Advanced Placement (AP), 22.06.2019 21:00 niah2699

This was a series of articles written to persuade new york to ratify the constitution.

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Answer from: Fintastichumanoid

The Federalist Papers was a series of Articles written to persuade New York to ratify the Constitution.

Explanation:

The Federalist Papers was a series of 85 essays aimed at persuade New York to ratify the Constitution of the United States, published in the journals The Independent Journal, The New York Packet and the Daily Advertiser between October 1787 and August 1788.

The Federalist Papers still remains a primary source for the interpretation of the United States Constitution, since the essays that make it up offer a clear and convincing vision of the new system of government that it had the objective of establishing. The authors of the Federalist Papers wanted both to support the vote in favor of ratifying the Constitution and to influence its future interpretations.

At the time of its publication, the identity of the authors of the articles was kept secret, but some of the more acute readers speculated that they had been written by Alexander Hamilton, James Madison and John Jay. This hypothesis was confirmed after Hamilton's death in 1804, when a list was made public that revealed the authorship of the individual essays.

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

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