answer:The divergence of the vector field vec{F}(x, y, z) = cosh(x)uvec{i} + e^yuvec{j} + z^5uvec{k} is found using the divergence theorem, which for a vector field vec{F} = f(x, y, z)uvec{i} + g(x, y, z)uvec{j} + h(x, y, z)uvec{k} is given by nabla cdot vec{F} = frac{partial f}{partial x} + frac{partial g}{partial y} + frac{partial h}{partial z}. Applying this to our vector field, we get: [ nabla cdot vec{F} = frac{partial}{partial x}(cosh(x)) + frac{partial}{partial y}(e^y) + frac{partial}{partial z}(z^5) ] [ = sinh(x) + e^y + 5z^4 ] Thus, the divergence of the vector field is sinh(x) + e^y + 5z^4.

answer:To solve the equation, we can first multiply both sides by the common denominator, which is (11 sqrt{3} x^2+4 sqrt{3} x+frac{10}{sqrt{3}}): 9 sqrt{3} x^2+8 sqrt{3} x-frac{38}{sqrt{3}} = 0 (9 sqrt{3} x^2+8 sqrt{3} x-frac{38}{sqrt{3}})(11 sqrt{3} x^2+4 sqrt{3} x+frac{10}{sqrt{3}}) = 0 This gives us a quadratic equation in terms of x: 99 sqrt{3} x^4+44 sqrt{3} x^3+frac{10}{sqrt{3}} x^2+36 sqrt{3} x^2+16 sqrt{3} x+frac{40}{sqrt{3}} = 0 99 sqrt{3} x^4+80 sqrt{3} x^3+frac{46}{sqrt{3}} x^2+16 sqrt{3} x+frac{40}{sqrt{3}} = 0 We can now solve this quadratic equation using the quadratic formula: x = frac{-b pm sqrt{b^2-4ac}}{2a} where a, b, and c are the coefficients of the quadratic equation. Plugging in the values, we get: x = frac{-80 sqrt{3} pm sqrt{(80 sqrt{3})^2-4(99 sqrt{3})(46 sqrt{3})}}{2(99 sqrt{3})} x = frac{-80 sqrt{3} pm sqrt{6400 cdot 3-18144 cdot 3}}{2(99 sqrt{3})} x = frac{-80 sqrt{3} pm sqrt{19200 cdot 3-18144 cdot 3}}{2(99 sqrt{3})} x = frac{-80 sqrt{3} pm sqrt{1056 cdot 3}}{2(99 sqrt{3})} x = frac{-80 sqrt{3} pm 32 sqrt{3}}{2(99 sqrt{3})} x = frac{-80 sqrt{3} pm 32 sqrt{3}}{198 sqrt{3}} x = frac{-40 sqrt{3} pm 16 sqrt{3}}{99 sqrt{3}} x = frac{-40 pm 16}{99} x = frac{-24}{99} quad text{or} quad x = frac{40}{99} x = -frac{8}{33} quad text{or} quad x = frac{40}{99} x = -frac{8}{33} quad text{or} quad x = frac{40}{33} x = -frac{8}{33} quad text{or} quad x = frac{4}{3} Therefore, the real solutions to the equation are x = -frac{8}{33} and x = frac{4}{3}. The answer is left{xto frac{1}{9} left(-4-sqrt{130}right), xto frac{1}{9} left(-4+sqrt{130}right)right}.

answer:To determine Suggett Corporation's free cash flow, we subtract its capital expenditures and cash dividends from the net cash provided by operating activities: Free cash flow = Net cash provided by operating activities - Capital expenditures - Cash dividends Free cash flow = 34 million - 24 million - 7 million After performing the calculation, we find that the free cash flow is: Free cash flow = 3 million Thus, the correct answer is: d. 3 million

answer:To find the limit of (1 + h)^{frac{1}{h}} as h approaches 0, we can use the following steps: 1. Rewrite the expression using logarithms: (1 + h)^{frac{1}{h}} = e^{frac{1}{h} ln(1 + h)} 2. Evaluate the limit of the exponent: lim_{h to 0} frac{1}{h} ln(1 + h) 3. Apply L'Hôpital's rule: Since the limit of the numerator and denominator is both 0, we can apply L'Hôpital's rule. Taking the derivative of the numerator and denominator, we get: lim_{h to 0} frac{frac{d}{dh} [ln(1 + h)]}{frac{d}{dh} [h]} = lim_{h to 0} frac{frac{1}{1 + h}}{1} = 1 4. Substitute the limit back into the original expression: lim_{h to 0} (1 + h)^{frac{1}{h}} = e^{lim_{h to 0} frac{1}{h} ln(1 + h)} = e^1 = e Therefore, the limit of (1 + h)^{frac{1}{h}} as h approaches 0 is e.