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Evgen [1.6K]
3 years ago
14

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the fun

ction h (t) = 300 minus 16 t squared. Which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall?
h(3) – h(0)
h (three-thirds) minus h (zero-thirds)
StartFraction h (3) Over 3 EndFraction
StartFraction h (3) minus h (0) Over 3 EndFraction
Mathematics
2 answers:
Bogdan [553]3 years ago
4 0

Answer:

Average rate of fall = \frac{h(3)-h(0)}{3}

Step-by-step explanation:

Since they give you an expression for the position of the object as a function of time, one can derive the average velocity of the object during the first 3 seconds of the fall by estimating the position of the object at time 0, the position of the object at time 3 seconds, finding the displacement of the object (distance covered) in that interval of time (via a subtraction of these two quantities), and finally dividing this change by the elapsed time (3 seconds) thus using the concept that velocity is the quotient between distance covered over the elapsed time:

Average rate of fall = (h(3) - h(0))/ 3

Average rate of fall = \frac{h(3)-h(0)}{3}

erik [133]3 years ago
4 0

Answer:

D

Step-by-step explanation:

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Answer:

\dfrac{(a+b)}{ab}

Step-by-step explanation:

The given expression is :

\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}

It can be solved as follows :

\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}

So, the solution of the given expression is equal to \dfrac{(a+b)}{ab}.

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3 years ago
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Serjik [45]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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