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posledela
3 years ago
8

A flare was launched straight up from the ground with an initial velocity of 176 ft/s and returned to the ground after 11 s. The

height of the flare t seconds after launch is modeled by the function f(t)=−16t2+176t . What is the maximum height of the flare, in feet?
Mathematics
2 answers:
Rufina [12.5K]3 years ago
8 0
The answer is 484 just took the test 
photoshop1234 [79]3 years ago
5 0

Answer:

484 feet is the answer.

Step-by-step explanation:

Maximum value  of f(t) = -16t^{2}+176t \\

is given by when t =  \frac{-176}{2(-16)}  \\

which give t = 5.5 seconds

maximum height is when we plug t = 5.5 seconds in equation

more over time of flight is 11 seconds (given )

it will be at its maximum when time is  half of 11 seconds.

that is 5.5 seconds

maximum height is obtained by plugging t = 5.5seconds.

f(5.5) = -16(5.5)(5.5) + 176(5.5)

       = -484 +968

       = 484  feet

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Answer:

4x^{3} y^{2} (\sqrt[3]{4 x y})

Step-by-step explanation:

Another complex expression, let's simplify it step by step...

We'll start by re-writing 256 as 4^4

\sqrt[3]{256 x^{10} y^{7} } = \sqrt[3]{4^{4} x^{10} y^{7} }

Then we'll extract the 4 from the cubic root.  We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.

\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }

Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.

4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }

For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.

4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}

The answer is then:

4x^{3} y^{2} \sqrt[3]{4 x y} = 4x^{3} y^{2} (\sqrt[3]{4 x y})

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