answer:1. To rewrite the equation in standard form, we complete the square for the x-term: -3 x^2-5 x+4 y^2-7=0 -3 left(x^2+frac{5}{3} xright)+4 y^2-7=0 -3 left(x^2+2cdotfrac{5}{6} x+frac{25}{36}right)+4 y^2-7=0+frac{25}{12} -3 left(x+frac{5}{6}right)^2+4 y^2=frac{59}{12} 4 y^2-3 left(x+frac{5}{6}right)^2=frac{59}{12} 2. The equation is now in standard form for a hyperbola. 3. The center of the hyperbola is left(-frac{5}{6},0right). 4. The foci are located at left(-frac{5}{6}, pm frac{sqrt{413}}{12}right). 5. The eccentricity of the hyperbola is sqrt{frac{7}{3}}. 6. The asymptotes of the hyperbola are y=-frac{sqrt{3} x}{2}-frac{5}{4 sqrt{3}} and y=frac{sqrt{3} x}{2}+frac{5}{4 sqrt{3}}. Classification: Hyperbola Equation: 4 y^2-3 left(x+frac{5}{6}right)^2=frac{59}{12} Foci: left( begin{array}{cc} -frac{5}{6} & -frac{sqrt{413}}{12} -frac{5}{6} & frac{sqrt{413}}{12} end{array} right) Eccentricity: sqrt{frac{7}{3}} Center: left{-frac{5}{6},0right} Asymptotes: left{y=-frac{sqrt{3} x}{2}-frac{5}{4 sqrt{3}},y=frac{sqrt{3} x}{2}+frac{5}{4 sqrt{3}}right}

answer:The payback period for Project X is 2.31 years. Explanation: The payback period is calculated as follows: 1. Start with the initial outlay: -8,682. 2. Add the cash flows for each year: Year 1: +3,764 Year 2: +3,764 Year 3: +3,685 Year 4: +3,783 3. Calculate the cumulative cash flows: Year 0: -8,682 Year 1: -4,918 (8,682 - 3,764) Year 2: -1,154 (4,918 - 3,764) Year 3: 2,531 (1,154 - 3,685) 4. Since the cumulative cash flows turn positive in Year 3, the payback period is the number of years until then (2 years) plus the remaining amount to be recovered divided by the next year's cash flow: Remaining amount: 1,154 Next year's cash flow: 3,685 5. Payback period = 2 years + (1,154 / 3,685) Payback period = 2.31 years (rounded to two decimal places)

answer:Given: - Temperature of the gas, ( T = 20°C = (20 + 273) K = 293 K ) - Pressure of the gas, ( P = 1 atm ) - Volume of the gas, ( V = 6 L ) Using the ideal gas law, which is ( PV = nRT ), we can determine the number of moles ( n ) of the gas, where ( R ) is the gas constant (approximately ( 0.0821 L atm K^{-1} mol^{-1} )). [ begin{align*} PV &= nRT 1 atm times 6 L &= n times 0.0821 (L atm K^{-1} mol^{-1}) times 293 K n &= frac{1 times 6}{0.0821 times 293} n &= 0.25 mol end{align*} ] Therefore, there are 0.25 moles of the ideal gas in the given conditions.

answer:To find the equation of the plane, we first need to determine the normal vector mathbf{n} to the plane. We can do this by taking the cross product of two vectors lying in the plane, which we can obtain by subtracting the coordinates of two points. Let's choose P_1P_2 and P_1P_3 as the vectors. Vector P_1P_2 = left(-3 - left(-frac{1}{2}right), -4 - left(-frac{1}{2}right), -frac{7}{2} - frac{5}{2}right) = left(-frac{5}{2}, -frac{7}{2}, -6right) Vector P_1P_3 = left(-frac{5}{2} - left(-frac{1}{2}right), 0 - left(-frac{1}{2}right), 1 - frac{5}{2}right) = left(-2, frac{1}{2}, -frac{3}{2}right) Now, calculate the cross product: mathbf{n} = P_1P_2 times P_1P_3 = left| begin{array}{ccc} mathbf{i} & mathbf{j} & mathbf{k} -frac{5}{2} & -frac{7}{2} & -6 -2 & frac{1}{2} & -frac{3}{2} end{array} right| Calculating the determinant, we get: mathbf{n} = left(frac{7}{2} cdot left(-frac{3}{2}right) - left(-frac{7}{2}right) cdot left(-2right)right) mathbf{i} + left(-frac{5}{2} cdot left(-frac{3}{2}right) - left(-6right) cdot left(-2right)right) mathbf{j} + left(-frac{5}{2} cdot left(frac{1}{2}right) - left(-2right) cdot left(-frac{7}{2}right)right) mathbf{k} mathbf{n} = left(-frac{21}{4} + frac{14}{2}right) mathbf{i} + left(frac{15}{4} - 12right) mathbf{j} + left(-frac{5}{4} - 7right) mathbf{k} mathbf{n} = left(-frac{21}{4} + frac{28}{4}right) mathbf{i} + left(frac{15}{4} - frac{48}{4}right) mathbf{j} + left(-frac{5}{4} - frac{28}{4}right) mathbf{k} mathbf{n} = frac{7}{4} mathbf{i} - frac{33}{4} mathbf{j} - frac{33}{4} mathbf{k} mathbf{n} = left(frac{7}{4}, -frac{33}{4}, -frac{33}{4}right) Now, we can use the point-normal form to find the equation of the plane: mathbf{n} cdot (mathbf{r} - mathbf{r}_0) = 0 where mathbf{r}_0 is a point on the plane, which we can choose as P_1. So, left(frac{7}{4}, -frac{33}{4}, -frac{33}{4}right) cdot left(x - left(-frac{1}{2}right), y - left(-frac{1}{2}right), z - frac{5}{2}right) = 0 Expanding and simplifying, we get the equation of the plane: frac{7}{4}(x + frac{1}{2}) - frac{33}{4}(y + frac{1}{2}) - frac{33}{4}(z - frac{5}{2}) = 0 7x + frac{7}{2} - 33y - frac{33}{2} - 33z + frac{165}{2} = 0 14x - 66y - 66z + 151 = 0 Multiplying through by 2 to clear the fractions: 28x - 132y - 132z + 302 = 0 Dividing through by the greatest common divisor (12) to simplify further: 2x - 11y - 11z + frac{151}{6} = 0 Finally, converting to standard form: 2x - 11y - 11z = -frac{151}{6} The equation of the plane in standard form is: 12x - 66y - 66z = -151 Note: The original answer was incorrect, as the equation provided does not pass through the given points. The revised answer has been calculated to ensure accuracy.