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3 years ago

Please Help pt 2 I don't understand any of this

Mathematics

1 answer:

dedylja [7]3 years ago

7 0

Answer: $\bold{(D)\ \dfrac{3}{4}-\dfrac{1}{4}x}$

<u>Step-by-step explanation:</u>

$\ \ \dfrac{3}{4}-x\bigg(\dfrac{1}{2}-\dfrac{5}{8}\bigg)+\bigg(-\dfrac{3}{8}x\bigg)\\\\\\=\dfrac{3}{4}\bigg(\dfrac{2}{2}\bigg)-x\bigg[\dfrac{1}{2}\bigg(\dfrac{4}{4}\bigg)-\dfrac{5}{8}\bigg]+\bigg(-\dfrac{3}{8}x\bigg)\\\\\\=\dfrac{6}{8}-x\bigg(\dfrac{4}{8}-\dfrac{5}{8}\bigg)-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}-x\bigg(-\dfrac{1}{8}\bigg)-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}+\dfrac{1}{8}x-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}-\dfrac{2}{8}x\\\\\\=\dfrac{3}{4}-\dfrac{1}{4}x\quad \text{(reduced both fractions)}$

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Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

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Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

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The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

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