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ankoles [38]
2 years ago
6

SORRY BUT I HAVE ANOTHER PROBLEM!!!!! WILL GIVE BRAINLIEST & 50 POINTS

Mathematics
2 answers:
11111nata11111 [884]2 years ago
8 0

Just factor out

\\ \rm\hookrightarrow \dfrac{x^2-6x+9}{6x-18}

\\ \rm\hookrightarrow \dfrac{x^2-3x-3x+9}{6(x-3)}

\\ \rm\hookrightarrow \dfrac{x(x-3)-3(x-3)}{6(x-3)}

\\ \rm\hookrightarrow \dfrac{(x-3)(x-3)}{6(x-3)}

\\ \rm\hookrightarrow \dfrac{x-3}{6}

liberstina [14]2 years ago
7 0

Answer:

The correct option is G.Hope this help!!

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Papessa [141]
<span>the particle's initial position is at t=0, x = 0 - 0 + 4 = 4m

velocity is rate of change of displacement = dx/dt = d(t^3 - 9t^2 +4)/dt
= 3t^2 - 18t

acceleration is rate of change of velocity = d(3t^2 -18t)/dt
= 6t - 18

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</span>3t*(t - 6) = 0
t = 0 or t = 6s

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