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valkas [14]
3 years ago
13

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

e given equation. Using the equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second y=-16x^2+235x+133
Mathematics
1 answer:
jeka943 years ago
6 0

Answer:

Step-by-step explanation:

Given tq equation of the height reached by the rocket as y=-16x^2+235x+133

The velocity of the object at its maximum height is zero

v = dy/dx = 0(at max height)

v =-32x +235

0 =-32x+235

x = 235/32

x = 7.34

Hence the rocket will reach its maximum height after 7.34seconds

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Lunna [17]

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Step-by-step explanation:

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2 years ago
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7 0
2 years ago
A young couple purchases their first new home in 2011 for​ $95,000. They sell it to move into a bigger home in 2018 for​ $105,00
mart [117]

Given:

The value of home in 2011 is $95,000.

The value of home in 2018 is $105,000.

To find:

The exponential model for the value of the home.

Solution:

The general exponential model is

y=ab^x       ...(i)

where, a is initial value and b is growth factor.

Let 2011 is initial year and x be the number of years after 2011.

So, initial value of home is 95,000, i.e., a=95,000.

Put a=95000 in (i).

y=95000b^x       ...(ii)

The value of home in 2018 is $105,000. It means the value of y is 105000 at x=7.

105000=95000b^7

\dfrac{105000}{95000}=b^7

\dfrac{21}{19}=b^7

Taking 7th root on both sides, we get

\left(\dfrac{21}{19}\right)^{\frac{1}{7}}=b

Put b=\left(\dfrac{21}{19}\right)^{\frac{1}{7}} in (ii).

y=95000\left(\left(\dfrac{21}{19}\right)^{\frac{1}{7}}\right)^x

y=95000\left(\dfrac{21}{19}\right)^{\frac{x}{7}}

Therefore, the required exponential model for the value of home is y=95000\left(\dfrac{21}{19}\right)^{\frac{x}{7}}, where x is the number of years after 2011.

5 0
2 years ago
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