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Pollution. In a large city, the amount of sulfur dioxide pollutant released into the atmosphere due to the burning of coal and oil for heating purposes is given approximately by
where P(n) is the amount of sulfur dioxide (in tons) released during the nth week after January 1.
(A) How many tons of pollutants were emitted into the atmosphere over the 2-year period?
(B) How many tons of pollutants were emitted into the atmosphere from n = 13 to n = 52?
(C) Illustrate part (B) graphically with an appropriate shaded region representing the total tons of pollutants emitted into the atmosphere.
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Chapter 8 Solutions
Pearson eText for Calculus for Business, Economics, Life Sciences, and Social Sciences, Brief Version -- Instant Access (Pearson+)
- Both in images okk. Instructions. "I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."arrow_forwardQuestion 1: If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p = 13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts) Solution: A barometer works based on the principle that the pressure exerted by the liquid column balances atmospheric pressure. The pressure is given by: P = pgh Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the liquid column is inversely proportional to its density: Step 1: Given Data PHg hol=hgx Poil • Density of mercury: PHg = 13.6 g/cm³ Density of oil: Poil = 0.92 g/cm³ • Standard height of mercury at 1.00 atm: hμg Step 2: Compute Height of Oil = 760 mm = 0.760 m 13.6 hoil = 0.760 x 0.92 hoil = 0.760 × 14.78 hoil = 11.23 m Step 3: Compare Heights Since oil is less dense than mercury, the column of oil must be much taller than that of mercury. The factor by which it is taller is: Final…arrow_forwardQuestion 3: A sealed flask at room temperature contains a mixture of neon (Ne) and nitrogen (N2) gases. Ne has a mass of 3.25 g and exerts a pressure of 48.2 torr. . N2 contributes a pressure of 142 torr. • What is the mass of the N2 in the flask? • Atomic mass of Ne = 20.1797 g/mol • Atomic mass of N = 14.0067 g/mol Solution: We will use the Ideal Gas Law to determine the number of moles of each gas and calculate the mass of N2. PV = nRT where: • P = total pressure • V volume of the flask (same for both gases) n = number of moles of gas • R 0.0821 L atm/mol K • T = Room temperature (assume 298 K) Since both gases are in the same flask, their partial pressures correspond to their mole fractions. Step 1: Convert Pressures to Atmospheres 48.2 PNe = 0.0634 atm 760 142 PN2 = = 0.1868 atm 760 Step 2: Determine Moles of Ne nNe = mass molar mass 3.25 nNe 20.1797 nne 0.1611 mol Step 3: Use Partial Pressure Ratio to Find narrow_forward
- "I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."arrow_forwardNo chatgpt pls will upvote Already got wrong chatgpt answerarrow_forwardConstruct a table of values for all the nonprincipal Dirichlet characters mod 16.arrow_forward
- MI P X /courses/segura10706/products/171960/pages/611?locale=&platformId=1030&lms=Y ☆ Finish Part I: Mathematics for Elementary and Middle School Teachers Continue in the app JJ 576 Chapter 12. Area of Shapes 9. Determine the area of the shaded shapes in Figure 12.48. Explain your reasoning. 1 unit S Figure 12.48 1 unit unit and the yarn for thearrow_forwardSuppose p > 3 is a prime. Show that (p − 3)!= − P+1 (mod p). Hint: Use Wilson's theorem.arrow_forwardWhich graph represents f(x) = √x-2+3?arrow_forward
- Suppose a = p²¹...p be the canonical factorization. Then the sum of all the factors of a, denoted by σ(a) is given by o(a) = II + k₂+1 P -1 Pi - 1 (you don't need to prove this). (a) Let a = 2³ × 7². Find σ(a), which the sum of all the factors a.arrow_forwardEvaluate the Legendre symbol (999|823). (Note that 823 is prime.)arrow_forwardIf p = 7 (mod 8), where p is prime, show that p divides 2(p-1)/2 — 1. Deduce that 275 - 1 and 2155 -1 are composite.arrow_forward
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