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adelina 88 [10]
3 years ago
9

No ruler took more liberties with his religion than Akbar, the greatest of the Mughals, the Muslim dynasty that dominated India

between the early 16th and 18th centuries. Like Ashoka and Gandhi, Akbar constructed a religious ideology that served to hold together a diffuse polity as it fed his own soul. It began with pragmatic policies of tolerance. Akbar had inherited the throne, at the age of 13, in 1556. In 1579 he abolished the jiziya, a tax imposed on all but the poorest non-Muslims. This was the most notable in a series of measures to recruit the Hindu majority and others to the cause of unifying and expanding his empire. He could be ruthless: his troops massacred 20,000–25,000 non-combatants after a four-month siege of Chitor, a nearly impregnable Hindu fortress in Rajasthan. But he preferred incentives to coercion. He defeated the war-like Rajputs, but gave them rank and married their princesses, who were permitted to conduct Hindu rites in the harem. The Mughal-Rajput alliance was a bulwark of his empire. "Multicultural Akbar,” The Economist, 1999.
Explain ONE specific political development that resulted from the conditions created by the religious policies described in the passage?
Advanced Placement (AP)
1 answer:
babymother [125]3 years ago
3 0

Answer:

Political Jealousy

Explanation:

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7 0
2 years ago
Use midpoints to approxiamte the area under the curve y=f(x) 5 sin(xx)!+ 2.5 cos(4xx) on the interval [0,1] using 10 equal subdi
alisha [4.7K]

Subdividing [0, 1] into 10 equally spaced intervals of length \Delta x=\frac{1-0}{10}=\frac1{10} gives the partition

[0,1] = \left[0,\dfrac1{10}\right] \cup \left[\dfrac1{10},\dfrac2{10}\right] \cup \cdots \cup \left[\dfrac9{10},1\right]

The i-th subinterval has left and right endpoints, respectively, given by

\ell_i = \dfrac{i-1}{10} \text{ and } r_i = \dfrac i{10}

where i\in\{1,2,3,\ldots,10\}.

The midpoint of the i-th interval is the average of these,

m_i = \dfrac{\ell_i+r_i}2 = \dfrac{2i-1}{20} \in \left\{\dfrac1{20},\dfrac3{20},\dfrac5{20},\ldots,\dfrac{19}{20}\right\}

We approximate the area under f(x) over [0, 1] by the Riemann sum,

\displaystyle \int_0^1 f(x) \, dx \approx \sum_{i=1}^{10} f(m_i) \Delta x \\\\ ~~~~~~~~ = \frac1{10} \sum_{i=1}^{10} \bigg(5\sin(\pi m_i) + 2.5 \cos(4\pi m_i)\bigg) \\\\ ~~~~~~~~ = \frac{\sin\left(\frac\pi{20}\right) + \sin\left(\frac{3\pi}{20}\right) + \cdots + \sin\left(\frac{19\pi}{20}\right)}2 \\\\ ~~~~~~~~~~~~ + \dfrac{\cos\left(\frac\pi5\right) + \cos\left(\frac{3\pi}5\right) + \cdots + \cos\left(\frac{19\pi}5\right)}4 \\\\ ~~~~~~~~ \approx 3.19623

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2 years ago
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