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
This is the end of the preview. Sign up
access the rest of the document.