answer:First, we can simplify the equation by multiplying both sides by the least common multiple of the denominators, which is 5: 5left(frac{12x^2-frac{44x}{5}-frac{22}{5}}{-frac{59x}{5}-frac{52}{5}}right)=5(0) 60x^2-44x-22=-59x-52 60x^2+5x-30=0 Now, we can use the quadratic formula to find the solutions: x=frac{-bpmsqrt{b^2-4ac}}{2a} x=frac{-5pmsqrt{5^2-4(60)(-30)}}{2(60)} x=frac{-5pmsqrt{25+7200}}{120} x=frac{-5pmsqrt{7225}}{120} x=frac{-5pm85}{120} x=frac{1}{30} left(11pmsqrt{451}right) Therefore, the solutions are x=frac{1}{30} left(11-sqrt{451}right) and x=frac{1}{30} left(11+sqrt{451}right). The answer is left{xto frac{1}{30} left(11-sqrt{451}right), xto frac{1}{30} left(11+sqrt{451}right)right}.

answer:A positive coefficient of the leading term in a cubic function indicates that the function tends towards positive infinity as x approaches positive infinity, and towards negative infinity as x approaches negative infinity. Conversely, a negative coefficient indicates that the function tends towards negative infinity as x approaches positive infinity, and towards positive infinity as x approaches negative infinity.

answer:To find the equation of the plane, we can use the following steps: 1. Find two vectors that lie in the plane. We can do this by subtracting the coordinates of two of the points: overrightarrow{v_1} = langle -3 - 4, 1 - 1, -3 - 2 rangle = langle -7, 0, -5 rangle overrightarrow{v_2} = langle -2 - 4, 5 - 1, -3 - 2 rangle = langle -6, 4, -5 rangle 2. Find the cross product of the two vectors: overrightarrow{v_1} times overrightarrow{v_2} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} -7 & 0 & -5 -6 & 4 & -5 end{vmatrix} = (0 + 20) mathbf{i} - (-35 - 30) mathbf{j} + (28 - 0) mathbf{k} = 20 mathbf{i} + 65 mathbf{j} + 28 mathbf{k} 3. The cross product is a vector that is perpendicular to both overrightarrow{v_1} and overrightarrow{v_2}, and therefore perpendicular to the plane. The equation of the plane can be written in the form a x + b y + c z + d = 0 where a, b, and c are the components of the cross product vector, and d is a constant. 4. To find the value of d, we can substitute the coordinates of one of the points into the equation: 20(4) - 5(1) - 28(2) + d = 0 80 - 5 - 56 + d = 0 19 + d = 0 d = -19 5. Therefore, the equation of the plane is 20x - 5y - 28z - 19 = 0 The equation of the plane is 20x - 5y - 28z - 19 = 0.

answer:The curl of the given vector field is calculated as follows: nabla times left(fuvec{i} + guvec{j} + huvec{k}right) = 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} Substituting the given functions, we get: nabla times left(fuvec{i} + guvec{j} + huvec{k}right) = left(-3y^2zsinleft(zleft(x^2 + y^3right)right)right)uvec{i} - left(2xzsinleft(zleft(x^2 + y^3right)right)right)uvec{j} + left(-frac{4x}{left(x^2 + y^3right)^3}right)uvec{k} Thus, the curl of the vector field is: left{-3y^2zsinleft(zleft(x^2 + y^3right)right), -2xzsinleft(zleft(x^2 + y^3right)right), -frac{4x}{left(x^2 + y^3right)^3}right}