Select all the correct answers.Why is it important to study the burial rituals of hunting and gathering communities of the Paleolithic age?to understand

Select all the correct answers. Why is it important to study the burial rituals of hunting and gathering communities of the Paleolithic age?
to understand their beliefs about death
to emulate their rituals in modern society
to find out who buried the individual and why
to prove that their burial rituals were unethical
to find out which objects were considered sacred

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  1. You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule.    [tex]\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}[/tex]set the derivative to equal -1 and solve for x.      [tex]\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}[/tex]take log base 10 of both sides. also recall that  [tex]\log_c(m^n) = n \log_c m[/tex].    [tex]\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}[/tex]so the x-coord of the point of tangency is [tex]x \approx 0.2356[/tex].the y-coordinate of this point of tangency can be calculated using the original function:     [tex]\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}[/tex]using point-slope form, the equation of the tangent line is      [tex]y - 0.7213 = -1(x-0.2356)[/tex]the x-intercept of this tangent line is when the graph is at y=0.    [tex]\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}[/tex]this rounds off to  [tex]x & = 0.957[/tex].choice c.

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