefore (d / 5) hours. On the return trip, he travels by bus at a speed of 21 miles per hour. The time it takes for the bus to bring him back home is (d / 21) hours. The total time for both trips is 1 hour, so we can write the equation: [ frac{d}{5} + frac{d}{21} = 1 ] To solve for (d), we need to find a common denominator for the fractions, which is 105. We can rewrite the equation as: [ frac{21d}{105} + frac{5d}{105} = 1 ] Combining the fractions, we get: [ frac{26d}{105} = 1 ] Now, we solve for (d): [ 26d = 105 ] [ d = frac{105}{26} ] [ d = 4.03846153846 ] Since we're looking for a practical answer, we can round this to a reasonable number of decimal places. So, the distance to the school is approximately boxed{4.04} miles.

answer:Since y=f(x) is an odd function, we have f(-2)-f(-3)=f(3)-f(2)=1. Therefore, the answer is boxed{1}.

answer:1. **Initial Setup and Assumptions:** - Let ( triangle ABC ) be a triangle with ( angle A = 60^circ ). - Points ( E ) and ( F ) are the feet of the angle bisectors of vertices ( B ) and ( C ) respectively. - Points ( P ) and ( Q ) are such that quadrilaterals ( BFPE ) and ( CEQF ) are parallelograms. - We need to prove that ( angle PAQ > 150^circ ). 2. **Reflecting the Problem:** - Assume without loss of generality that ( AB < AC ). - Let ( I ) be the incenter of ( triangle ABC ). - Let ( T ) be the point such that ( AETF ) is a parallelogram. - Reflect the problem over the midpoint of ( EF ). We need to show that ( angle BTC > 150^circ ), where ( angle BTC ) is the angle containing only a finite area of points on the same side of ( BC ) as ( A ). 3. **Reflection and Angle Chasing:** - Let ( D ) be the reflection of ( F ) over ( BI ) (also the reflection of ( E ) over ( CI )). - Let ( D' ) be the point on ( BC ) such that ( ID = ID' ) and ( D neq D' ). - By angle chasing, ( D' ) is in the interior of segment ( BC ). - We claim that ( T ) is the reflection of ( D' ) over ( ID ). 4. **Observations and Cyclic Quadrilaterals:** - By angle chasing, ( AEIF ) is cyclic, ( triangle DEF ) is equilateral, and ( I ) is the center of ( triangle DEF ). - Since ( angle ETF = angle EAF = 60^circ = angle EDF ), ( ETDF ) is cyclic. - This implies ( IT = ID = ID' ). - Furthermore, ( angle DIT = 2 angle DFT ). 5. **Parallel Lines and Perpendicular Bisectors:** - Since ( AE parallel FT ) and ( BI perp DF ), we have: [ angle DIT = 2(90^circ - angle BEA) = 2left(frac{angle B}{2} - 30^circright) = frac{angle B - angle C}{2} ] 6. **Angle Calculation:** - To prove ( angle DID' = angle DIT ), we calculate: [ angle DID' = 180^circ - 2 angle IDD' = 2(90^circ - angle BDI) = 2(angle DIB + angle IBD - 90^circ) = 2left(frac{angle B}{2} - 30^circright) = frac{angle B - angle C}{2} ] - This confirms our claim. 7. **Equilateral Triangle and Cyclic Quadrilateral:** - Let ( X ) be the point such that ( triangle BCX ) is an equilateral triangle, and ( A ) and ( X ) are on opposite sides of ( BC ). - Observe that ( BICX ) is cyclic, and since ( BX = XC ), ( D in IX ). 8. **Circle and Interior Points:** - Consider the circle ( omega ) centered at ( X ) with radius ( XB ). - Since segment ( BC ) is entirely within ( omega ), ( D' ) is in the interior of ( omega ). - Since ( D' ) and ( T ) are reflections over ( IX ), ( XT = XD' < XB ). - Thus, ( T ) is also in the interior of ( omega ). 9. **Conclusion:** - This implies that ( angle BTC > 150^circ ), and hence ( angle PAQ > 150^circ ). (blacksquare)