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alisha [4.7K]
3 years ago
6

An object is launched straight up into

Mathematics
1 answer:
DiKsa [7]3 years ago
8 0

Answer:

  t ∈ {1, 3}

Step-by-step explanation:

You want to find t such that ...

  h = 27

  27 = -8t^2 +32t +3 . . . . . . substitute the expression for h

  24 = -8t^2 +32t . . . . . . . . . subtract 3

  -3 = t^2 -4t . . . . . . . . . . . . . divide by -8

  1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square

  ±√1 = t -2 . . . . . . . . . . . . . . take the square root

  t = 2 ± 1 . . . . . . . . . . . . . . . . add 2

  t = 1 or 3

The object is 27 ft off the ground at t = 1 and again at t = 3.

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<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

(8a^{-3})^{\frac{-2}{3}

We have to write in simplest form

<em><u>Use the following law of exponent</u></em>

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Using this, simplify the given expression

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Step-by-step explanation:

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