answer:Dian Fossey and Jane Goodall did have a professional relationship. They first met in 1966 when Fossey visited Goodall's research site in Gombe, Tanzania. They shared a common interest in studying primates and exchanged ideas and research findings. Although they worked in different locations and studied different species, they maintained a cordial and supportive relationship throughout their careers.

answer:Given: Initial velocity of the projectile: {eq}u = 50 ms^{-1} {/eq} Angle at which the projectile is launched: {eq}theta = 30^circ {/eq} with respect to horizontal Horizontal distance traveled by the projectile: {eq}d = 100 m {/eq} Initial velocity along horizontal direction is: {eq}u_x = u times cos theta = 50 times cos 30^circ = 43.301 ms^{-1} {/eq} As horizontal component of velocity remains constant, total time taken to travel the horizontal distance is: {eq}t = dfrac{d}{u_x} = dfrac{100}{43.301} = 2.309 s {/eq} Time taken for the projectile to reach maximum height is: {eq}t_{max} = dfrac{u_y}{g} = dfrac{50 times sin 30^circ}{9.8} = 2.551 s {/eq} As {eq}t_{max} < t {/eq}, the maximum height reached by the projectile is: {eq}h_{max} = u_y times t_{max} - dfrac{1}{2} times g times t_{max}^2 {/eq} Substituting the values, {eq}h_{max} = 50 times sin 30^circ times 2.551 - 0.5 times 9.8 times 2.551^2 implies h_{max} = 31.888 m {/eq} Therefore, the maximum height reached by the projectile is 31.888 m.

answer:We can evaluate the triple integral as follows: {eq}begin{align*} iiint_E y dV &= int_0^3 int_0^x int_{x-y}^{x+y} y dz dy dx &= int_0^3 int_0^x y(x+y - (x-y)) dy dx &= int_0^3 int_0^x 2y^2 dy dx &= int_0^3 left [ frac23 y^3 right ]_0^x dx &= int_0^3 frac23 x^3 dx &= left [ frac16 x^4 right ]_0^3 &= frac16 (3)^4 &= frac{27}{2} &= 13.5 end{align*} {/eq} Therefore, the value of the triple integral is 13.5.

answer:We can use the following trigonometric substitution: x = sqrt{10} tan theta dx = sqrt{10} sec^2 theta dtheta Substituting into the integral, we get: int frac {dx}{sqrt {x^2+10}} = int frac {sqrt{10} sec^2 theta dtheta}{sqrt{(sqrt{10} tan theta)^2+10}} = int frac {sqrt{10} sec^2 theta dtheta}{sqrt{10 tan^2 theta+10}} = int frac {sqrt{10} sec^2 theta dtheta}{sqrt{10 (tan^2 theta+1)}} = int frac {sqrt{10} sec^2 theta dtheta}{sqrt{10 sec^2 theta}} = int frac {sqrt{10} sec^2 theta dtheta}{sqrt{10} sec theta} = int sec theta dtheta = ln |sec theta + tan theta| + C Substituting back for (theta), we get: int frac {dx}{sqrt {x^2+10}} = ln left|sec left(tan^{-1} frac{x}{sqrt{10}}right) + tan left(tan^{-1} frac{x}{sqrt{10}}right)right| + C = ln left|frac{sqrt{x^2+10}}{x} + frac{x}{sqrt{x^2+10}}right| + C = ln left|frac{x^2+10+x^2}{xsqrt{x^2+10}}right| + C = ln left|frac{2x^2+10}{xsqrt{x^2+10}}right| + C = ln |x+sqrt{x^2+10}| + C The answer is (int frac {dx}{sqrt {x^2+10}} = ln |x+sqrt{x^2+10}| + C)