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2 years ago

Please help pls pls help pls belp

Mathematics

1 answer:

Delvig [45]2 years ago

8 0

Answer:

After 1 second, the ball will reach a maximum height of 16 feet

Step-by-step explanation:

The height of the ball after t seconds: h(t) = -16t^2 + 32t

The graph of this quadratic function is parabola which opens downwards. The vertex of a quadratic equation is the maximum or minimum point on the equation's parabola

t = -b/2a = -(32)/2(-16) = -32/-32 = 1 second

then

h(t) = -16(1)^2 + 32(1) = -16 + 32 = 16

After 1 second, the ball will reach a maximum height of 16 feet

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Answer:

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Since 30 divided by .15= 200. And there is a .15 cent cost per minute meaning Javier would have to talk for 200 minutes for the prices to be equal.

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a band of 45 ewoks crash-landed in the forest last night. this sounds like a small problem, but the population will grow at the

kogti [31]

Given Information:

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rate of growth = 22%

Required Information:

Population every five years from this year to the year 2050 = ?

Answer:

$Year \: 2020 = P(0) = 45e^{0} = 45\\\\Year \: 2025 = P(5) = 45e^{0.22*5} = 135\\\\Year \: 2030 = P(10) = 45e^{0.22*10} = 406\\\\Year \: 2035 = P(15) = 45e^{0.22*15} = 1,220\\\\Year \: 2040 = P(20) = 45e^{0.22*20} = 3,665\\\\Year \: 2045 = P(25) = 45e^{0.22*25} = 11,011\\\\Year \: 2050 = P(30) = 45e^{0.22*30} = 33,079\\\\$

Step-by-step explanation:

The population growth can be modeled as an exponential function,

$P(t) = P_0e^{rt}$

Where P₀ is the starting population, r is the rate of growth of the population and t is the time in years.

We are given that starting population of 45 and growth rate of 22%

$P(t) = 45e^{0.22t}$

Assuming that the starting year is 2020,

Therefore, the starting population of ewoks was 45 in 2020 and increased to 33,079 by 2050 in a time span of 30 years.

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3 years ago