answer:The initial assumption that the friction force f between the blocks is equal to 10N is incorrect. To resolve the paradox, we must determine the actual friction force that ensures both blocks have the same acceleration. Let the common acceleration of both blocks be a, and the friction force acting on m_1 by m_2 be f. We can set up two equations based on Newton's second law: For block m_2 (the bottom block): [ F - f = m_2 cdot a ] For block m_1 (the top block): [ f = m_1 cdot a ] Solving these equations simultaneously, we find: [ f = frac{50}{8}N ] Now, using this friction force, the acceleration a can be calculated for both blocks: [ a = frac{F}{m_1 + m_2} = frac{10}{5 + 3} = frac{10}{8} m/s^2 ] This acceleration is consistent with the original calculation and eliminates the paradox. The difference in the initial approach was due to the incorrect assumption of the friction force between the blocks.

answer:To evaluate the function ( f(x) ) at ( x = -7 ), we will substitute ( x ) with ( -7 ) in the expression: [ f(-7) = -tan(6) tanleft(1 - 2(-7)^4right) ] First, calculate the term inside the tangent: [ 1 - 2(-7)^4 = 1 - 2(2401) = 1 - 4802 = -4801 ] Now, evaluate the function: [ f(-7) = -tan(6) tan(-4801) ] Since (tan(-theta) = -tan(theta)), the tangent of ( -4801 ) is the negative of the tangent of ( 4801 ): [ f(-7) = -tan(6) (-tan(4801)) ] [ f(-7) = tan(6) tan(4801) ] The calculated value is: [ tan(6) tan(4801) approx 0.22 ] So, the revised answer is: [ f(-7) approx 0.22 ]

answer:The coordinates of the points defining the lines are as follows: Cevian: The point where the cevian intersects the side opposite the 9.1 unit side is left( begin{array}{cc} 10.25 & 0. 0.78 & 4.48 end{array} right). Symmedian: The point where the symmedian intersects the side opposite the 12.48 unit side is left( begin{array}{cc} 10.25 & 0. 0.63 & 3.61 end{array} right). Altitude: The point where the altitude from the vertex opposite the 12.48 unit side intersects the side opposite the 9.1 unit side is left( begin{array}{cc} 10.25 & 0. 0.3 & 1.74 end{array} right). Median: The point where the median intersects the side opposite the 9.1 unit side is the same as the cevian, which is left( begin{array}{cc} 10.25 & 0. 0.78 & 4.48 end{array} right). Note: These coordinates are based on the assumption that the triangle's vertices are labeled such that the 12.48 unit side is adjacent to the angle of 54 {}^{circ}.

answer:To compute the limit, we can use polar coordinates. Let x=rcostheta and y=rsintheta. Then, the region R becomes R={(r,theta): 0leq rleq 1, 0leq thetaleq frac{pi}{4}}cup{(r,theta): 0leq rleq 1, frac{3pi}{4}leq thetaleq pi} Therefore, the integral becomes begin{split} lim_{c rightarrow 0} iint_{R} frac{1}{(x^2+y^2)^{3/4}},dA &= lim_{c rightarrow 0} int_0^{frac{pi}{4}} int_0^1 frac{1}{r^{3/2}},r,dr,dtheta + lim_{c rightarrow 0} int_{frac{3pi}{4}}^{pi} int_0^1 frac{1}{r^{3/2}},r,dr,dtheta &= lim_{c rightarrow 0} left[ int_0^{frac{pi}{4}} left(-frac{2}{r^{1/2}}right)bigg|_{r=0}^{r=1} dtheta + int_{frac{3pi}{4}}^{pi} left(-frac{2}{r^{1/2}}right)bigg|_{r=0}^{r=1} dtheta right] &= lim_{c rightarrow 0} left[ int_0^{frac{pi}{4}} (-2) dtheta + int_{frac{3pi}{4}}^{pi} (-2) dtheta right] &= lim_{c rightarrow 0} left[ -2left(frac{pi}{4}right) -2left(frac{pi}{4}right) right] &= -2pi end{split}