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question:Simplify the mathematical expression 6 + (3 times 8).

answer:To simplify the expression, first perform the multiplication inside the parentheses: 3 times 8 = 24 Now, add this result to 6: 6 + 24 = 30 Therefore, the expression simplifies to 30.

question:Determine the equations of the lines containing the altitude, cevian, median, and symmedian of the angle-side-angle triangle described in the original question.

answer:* Altitude: y = -frac{9}{4} cos left(frac{2 pi }{9}right) * Cevian: y = frac{sin left(frac{pi }{36}right) sin left(frac{5 pi }{36}right)}{cos left(frac{5 pi }{36}right)} left(x-frac{9}{2}right) * Median: y = frac{sin left(frac{pi }{36}right) sin left(frac{5 pi }{36}right)}{cos left(frac{5 pi }{36}right)} left(x-frac{9}{2}right) * Symmedian: y = -frac{sin left(frac{pi }{36}right) sin left(frac{pi }{9}right) sin left(frac{5 pi }{36}right)}{left(-2+sin left(frac{2 pi }{9}right)+cos left(frac{2 pi }{9}right)right) cos left(frac{5 pi }{36}right)} left(x-frac{9}{2}right)

question:Your friend has two children. You know that at least one of them is a girl who is not named Florida. What is the probability that the other child is also a girl?

answer:{eq}begin{align*} P(text{Both are girls}|text{At least one is girl and not Florida}) & = dfrac{text{Both are girls}}{P(text{At least one is girl and not Florida})} [2ex] & = dfrac{P(G_1 cap G_2)}{P(G_1 cup G_2) - P(text{Florida})} [2ex] & = dfrac{P(G_1 cap G_2)}{P(G_1) + P(G_2) - P(G_1 cap G_2) - P(text{Florida})} [2ex] & = dfrac{0.25}{0.5 + 0.5 - 0.25 - 0.5} [2ex] & = dfrac{1}{2} end{align*} {/eq}

question:A uniformly charged sphere has a surface potential of 410 V. At a radial distance of 0.2 m from the surface, the potential drops to 150 V. Given the Coulomb constant, {eq}k = 8.99 times 10^9 text{N} cdot text{m}^2/text{C}^2 {/eq}, what is the radius of the sphere in meters?

answer:Using the given data: Surface potential, {eq}V_1 = 410 V {/eq} Potential at 0.2 m away, {eq}V_2 = 150 V {/eq} The potential at any point outside a uniformly charged sphere is given by the formula: {eq}V = frac{kq}{r} {/eq} where: - {eq}V {/eq} is the potential - {eq}k {/eq} is the Coulomb constant - {eq}q {/eq} is the charge of the sphere - {eq}r {/eq} is the distance from the center of the sphere For the surface, {eq}r = R {/eq}, the radius of the sphere, and at 0.2 m away, {eq}r = R + 0.2 {/eq}. So, we have two equations: 1. {eq}V_1 = frac{kq}{R} {/eq} 2. {eq}V_2 = frac{kq}{R + 0.2} {/eq} Dividing these two equations, we get: {eq}frac{V_1}{V_2} = frac{R + 0.2}{R} {/eq} Substituting the values: {eq}frac{410}{150} = frac{R + 0.2}{R} {/eq} Solving for {eq}R {/eq}: {eq}410R = 150R + 30 {/eq} {eq}260R = 30 {/eq} {eq}R = frac{30}{260} {/eq} {eq}R = 0.115 m {/eq} Therefore, the radius of the sphere is 0.115 meters.