answer:Given {eq}displaystyle f(x) = x^{-2} {/eq} Differentiating with respect to {eq}displaystyle x {/eq} we get {eq}displaystyle frac{d}{dx}f(x) = frac{d}{dx}x^{-2} f'(x) = (-2)x^{-3} {/eq} Again differentiating with respect to {eq}displaystyle x {/eq} we get {eq}displaystyle frac{d}{dx}f'(x) = frac{d}{dx}(-2)x^{-3} f''(x) = (-2)(-3)x^{-4} {/eq} Again differentiating with respect to {eq}displaystyle x {/eq} we get {eq}displaystyle frac{d}{dx}f''(x) = frac{d}{dx}(-2)(-3)x^{-4} f'''(x) = (-2)(-3)(-4)x^{-5} {/eq} By observation we get {eq}displaystyle f^n(x) = frac{(-2)(-3)(-4)..........(-(n+1))}{x^{n+2}} {/eq} That gives {eq}displaystyle f^n(x) = frac{(-1)^n (n+1)!}{x^{n+2}} {/eq} So {eq}displaystyle f^n(x) {/eq} is given by {eq}displaystyle f^n(x) = frac{(-1)^n (n+1)!}{x^{n+2}} {/eq}

answer:The magnitude (norm) of a complex number a + bi is given by sqrt{a^2 + b^2}. For the given complex number frac{54}{7} left(cos left(frac{19 pi }{90}right)-i sin left(frac{19 pi }{90}right)right), where a = frac{54}{7} cos left(frac{19 pi }{90}right) and b = -frac{54}{7} sin left(frac{19 pi }{90}right), the norm is: text{Norm} = sqrt{left(frac{54}{7} cos left(frac{19 pi }{90}right)right)^2 + left(-frac{54}{7} sin left(frac{19 pi }{90}right)right)^2} Simplifying, we get: text{Norm} = frac{54}{7} sqrt{cos^2 left(frac{19 pi }{90}right) + sin^2 left(frac{19 pi }{90}right)} Since cos^2 theta + sin^2 theta = 1 for any angle theta, the norm is: text{Norm} = frac{54}{7} The argument (phase angle) of the complex number is the angle formed by the vector representing the complex number with the positive real axis. The argument is given by the angle frac{19 pi }{90}, but since the imaginary part is negative, we need to subtract this angle from 2pi to get the angle in the standard position: text{Argument} = 2pi - frac{19 pi }{90} = frac{180pi}{90} - frac{19pi}{90} = frac{161pi}{90} However, the angle frac{19pi}{90} is already in the fourth quadrant where cosine is positive and sine is negative, so no adjustment is needed: text{Argument} = frac{19 pi }{90} Therefore, the norm is frac{54}{7} and the argument is frac{19 pi }{90}.

answer:Given data: Mass of the Ferrari: m1 = 1,207 kg Upward acceleration of the cars: a = 1.25 m/s^2 Mass of the BMW car: m2 = 1,461 kg Let T be the tension in the cable above the Ferrari. Applying Newton's second law to the Ferrari, we have: T - m1g = m1a Substituting the given values, we get: T - (1,207 kg)(9.81 m/s^2) = (1,207 kg)(1.25 m/s^2) Simplifying the equation, we obtain: T = 1,207 kg(9.81 m/s^2 + 1.25 m/s^2) T = 1,207 kg(11.06 m/s^2) T = 13,350 N Therefore, the tension in the cable above the Ferrari is 13,350 N.

answer:The given series is an infinite geometric series with the first term a = -frac{47}{76} and the common ratio r = frac{1}{5}. To find the sum, we use the formula for the sum of an infinite geometric series: S = frac{a}{1 - r} Plugging in the values: S = frac{-frac{47}{76}}{1 - frac{1}{5}} S = frac{-frac{47}{76}}{frac{4}{5}} S = -frac{47}{76} cdot frac{5}{4} S = -frac{235}{304} Therefore, the sum of the infinite geometric series is -frac{235}{304}.