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Westkost [7]
3 years ago
5

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function

h = –16t2+ 36t + 9.
In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.
What is the ball’s maximum height?
Mathematics
2 answers:
nirvana33 [79]3 years ago
7 0

Answer:

After 5 seconds (c)

Step-by-step explanation:

FinnZ [79.3K]3 years ago
6 0

Answer:

The ball reaches a height of 29.25 ft after 1.125 seconds

Step-by-step explanation:

The maximum height of a parabola can always be found by looking for the vertex. You can find the x value (or in this case the t value) of a vertex by using -b/2a in which a is the coefficient of x^2 and b is the coefficient of x.

-b/2a

-(36)/2(-16)

-36/-32

1.125 seconds

Now to find the height, we input that value in for t

h = -16t^2 + 36t + 9

h = -16(1.125)^2 + 26(1.125) + 9

29.25 feet

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Given:

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The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

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Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

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Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

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Step-by-step explanation:

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