180=-16t^2+112t
2007-11-23 15:42:19 · 5 answers · asked by too_b2000 1 in Science & Mathematics ➔ Mathematics
16 t ² - 112 t + 180 = 0
4t ² - 28 t + 45 = 0
t = [ 28 ± √(784 - 720) ] / 8
t = [ 28 ± √(64) ] / 8
t = [ 28 ± (8) ] / 8
t = 36/8 , t = 20/8
t = 4.5 , t = 2.5
Also by factorising:-
(2t - 9) (2t - 5) = 0
t = 4.5 , t = 2.5
2007-11-23 21:33:08 · answer #1 · answered by Como 7 · 3⤊ 1⤋
Move it around a bit to get
16 t2 -112 t +180=0
You can factor out 4 to get:
4 t2 - 28 t + 45 =0
You can factor this to (2t-9)(2t-5)=0
Etc.
2007-11-23 23:51:40 · answer #2 · answered by cattbarf 7 · 1⤊ 0⤋
180 = - 16t^2 + 112t
45 = - 4t^2 + 28t
4t^2 - 28t + 45 = 0
(2t - 9)(2t - 5) = 0
2t - 9 = 0, 2t = 9, t = 4.5
2t - 5 = 0, 2t = 5, t = 2.5
Answer: t = 4.5, 2.5
Proof (t = 4.5):
180 = - 16(4.5^2) + 112(4.5)
180 = - 16(20.25) + 504
180 = - 324 + 504
180 = 180
Proof (t = 2.5):
180 = - 16(2.5^2) + 112(2.5)
180 = - 16(6.25) + 112(6.25)
180 = - 100 + 280
180 = 180
2007-11-24 00:53:25 · answer #3 · answered by Jun Agruda 7 · 4⤊ 2⤋
180 = -16t^2 + 112t
180 = -4(4t^2 - 28t)
45 = 4t^2 - 28t
4t^2 - 28t - 45 = 0
t = (-b ± sqrt(b^2 - 4ac))/(2a)
t = (-(-28) ± sqrt((-28)^2 - 4(4)(45)))/(2(4))
t = (28 ± sqrt(784 - 720))/8
t = (28 ± sqrt(64))/8
t = (28 ± 8)/8
t = (20/8) or (36/8)
t = (5/2) or (9/2)
in factored form it would be (2t - 5)(2t - 9)
2007-11-24 00:19:55 · answer #4 · answered by Sherman81 6 · 1⤊ 0⤋
180= -16t^2+112t
<=> 16t^2-112t+180=0
<=>4(4t^2-28t+45)=0
<=>4t^2-18t-10t+45=0
<=>2t(2t-9)-5(2t-9)=0
<=>(2t-5)(2t-9)=0
<=>t=5/2 or t=9/2!
2007-11-24 12:15:49 · answer #5 · answered by kami 5 · 0⤊ 3⤋