Chapter 7: Problem 26
For exercises \(5-48\), simplify. $$ \frac{k}{k^{2}+49}-\frac{7}{k^{2}+49} $$
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Chapter 7: Problem 26
For exercises \(5-48\), simplify. $$ \frac{k}{k^{2}+49}-\frac{7}{k^{2}+49} $$
These are the key concepts you need to understand to accurately answer the question.
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For exercises \(65-68\), evaluate. $$ \sqrt{16} $$
$$ \text { Solve: } 0.75=\frac{k}{60} $$
For exercises \(35-36, T=\frac{336 \mathrm{gm}}{R}\) represents the relationship of tire diameter, \(T\); gear ratio, \(g\); speed, \(m\); and revolutions of the tire per minute, \(R\). Is the relationship of the given variables a direct variation or an inverse variation? $$ g \text { and } R \text { are constant; the relationship of } T \text { and } m \text {. } $$
For exercises 37-38, \(T=\frac{R}{A}\) represents the relationship of the asset turnover ratio, \(T\); the sales revenue of a company, \(R\); and the total revenues of a company, \(A\). Is the relationship of the given variables a direct variation or an inverse variation? $$ A \text { is constant; the relationship of } R \text { and } T \text {. } $$
When the top of a cone is removed, the formula for the volume of the remaining cone (the frustrum) is \(V=\frac{1}{3} \pi\left(R^{2}+R r+r^{2}\right) h\), where \(r\) is the radius of the circle at the top of the frustrum and \(R\) is the radius of the circle at the bottom of the frustrum. In 1856, an American army officer, Henry Hopkins Sibley, invented and received a patent for the design of a conical tent that could sleep 12 soldiers. (The apex is the diameter of the top of the frustrum.) Find the volume of the tent in cubic feet. Use \(\pi \approx 3.14\). Round to the nearest whole number. Be it known that I, H.H. Sibley, United States Army, have invented a new and improved Conical Tent ... the tent is in shape the frustrum of a cone; the base 18 feet; the height 12 feet; the apex 1 foot 6 inches [1.5 ft]. (Source: patimg1.uspto.gov)
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