answer:(I) Since the polar coordinate equation of curve C_{1} is rho =2, the rectangular coordinate equation of curve C_{1} is {{x}^{2}}+{{y}^{2}}=4. Since the vertices of the positive triangle ABC are all on C_{1}, and the coordinates of points A, B, and C are arranged in counterclockwise order, with point A having coordinates (2,0), the coordinates of point B are (2cos 120{}^circ ,2sin 120{}^circ ), i.e., B(-1,sqrt{3}), and the coordinates of point C are (2cos 240{}^circ ,2sin 240{}^circ ), i.e., C(-1,-sqrt{3}). (II) Since the circle C_{2}: {{x}^{2}}+{{(y+{{sqrt{3}}})}^{2}}=1, the parametric equation of circle C_{2} is begin{cases} x=cos alpha y=-{{sqrt{3}}}+sin alpha end{cases},0le alpha <2pi , Let point P(cos alpha ,-{{sqrt{3}}}+sin alpha ), 0le alpha <2pi , |PB{{|}^{2}}+|PC{{|}^{2}}={{(x+1)}^{2}}+{{(y-2{{sqrt{3}}})}^{2}}+{{(x+1)}^{2}}+{{y}^{2}} =16+4cos alpha -4{{sqrt{3}}}sin alpha =16+8cos (alpha +frac{pi }{3}), Thus, the range of |PB{{|}^{2}}+|PC{{|}^{2}} is boxed{[8,24]}.

answer:To solve the given problem, we will break it down into detailed steps: 1. Recognize the difference of squares and other properties: [ begin{align*} left(lg 5right)^2 - left(lg 2right)^2 + 8^{frac{2}{3}} times lg sqrt{2} - 0.6^0 + 0.2^{-1} &= left(lg 5 + lg 2right)left(lg 5 - lg 2right) + left[2^3right]^{frac{2}{3}} times frac{1}{2} lg 2 - 1 + left(frac{1}{5}right)^{-1} &= lg 10 cdot left(lg 5 - lg 2right) + 2 times frac{1}{2} lg 2 - 1 + 5 &= lg 10 cdot left(lg frac{5}{2}right) + lg 2 - 1 + 5 end{align*} ] 2. Simplify using logarithm properties and arithmetic operations: [ begin{align*} &= lg 10 + lg frac{5}{2} + lg 2 - 1 + 5 &= lg 10 + lg 5 - 1 + 5 &= 1 + 4 &= 5 end{align*} ] Therefore, the final answer is boxed{5}.

answer:1. **Identify and Analyze Given Information**: - XYZ is a triangle with XY = 10 inches. - PQR is parallel to XY, where P is on XZ, Q is on YZ, and PQ= 7 inches. - XQ extended bisects angle PRY. 2. **Using Properties of Parallel Lines**: - The parallelism of PQR and XY implies triangle XYZ sim triangle PZR (AA similarity: corresponding angles are equal). 3. **Applying the Angle Bisector Theorem**: - The angle bisector XQ extended makes angle PRX = angle XRY. - Let angle PRY = 2alpha. Thus, angle PRX = alpha and angle XRY = alpha. 4. **Setting Up the Proportion from Similar Triangles**: - From triangle XYZ sim triangle PZR, we have: [ frac{XY}{PQ} = frac{YZ}{RY} = frac{XZ}{PZ} ] - Let RY = y. Given XY = 10 inches and PQ = 7 inches, the ratio becomes: [ frac{10}{7} = frac{10 + y}{y} ] 5. **Solving for y**: - Cross multiplying gives: [ 10y = 7(10 + y) Rightarrow 10y = 70 + 7y Rightarrow 3y = 70 Rightarrow y = frac{70}{3} ] 6. **Conclusion**: - The length of RY is frac{70{3}} inches. The final answer is boxed{frac{70}{3}}.

answer:Given that m lt 2sqrt{3} lt m+1, and m is an integer, we need to find the value of m. First, let's consider the inequality in terms of perfect squares to find a range that fits 2sqrt{3}: 9 lt 12 lt 16 This translates to: 3^2 lt 2^2 cdot 3 lt 4^2 Taking the square root of all parts of the inequality gives: 3 lt 2sqrt{3} lt 4 This means 2sqrt{3} is greater than 3 but less than 4. Given that m is an integer and fits within the inequality m lt 2sqrt{3} lt m+1, the only integer value m can take that satisfies this condition is 3. Therefore, the value of m is boxed{3}.