Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition

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DIAGNOSTIC TESTSTest AAlgebra1.(a)(−3)4= (−3)(−3)(−3)(−3) = 81(b)−34=−(3)(3)(3)(3) =−81(c)3−4=134=181(d)523521= 523−21= 52= 25(e)23−2=322=94(f )16−34=11634=14√163=123= 182.(a) Note that√200 =√100·2 = 10√2and√32 =√16·2 = 4√2. Thus√200− √32 = 10√2−4√2 = 6√2.(b)(333)(42)2= 3331624= 4857(c)33232−12−2=2−1233232= (2−12)2(3323)2=4−1936=4936=973.(a)3(+ 6) + 4(2−5) = 3+ 18 + 8−20 = 11−2(b)(+ 3)(4−5) = 42−5+ 12−15 = 42+ 7−15(c)√+√ √− √=√2− √√+√√−√2=−Or:Use the formula for the difference of two squares to see that√+√√− √=√2−√2=−.(d)(2+ 3)2= (2+ 3)(2+ 3) = 42+ 6+ 6+ 9 = 42+ 12+ 9.Note:A quicker way to expand this binomial is to use the formula(+)2=2+ 2+2with= 2and= 3:(2+ 3)2= (2)2+ 2(2)(3) + 32= 42+ 12+ 9(e) See Reference Page 1 for the binomial formula(+)3=3+ 32+ 32+3. Using it, we get(+ 2)3=3+ 32(2) + 3(22) + 23=3+ 62+ 12+ 8.4.(a) Using the difference of two squares formula,2−2= (+)(−), we have42−25 = (2)2−52= (2+ 5)(2−5).(b) Factoring by trial and error, we get22+ 5−12 = (2−3)(+ 4).(c) Using factoring by grouping and the difference of two squares formula, we have3−32−4+ 12 =2(−3)−4(−3) = (2−4)(−3) = (−2)(+ 2)(−3).(d)4+ 27=(3+ 27) =(+ 3)(2−3+ 9)This last expression was obtained using the sum of two cubes formula,3+3= (+)(2−+2)with=and= 3. [See Reference Page 1 in the textbook.](e) The smallest exponent onis−12, so we will factor out−12.332−912+ 6−12= 3−12(2−3+ 2) = 3−12(−1)(−2)(f )3−4=(2−4) =(−2)(+ 2)

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Biocalculus: Calculus For Life Scie...

Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition

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