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laiz [17]
4 years ago
7

An acrobat was launched toward a water tank, with an upward velocity of 48 feet per second. The launch point was 32 feet below t

he water level in the tank. Use the quadratic equation 0 = -16t2 + 48t – 32 to find how much time, t, it took for her to reach the water level in the tank.
Mathematics
1 answer:
ludmilkaskok [199]4 years ago
3 0
Hello. How are you. Do u take K12
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B

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f(t) =  -16t^{2} +48t + 100
Average rate of change from t = 3 seconds to t = 5 seconds = \frac{f(5)-f(3)}{5-3} =  \frac{(-16( 5^{2})+48(5)+100)-(-16( 3^{2})+48(3)+100  }{2}=\frac{(-16(25)+240+100)-(-16(9)+144+100)}{2} = \frac{(-400+240+100)-(-144+144+100)}{2}= \frac{-60-100}{2} = \frac{-160}{2} = -80
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3 years ago
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18 and 0

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