Arace car driving under the caution flag at 80 feet per second begins to accelerate at a constant rate after the warning flag. the distance traveled since the warning flag in feet is characterized by 30t 2 + 80t, where t is the time in seconds after the car starts accelerating again. how long does it take the car to travel 30 feet after it begins accelerating?

1/3 sec or 0.333 sec

Step-by-step explanation:

d = 30*t^2 + 80*t = 30

putting into quadratic form, a*t^2 + b*t + c = 0

30*t^2 + 80*t - 30 = 0 where: a = 30 b = 80 c = -30

solve quadratic:

t = [-b + sqrt( b^2 - 4*a*c)]/2*a

substitute in values above

therefore, t = 1/3 sec or 0.333 sec

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.

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step-by-step explanation: