answer:In languages like propositional logic and first-order number theory, there are specific rules (syntax) for constructing well-formed formulas (wffs). These rules do not include provisions for concatenating two wffs without an intervening connective or operator. This is because the resulting expression would not adhere to the established syntax and would therefore not be considered a valid wff. For instance, in propositional logic, wffs are constructed using sentential variables and connectives like "and" and "or," but there is no rule allowing for the juxtaposition of two wffs without a connective.

answer:Given, {eq}bullet Electric field strength, E=2.4 times 10^5 N/C bullet Diameter of the disk, d = 20 cm = 0.2 m {/eq} Radius of the disk, {eq}r = frac{d}{2} = 0.1 m{/eq} Surface area of the disk, {eq}A = pi r^2 = pi (0.1)^2 = 0.0314 m^2{/eq} Electric field in a parallel plate capacitor, {eq}E = frac{sigma}{epsilon_0}{/eq} Surface charge density, {eq}sigma = E epsilon_0 = (2.4 times 10^5) (8.85 times 10^{-12}) = 2.124 times 10^{-6} C/m^2{/eq} Total charge on either disk, {eq}Q = sigma A = (2.124 times 10^{-6}) (0.0314) = 6.67 times 10^{-8} C{/eq} Number of electrons transferred, {eq}n = frac{Q}{|e|} = frac{6.67 times 10^{-8}}{1.6 times 10^{-19}} = 4.17 times 10^{11}{/eq} electrons

answer:To find the patch's area in square kilometers, we'll convert the area from square centimeters to square kilometers. Given: 1 cm = 10^(-5) km Squaring both sides: 1 cm² = (10^(-5))² km² = 10^(-10) km² Now, the patch area in square kilometers is: 19.0 cm² = 19.0 × 10^(-10) km² For part (b), we need to convert the area from square centimeters to square inches and then calculate the cost. Conversion factor: 1 cm = 0.39 in 1 cm² = (0.39 in)^2 = 0.1521 in² So, the patch area in square inches is: 19.0 cm² × 0.1521 in²/cm² = 2.89 in² The cost of the patch per square inch is 2.47. Thus, the total cost is: 2.89 in² × 2.47/in² = 7.14 Therefore, the patch area in square kilometers is approximately 19.0 × 10^(-10) km², and the cost of the patch is 7.14.

answer:1. To rewrite the equation in standard form, we complete the square for both x and y terms: -8 x^2-2 x+9 y^2+6 y+3=0 Rightarrow -8 left(x^2+frac{1}{4}xright)+9 left(y^2+frac{2}{3}yright)=-3 Rightarrow -8 left(x^2+2cdotfrac{1}{8}x+left(frac{1}{8}right)^2right)+9 left(y^2+2cdotfrac{1}{6}y+left(frac{1}{6}right)^2right)=-3+frac{1}{64}+frac{1}{4} Rightarrow -8 left(x+frac{1}{8}right)^2+9 left(y+frac{1}{3}right)^2=-frac{17}{8} 2. The equation is now in standard form, and we can identify it as a hyperbola because the coefficients of x^2 and y^2 have opposite signs. 3. The center of the hyperbola is left(-frac{1}{8},-frac{1}{3}right). 4. The foci are located at left(-frac{5}{6},-frac{1}{3}right) and left(frac{7}{12},-frac{1}{3}right). 5. The eccentricity of the hyperbola is frac{sqrt{17}}{3}. 6. The asymptotes of the hyperbola are y=frac{2 sqrt{2} x}{3}+frac{1}{12} left(sqrt{2}-4right) and y=frac{1}{12} left(-4-sqrt{2}right)-frac{2 sqrt{2} x}{3}. Classification: Hyperbola Equation: 9 left(y+frac{1}{3}right)^2-8 left(x+frac{1}{8}right)^2=-frac{17}{8} Foci: left( begin{array}{cc} -frac{5}{6} & -frac{1}{3} frac{7}{12} & -frac{1}{3} end{array} right) Eccentricity: frac{sqrt{17}}{3} Center: left{-frac{1}{8},-frac{1}{3}right} Asymptotes: left{y=frac{2 sqrt{2} x}{3}+frac{1}{12} left(sqrt{2}-4right),y=frac{1}{12} left(-4-sqrt{2}right)-frac{2 sqrt{2} x}{3}right}