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katovenus [111]
3 years ago
9

a race car is driving under the caution flag at 440 feet per second begins to accelerate at a constant rate after the warning fl

ag. the distance traveled since the warning flag in feet is characterized by 30t^2+440t, where t is the time in seconds after the car starts accelerating again. How long will the car take to travel 150 feet?
Mathematics
1 answer:
slava [35]3 years ago
6 0

Answer: \frac{1}{3}\ s

Step-by-step explanation:

Given

Distance traveled is s=30t^2+440t

for s=150 ft

\Rightarrow 30t^2+440t=150

\Rightarrow 30t^2+440t-150=0\\\Rightarrow 3t^2+44t-15=0\\\Rightarrow (3t-1)(t+15)=0\\\Rightarrow t=\dfrac{1}{3}\ s

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A monthly inflation of $ 0 is assumed

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For the Part A.

The inequality that this situation represents

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So:

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For the Part B.

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answer:

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bogdanovich [222]
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