What does 3x+2(4+6x)+113 equael

1 answer:

7 0

<span>Simplifying 3x + 2(4 + 6x) = 113 3x + (4 * 2 + 6x * 2) = 113 3x + (8 + 12x) = 113 Reorder the terms: 8 + 3x + 12x = 113 Combine like terms: 3x + 12x = 15x 8 + 15x = 113 Solving 8 + 15x = 113 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-8' to each side of the equation. 8 + -8 + 15x = 113 + -8 Combine like terms: 8 + -8 = 0 0 + 15x = 113 + -8 15x = 113 + -8 Combine like terms: 113 + -8 = 105 15x = 105 Divide each side by '15'. x = 7 Simplifying x = 7</span>

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A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

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

The distance is 106.109 m

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 :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

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