Chapter 5: Problem 48
A building that is 250 feet high casts a shadow 40 feet long. Find the angle of elevation, to the nearest tenth of a degree, of the Sun at this time.
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Chapter 5: Problem 48
A building that is 250 feet high casts a shadow 40 feet long. Find the angle of elevation, to the nearest tenth of a degree, of the Sun at this time.
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Let \(f(x)=\left\\{\begin{array}{ll}{x^{2}+2 x-1} & {\text { if } x \geq 2} \\\ {3 x+1} & {\text { if } x<2}\end{array}\right.\) Find \(f(5)-f(-5)\)
For years, mathematicians were challenged by the following problem: What is the area of a region under a curve between two values of \(x ?\) The problem was solved in the seventeenth century with the development of integral calculus. Using calculus, the area of the region under \(y=\frac{1}{x^{2}+1},\) above the \(x\) -axis, and between \(x=a\) and \(x=b\) is \(\tan ^{-1} b-\tan ^{-1} a\). Use this result, shown in the figure, to find the area of the region under \(y=\frac{1}{x^{2}+1}\) above the \(x\) -axis, and between the values of a and b given in Exercises \(97-98\). (GRAPH CANNOT COPY) \(a=0\) and \(b=2\)
Determine whether each statement makes sense or does not make sense, and explain your reasoning. Determine the range of each of the following functions Then give a viewing rectangle, or window, that shows two periods of the function's graph. a. $$f(x)=3 \sin \left(x+\frac{\pi}{6}\right)-2$$ b. $$g(x)=\sin 3\left(x+\frac{\pi}{6}\right)-2$$
For years, mathematicians were challenged by the following problem: What is the area of a region under a curve between two values of \(x ?\) The problem was solved in the seventeenth century with the development of integral calculus. Using calculus, the area of the region under \(y=\frac{1}{x^{2}+1},\) above the \(x\) -axis, and between \(x=a\) and \(x=b\) is \(\tan ^{-1} b-\tan ^{-1} a\). Use this result, shown in the figure, to find the area of the region under \(y=\frac{1}{x^{2}+1}\) above the \(x\) -axis, and between the values of a and b given in Exercises \(97-98\). (GRAPH CANNOT COPY) \(a=-2\) and \(b=1\)
The angular speed of a point on Earth is \(\frac{\pi}{12}\) radian per hour. The Equator lies on a circle of radius approximately 4000 miles. Find the linear velocity, in miles per hour, of \(\overline{\mathbf{a}}\) point on the Equator.
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