1. Q1 a) Problem statement: Solve the equation $\frac{m}{2}+\frac{m}{3}+3=2+\frac{m}{6}$. 2. Q1 a) Step 1: Get a common denominator of 6 for the rational terms. 3. Q1 a) Step 2: Rewrite each term with denominator 6 and simplify. 4. Q1 a) Work: $\frac{m}{2}=\frac{3m}{6}$. $\frac{m}{3}=\frac{2m}{6}$. So the left side becomes $\frac{3m}{6}+\frac{2m}{6}+3=\frac{5m}{6}+3$. 5. Q1 a) Continue solving: $\frac{5m}{6}+3=2+\frac{m}{6}$. Subtract $\frac{m}{6}$ from both sides to isolate $m$ terms. $\frac{5m}{6}-\frac{m}{6}+3=2$. This gives $\frac{4m}{6}+3=2$. Simplify the fraction to get $\frac{2m}{3}+3=2$. Subtract 3 from both sides to obtain $\frac{2m}{3}=-1$. Multiply both sides by $\frac{3}{2}$ to get $m=-\frac{3}{2}$. 6. Q1 a) Final answer: $m=-\frac{3}{2}=-1.5$. 7. Q1 b) Problem statement: Evaluate $\sqrt{\dfrac{(15.62)^2}{29.21\times\sqrt{10.52}}}$. 8. Q1 b) Step 1: Compute the numerator $(15.62)^2$. 9. Q1 b) Work: $(15.62)^2=243.9844$. 10. Q1 b) Step 2: Compute the inner square root $\sqrt{10.52}$ approximately. 11. Q1 b) Work: $\sqrt{10.52}\approx3.2438$. 12. Q1 b) Step 3: Compute the denominator $29.21\times\sqrt{10.52}$. 13. Q1 b) Work: $29.21\times3.2438\approx94.756$. 14. Q1 b) Step 4: Form the quotient and take the outer square root. 15. Q1 b) Work: $\dfrac{243.9844}{94.756}\approx2.5740$. $\sqrt{2.5740}\approx1.604$. 16. Q1 b) Final answer: $\sqrt{\dfrac{(15.62)^2}{29.21\times\sqrt{10.52}}}\approx1.604$. 17. Q1 c) Problem statement: Solve the system $y=x+2$ and $x^{2}+y^{2}=28$. 18. Q1 c) Step 1: Substitute $y=x+2$ into the circle equation. 19. Q1 c) Work: $x^{2}+(x+2)^{2}=28$. Expand the square to get $x^{2}+x^{2}+4x+4=28$. Combine like terms to obtain $2x^{2}+4x+4=28$. Subtract 28 from both sides to get $2x^{2}+4x-24=0$. Divide every term by 2 to simplify to $x^{2}+2x-12=0$. 20. Q1 c) Step 2: Solve the quadratic using the quadratic formula. 21. Q1 c) Work: $x=\dfrac{-2\pm\sqrt{(2)^{2}-4\cdot1\cdot(-12)}}{2\cdot1}$. Compute the discriminant: $4+48=52$. So $x=\dfrac{-2\pm\sqrt{52}}{2}=\dfrac{-2\pm2\sqrt{13}}{2}=-1\pm\sqrt{13}$. 22. Q1 c) Step 3: Find corresponding $y$ values by $y=x+2$. 23. Q1 c) Work: If $x=-1+\sqrt{13}$ then $y=1+\sqrt{13}$. If $x=-1-\sqrt{13}$ then $y=1-\sqrt{13}$. 24. Q1 c) Final answers: $(x,y)=\left(-1+\sqrt{13},\;1+\sqrt{13}\right)$ and $(x,y)=\left(-1-\sqrt{13},\;1-\sqrt{13}\right)$. 25. Q2 a) Problem statement: Determine $y$ from $\sqrt{\dfrac{y+2}{3-y}}=-15+10$. 26. Q2 a) Step 1: Simplify the right side and examine the equation. 27. Q2 a) Work: The right side is $-15+10=-5$. So the equation becomes $\sqrt{\dfrac{y+2}{3-y}}=-5$. 28. Q2 a) Step 2: Reason about square root values. 29. Q2 a) Work: The principal square root symbol $\sqrt{\cdot}$ yields a nonnegative value for real inputs. Therefore it cannot equal $-5$ for any real argument. 30. Q2 a) Final answer: There is no real solution because a square root cannot be negative. 31. Q2 b) Problem statement: Compute $F=G\dfrac{m_{1}m_{2}}{d^{2}}$ with $G=6.67\times10^{-11}$, $m_{1}=7.36$, $m_{2}=15.5$, $d=22.6$ and express the result in standard form to three decimal places. 32. Q2 b) Step 1: Multiply the masses $m_{1}m_{2}$. 33. Q2 b) Work: $7.36\times15.5=114.08$. 34. Q2 b) Step 2: Multiply by $G$ to get the numerator. 35. Q2 b) Work: $6.67\times10^{-11}\times114.08=7.609136\times10^{-9}$. 36. Q2 b) Step 3: Compute $d^{2}$. 37. Q2 b) Work: $22.6^{2}=510.76$. 38. Q2 b) Step 4: Divide numerator by $d^{2}$. 39. Q2 b) Work: $F=\dfrac{7.609136\times10^{-9}}{510.76}\approx1.49004\times10^{-11}$. 40. Q2 b) Final answer (standard form to three decimal places): $F\approx1.490\times10^{-11}$. 41. Q3 a) Problem statement: Use Heron's formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ with $s=\dfrac{a+b+c}{2}$ to find the area for $a=3.60$, $b=4.00$, $c=5.20$. 42. Q3 a) Step 1: Compute the semiperimeter $s$. 43. Q3 a) Work: $s=\dfrac{3.60+4.00+5.20}{2}=\dfrac{12.80}{2}=6.40$. 44. Q3 a) Step 2: Compute the factors $s-a$, $s-b$, $s-c$. 45. Q3 a) Work: $s-a=6.40-3.60=2.80$. $s-b=6.40-4.00=2.40$. $s-c=6.40-5.20=1.20$. 46. Q3 a) Step 3: Multiply the factors and take the square root. 47. Q3 a) Work: $s(s-a)(s-b)(s-c)=6.40\times2.80\times2.40\times1.20=51.6096$. $A=\sqrt{51.6096}\approx7.184$. 48. Q3 a) Final answer: $A\approx7.184$ square centimeters. 49. Q3 b) Problem statement: Given $F=aL+b$ and the points $(L,F)=(8.0,5.6)$ and $(2.0,4.4)$, find $a$, $b$ and $F$ when $L=6.5$. 50. Q3 b) Step 1: Compute the slope $a$ from the two given points. 51. Q3 b) Work: $a=\dfrac{5.6-4.4}{8.0-2.0}=\dfrac{1.2}{6.0}=0.2$. 52. Q3 b) Step 2: Compute $b$ using one known point. 53. Q3 b) Work: Using $(2.0,4.4)$, $b=4.4-a\times2.0=4.4-0.2\times2.0=4.4-0.4=4.0$. 54. Q3 b) Step 3: Compute $F$ for $L=6.5$. 55. Q3 b) Work: $F=0.2\times6.5+4.0=1.3+4.0=5.3$. 56. Q3 b) Final answers: $a=0.2$, $b=4.0$, and $F=5.3$ when $L=6.5$. 57. Summary of final answers: Q1 a) $m=-\dfrac{3}{2}=-1.5$. Q1 b) $\approx1.604$. Q1 c) $(x,y)=\left(-1\pm\sqrt{13},\;1\pm\sqrt{13}\right)$ with matching signs. Q2 a) No real solution. Q2 b) $F\approx1.490\times10^{-11}$. Q3 a) $A\approx7.184$ cm$^{2}$. Q3 b) $a=0.2$, $b=4.0$, $F(6.5)=5.3$.