answer:1. To find the second derivative, we first need to find the first derivative of the function: f'(x) = frac{7}{2sqrt{7x-9}} - 5e^{-5x-4} 2. Then, we can find the second derivative by differentiating the first derivative: f''(x) = -frac{49}{4(7x-9)^{3/2}} + 25e^{-5x-4} Therefore, the second derivative of the given function is -frac{49}{4(7x-9)^{3/2}} + 25e^{-5x-4}. The answer is f''(x) = -frac{49}{4(7x-9)^{3/2}} + 25e^{-5x-4}

answer:1. Square both sides of the equation: left(sqrt{frac{45}{7}-frac{90 x}{7}}+sqrt{frac{39}{7}-frac{65 x}{7}}right)^2=left(frac{50}{7}right)^2 frac{45}{7}-frac{90 x}{7}+frac{39}{7}-frac{65 x}{7}+2sqrt{left(frac{45}{7}-frac{90 x}{7}right)left(frac{39}{7}-frac{65 x}{7}right)}=frac{2500}{49} frac{84}{7}-frac{155 x}{7}+2sqrt{left(frac{45}{7}-frac{90 x}{7}right)left(frac{39}{7}-frac{65 x}{7}right)}=frac{2500}{49} 2. Simplify the equation: frac{155 x}{7}-2sqrt{left(frac{45}{7}-frac{90 x}{7}right)left(frac{39}{7}-frac{65 x}{7}right)}=frac{2416}{49} frac{155 x}{7}-2sqrt{frac{1755}{49}-frac{3255 x}{49}+frac{3675 x^2}{49}}=frac{2416}{49} frac{155 x}{7}-2sqrt{frac{1755-3255 x+3675 x^2}{49}}=frac{2416}{49} frac{155 x}{7}-2sqrt{frac{(65-15 x)^2}{49}}=frac{2416}{49} frac{155 x}{7}-frac{2}{7}(65-15 x)=frac{2416}{49} frac{155 x}{7}-frac{130}{7}+frac{30 x}{7}=frac{2416}{49} frac{185 x}{7}-frac{130}{7}=frac{2416}{49} frac{185 x-130}{7}=frac{2416}{49} 3. Multiply both sides by 7: 185 x-130=2416 185 x=2546 x=frac{2546}{185} x=frac{1}{175} left(-15458+60 sqrt{65455}right) Therefore, the only real solution to the equation is x=frac{1}{175} left(-15458+60 sqrt{65455}right). The answer is left{xto frac{1}{175} left(-15458+60 sqrt{65455}right)right}

answer:The curvature {eq}kappa{/eq} is given by: {eq}displaystyle kappa=frac{left | T'(t)right |}{left | r'(t)right |} {/eq} From the original answer, we have: {eq}displaystyle left | T'(t)right |=sqrt{cos^2 2t+cos^2 t} {/eq} {eq}displaystyle left | r'(t)right |=sec t {/eq} At the point (1, 0, 0), which corresponds to {eq}t=0{/eq}, we have: {eq}displaystyle left | T'(0)right |=sqrt{2} {/eq} {eq}displaystyle left | r'(0)right |=1 {/eq} Therefore, the curvature at the point (1, 0, 0) is: {eq}displaystyle kappa=frac{sqrt{2}}{1}=boxed{sqrt{2}} {/eq}

answer:The result of multiplying the scalar frac{1}{2} with the vector is: [ left( begin{array}{c} -frac{7}{2} -1 1 frac{9}{2} end{array} right) ]