Tuesday,+February+16,+2010Set6

problem 2: Chris A. (0,105) (5,30) (10,21) (Time in Seconds, Feet in the Air) 105=0a+0b+c 30=25a+5b+c 21=100a+10b+c c=105 30=25a+5b+105 21=100a+10b+105 -75=25a+5b -84=100a+10b

100a+10b=-84 100a+10b=-84 25(1.32)+5b=-75 -2(25a+5b=-75) __-50a-10b+150__ 33+5b=-75 50a=66 a=1.32 5b=-108 b=-21.6 y=1.32x^2-21.6x+105

B. (4 seconds) y=1.32(4)^2-21.6(4)+105 (15 Seconds) y=1.32(15)^2-21.6(15)+105 y=21.12-86.4+105 y=297-319.5+105 y=39.72 39.72 Feet y=82.5 82.5 Feet

C. (Lowest Point) y=1.32x^2-21.6x+105 -b/2a 21.6/2.64=8.1818 y=1.32(8.1818)^2-21.6(8.1818)+105 y=88.363-176.727+105 y=16.6363 16.6363 Feet at 8.1818 Seconds

D. (200 Feet) 200=1.32x^2-21.6x+105 0=1.32x^2-21.6x-95 (21.6+-(466.56-4(1.32)(-95))^1/2)/2.64 (21.6+-(908.16)^1/2)/2.64 (21.6+-31.115)/2.64 19.9678 or -3.604 Only 19.9678 Seconds works Problem 3: Avi

therefore after 12 minutes he would hit 6.48 psi which is greater than the minimum 5 psi

Problem 4: Augustine a. (0,10) (2,30)

10=ab^0 a=10

30=10b^2 3=b^2 b=sq root 3

b. (?, 100) 100=10(rt3)^x 10=(rt3)^x log 10=x log rt3 log 10/log rt3=x x=4.19+2005 x=2009.19 years

c. 1=10(rt3)^x .1=rt3^x log .1= x log rt3 x=log.1/log rt3 x=-4.19 -4.19-2005 x=2000.81 years

d. (20, ?) y=10 rt3^20 y=10(59049) y=590,490 stalks