answer:The triangle's vertices are: 1. Vertex A: (0, 0) 2. Vertex B: (12, 0) 3. Vertex C: left(frac{21}{2}, frac{3 sqrt{15}}{2}right) The measures of the interior angles, labeled as angle A, angle B, and angle C (with AC = BC being the equal sides), are: 1. angle A = cos ^{-1}left(frac{7}{8}right) 2. angle B = angle C = cos ^{-1}left(frac{1}{4}right) Note that angle B and angle C are equal due to the isosceles property of the triangle.

answer:To solve the equation left| -23 x^2-4 xright| =23, we first need to isolate the absolute value expression. left| -23 x^2-4 xright| =23 -23 x^2-4 x = 23 quad text{or} quad -23 x^2-4 x = -23 Now, we can solve each of these equations separately. -23 x^2-4 x = 23 -23 x^2-4 x - 23 = 0 (23 x + 1)(x - 1) = 0 x = -frac{1}{23} quad text{or} quad x = 1 -23 x^2-4 x = -23 -23 x^2-4 x + 23 = 0 (23 x - 1)(x + 1) = 0 x = frac{1}{23} quad text{or} quad x = -1 Therefore, the solutions to the equation left| -23 x^2-4 xright| =23 are x = -frac{2}{23} pm frac{sqrt{533}}{23}. The answer is left{xto -frac{2}{23} pm frac{sqrt{533}}{23}right}

answer:The curl of the vector field vec{F}(x, y, z) is given by the following components: nabla times vec{F} = left(frac{partial h}{partial y} - frac{partial g}{partial z}right)uvec{i} - left(frac{partial h}{partial x} - frac{partial f}{partial z}right)uvec{j} + left(frac{partial g}{partial x} - frac{partial f}{partial y}right)uvec{k} Applying the derivatives: 1. frac{partial h}{partial y} = frac{partial}{partial y}e^{frac{x}{z}} = 0 2. frac{partial g}{partial z} = frac{partial}{partial z}tan(y) = 0 3. frac{partial h}{partial x} = frac{partial}{partial x}e^{frac{x}{z}} = frac{e^{frac{x}{z}}}{z} 4. frac{partial f}{partial z} = frac{partial}{partial z}sqrt{x} = 0 5. frac{partial g}{partial x} = frac{partial}{partial x}tan(y) = 0 6. frac{partial f}{partial y} = frac{partial}{partial y}sqrt{x} = 0 Substituting these derivatives into the curl formula: nabla times vec{F} = left(0 - 0right)uvec{i} - left(frac{e^{frac{x}{z}}}{z} - 0right)uvec{j} + left(0 - 0right)uvec{k} = left{0, -frac{e^{frac{x}{z}}}{z}, 0right} Therefore, the curl of the given vector field is left{0, -frac{e^{frac{x}{z}}}{z}, 0right}.

answer:The real solutions to the equation are given by: x = frac{1}{22} left(255 - sqrt{91909}right) and x = frac{1}{22} left(255 + sqrt{91909}right)