r=12rSπ;V=43πr3=43π·18µSπ±3/2=16πµSπ±3/2.Answer:V(S)=16πµSπ±3/2,05S<+∞.C01S01.039:To avoid decimals, we note that a change of 5◦C is the same as a change of 9◦F, so whenthe temperature is 10◦Citis32+18=50◦F; when the temperature is 20◦C then it is 32 + 2·18 = 68◦F.In general we get the Fahrenheit temperatureFby adding 32 to the product of110Cand 18, whereCis theCelsius temperature. That is,F=32+95C,and thereforeC=59(F−32). Answer:C(F59(F−32),F>−459.67.C01S01.040:Suppose that a rectangle has base lengthxand perimeter 100. Lethdenote the height ofsuch a rectangle. Then 2x+2h= 100, so thath=50−x. Becausex=0 andh=0, we see that 05x550.The areaAof the rectangle isxh, so thatA(xx(50−x),05x550.C01S01.041:Letydenote the height of such a rectangle. The rectangle is inscribed in a circle of diameter4, so the bottom sidexand the left sideyare the two legs of a right triangle with hypotenuse 4. Consequentlyx2+y2= 16, soy=√16−x2(not−√16−x2becausey=0). Becausex=0 andy=0, we see that05x54. The rectangle has areaA=xy,soA(xxp16−x2,05x54.C01S042.042:We take the problem to mean that current production is 200 barrels per day per well, that
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