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anygoal [31]
3 years ago
11

APML - ATAQ. Solve for x and QR HELP ASAP!!!!! PLZZZZZ

Mathematics
1 answer:
Roman55 [17]3 years ago
7 0

Answer:

x = 7

Step-by-step explanation:

(7x-9)(2x+2) = 35/14

98x - 126 = 70x + 70

28x = 196

x = 7

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A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t) = 2t3 - 21t2 + 72t - 53. At what time t is
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Step-by-step explanation:

The particle is at rest when the derivative of x(t), which is its velocity, is zero, i.e.,

\dfrac{d}{dt}[x(t)] = \dot{x}(t) = 0

Since x(t) = 2t^3 - 21t^2 + 72t - 53, its derivative is

\dot{x}(t) = 6t^2 - 42t + 72

Equating this to zero, we get

6t^2 - 42t + 72 = 0 \Rightarrow t^2 - 7t + 12 = 0

This is a familiar quadratic equation whose roots are

t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

\;\;\;= \dfrac{7 \pm \sqrt{(7)^2 - 4(1)(12)}}{2}

\;\;\;= \dfrac{7 \pm 1}{2}

\;\;\;= 3\;\text{and}\;4

This means that the particle will be at rest at t = 3 and t= 4.

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3 years ago
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