Advanced Placement (AP), 20.01.2021 23:10 Marisolv06

Explain the process by which the Industrial Revolution spread to one country in Europe other than Great Britain.

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It should be noted that the Industrial Revolution did not spread to Britain because British entrepreneurs forbade the export of machinery and skilled workers.

The Industrial Revolution was simply the transition to new manufacturing processes in continental Europe and the United States.

The Industrial Revolution did not spread to Britain because British entrepreneurs forbade the export of machinery and skilled workers.

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Answer from: xczapien782

The industrial revolution spread to other countries in Europe, because British professionals were hired by other countries.

Explanation:

The industrial revolution was a technological breakthrough that started in England and became something very profitable, making other countries want it too. However, the English government, trying to thwart competition, prohibited industrial machines from being exported to other countries, however, other European countries began to make very generous job offers to English professionals who knew how these machines work and thus the revolution industrialization started to spread.

Answer from: Quest

Does the meme have to be funny educational or both

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.

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