Advanced Placement (AP), 15.12.2020 03:10 sanji8442

Find the missing length of the tepee if the height is 12 in, and the radius is 9 in.

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Answer from: Quest

If you are talking about typical sand you would find at the beach or in a river bed here on earth, there are some differences. i would say the first biggest difference is in composition and mineral make up. although the moon is largely made of the same materials as earth’s mantle and surface, the moon experiences very little weathering (no liquid water and virtually no atmosphere). “earth” sand is typically composed mostly of quartz grains. this is because quartz is highly resistant to both physical and chemical weathering. most of the other minerals originally with the quartz are broken down into fine clay and carried away, chemically eroded/ dissolved, or some combination of the two. this usually leaves only the quartz behind, and so most of the time when you see/think of sand, you are talking about quartz. without these weathering mechanisms on the moon, the “sand” (which is likely more silt or clay sized grains anyways) would consist of mostly other minerals like anorthosite, olivine, and proxene.

: )

Answer from: Quest

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. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

Answer from: Quest

Ok i’m not good at this but imma try : )

Answer from: Quest

answer:

my friends lol btw if thts you in your profile pic i have to ur kinda cute

explanation: idk

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