
ROGER F. answered 06/26/15
Tutor
4.9
(127)
DR ROGER - TUTOR OF MATH, PHYSICS AND CHEMISTRY
A) I(0.5) = 2(O.5)2 - 6(0.5) + 7 = 4.50 A
B) 15 = 2t2 - 6t +7 REARRANGE THIS TO: 2t2 - 6t - 8 = 0 OR t2 - 3t - 4 = 0 (divided by 2)
THIS IS A QUADRATIC EQUATION THAT FACTORS INTO (t - 4)(t + 3) = 0
SO t = 4, or -3 YOU CAN'T HAVE A NEGATIVE TIME, SO t = 4 sec
C) 30 = 2t2 - 6t +7 REARRANGE THIS TO 2t2 - 6t - 23 = 0
THIS WON'T FACTOR, SO IT CAN BE SOLVED USING THE QUADRATIC FORMULA:
t = 6 +/- √(-6)2 - 4(2)(-23)/2(2) = (6 +/- √220)/ 4 = 5.21 SEC (OR -2.21 SEC, BUT IT CAN'T BE NEGATIVE)