answer:To find the period: {eq}T=2*Pi*sqrt{frac{l}{g}} T=2*Pi*sqrt{frac{3.550 m}{9.795 frac{m}{s^2}}} T=3.772 s {/eq} To find the frequency: {eq}f=frac{1}{T} f=frac{1}{3.772 s} f=0.265 Hz {/eq}

answer:begin{array}{l} begin{array}{l} text{Simplify the following}: -frac{2}{3}left( begin{array}{ccc} 6 & 0 & -3 -9 & 12 & 0 end{array} right) end{array} hline begin{array}{l} -frac{2}{3}left( begin{array}{ccc} 6 & 0 & -3 -9 & 12 & 0 end{array} right)=frac{-2left( begin{array}{ccc} 6 & 0 & -3 -9 & 12 & 0 end{array} right)}{3}: frac{-2left( begin{array}{ccc} 6 & 0 & -3 -9 & 12 & 0 end{array} right)}{3} end{array} begin{array}{l} -2left( begin{array}{ccc} 6 & 0 & -3 -9 & 12 & 0 end{array} right)=left( begin{array}{ccc} -2times 6 & -2times 0 & -2 (-3) -2 (-9) & -2times 12 & -2times 0 end{array} right): frac{fbox{left( begin{array}{ccc} -2times 6 & -2times 0 & -2 (-3) -2 (-9) & -2times 12 & -2times 0 end{array} right)}}{3} end{array} begin{array}{l} 6 (-2) text{= }-12: frac{1}{3}left( begin{array}{ccc} fbox{-12} & -2times 0 & -2 (-3) -2 (-9) & -2times 12 & -2times 0 end{array} right) end{array} begin{array}{l} -2times 0 text{= }0: frac{1}{3}left( begin{array}{ccc} -12 & fbox{0} & -2 (-3) -2 (-9) & -2times 12 & -2times 0 end{array} right) end{array} begin{array}{l} -2 (-3) text{= }6: frac{1}{3}left( begin{array}{ccc} -12 & 0 & fbox{6} -2 (-9) & -2times 12 & -2times 0 end{array} right) end{array} begin{array}{l} -2 (-9) text{= }18: frac{1}{3}left( begin{array}{ccc} -12 & 0 & 6 fbox{18} & -2times 12 & -2times 0 end{array} right) end{array} begin{array}{l} -2times 12 text{= }-24: frac{1}{3}left( begin{array}{ccc} -12 & 0 & 6 18 & fbox{-24} & -2times 0 end{array} right) end{array} begin{array}{l} -2times 0 text{= }0: frac{1}{3}left( begin{array}{ccc

answer:The derivative of the function ( f(x) ) with respect to ( x ) is: [ f'(x) = frac{d}{dx}left(sqrt{frac{15x^4}{2} - frac{7}{2}}right) + frac{d}{dx}(log(-6x)) ] Differentiate each term: [ f'(x) = frac{15x^3}{2sqrt{frac{15x^4}{2} - frac{7}{2}}} + frac{1}{-6x} cdot (-6) ] Simplify: [ f'(x) = frac{15x^3}{sqrt{15x^4 - 7}} + frac{1}{x} ] This derivative is correct and represents the rate of change of the original function with respect to ( x ).

answer:Trade secret. A trade secret is a confidential formula, process, or other business information that provides a company with a competitive advantage. It is legally protected, allowing businesses to keep their unique methods of production or innovation secret, thereby preventing competitors from replicating these techniques.