Advanced Placement (AP), 04.09.2020 08:01 anniebear

What was the treaty of Paris 1763

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The Treaty of Paris, often called the Peace of Paris, or the Treaty of 1763, was signed on February 10, 1763, by the kingdoms of Great Britain, France and Spain, with Portugal in agreement. Together with the Treaty of Hubertusburg, it ended the French and Indian War or the Seven Years' War. The treaties marked the beginning of an extensive period of British dominance outside of Europe.
While the bulk of conquered territories were restored to their pre-war owners, the British made some substantial overseas gains at the expense of France and, to a lesser extent, Spain. Preferring to keep Guadaloupe, France gave up New France and all of its claims to the territory east of the Mississippi River to Britain. Spain ceded Florida to the British, but later received New Orleans and French Louisiana from France; Manila and Cuba were restored to Spain. France retained Saint Pierre and Miquelon and recovered Guadeloupe, Martinique, and Saint Lucia in exchange for Dominica, Grenada, Saint Vincent and the Grenadines, and Tobago going to the British. In India, the French lost out to the British, receiving back its "factories" (trading posts), but agreeing to support the British client governments, as well as returning Sumatra and agreeing not to base troops in Bengal. The British garrison on the Mediterranean island of Minorca was returned to her control, having been captured by the French at the outbreak of hostilities in Europe.
Britain returned the slave station on the isle of Gorée to the French, but gained the Senegal River and its settlements. Britain agreed to demolish its fortifications in Honduras, but received permission from Spain to keep a logwood-cutting colony there. Britain confirmed in the treaty the rights of its new subjects to practice the Roman Catholic religion and received confirmation of the continuation of the British king's Hanoverian right as a Prince-elector of the Holy Roman Empire.
It is sometimes claimed that the British King George III renounced his claim to be King of France by the treaty. However, this a historical myth, and it is also falsely attributed to some of the treaties of the French Revolutionary Wars. Such a renunciation is nowhere in the text of the treaty, and, in fact, George III continued to be styled "King of France" and used the fleurs-de-lis as part of his arms until 1801, when Britain and Ireland united. It was dropped then because the claim was regarded as anachronistic.

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This metaphor uses symbolism when describing the “arch” of ulysses life. he is trapped in a time of life where adventure is moving as he tries to proceed to it. he will never reach it and will learn to accept it. ulysses symbolizes that arch in life, that no one can reach every goal, or every sight in life. even though this personality of curiosity and purposefulness exist in all mankind, the purpose of showing his strive to succeed symbolizes great leaders today. his curiosity represents the curiosity of humans that everyone desires to find answers and wont give up without a fight. ulysses is a symbol of the human race.

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Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)