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PolarNik [594]
3 years ago
6

A model airplane is shot up from a platform 1 foot above the ground with an initial upward velocity of 56 feet per second. The h

ight of the airplane above ground after t seconds is given by the equation h=-16t^2+56t+1 where h is the hight of the airplane in feet and t is the time in seconds after it is launched. Approximately how long does it take the airplane to reach its maximum height?
A)0.3 seconds
B)1.8 seconds
C)3.5 seconds
D)6.9 seconds
Please Help
Mathematics
2 answers:
docker41 [41]3 years ago
8 0

Answer:

The correct answer is B. 1.8 seconds. I just took the test!


Law Incorporation [45]3 years ago
4 0
H=-16t^2+56t+1, the airplane is shot up from a platform 1 foot above the ground, so 1= - 16t^2+56t+1, we solve this equation
(- 16t+56)t=0, (- 16t+56)=0 or t=o, - 16t+56=0, implies t =3.5s
so the answer is C)3.5 seconds 
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andre [41]

Answer:

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Step-by-step explanation:

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3 years ago
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Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m
Dovator [93]

Option a: \frac{m^{5} }{162n} is the equivalent expression.

Explanation:

The expression is \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } where m\neq 0, n\neq 0

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }

Cancelling the like terms, we have,

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This equation can also be written as,

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Multiplying the terms in denominator, we have,

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3 years ago
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