A) 10s + 7t = 149 5s + 7t = 99

Mathematics

2 answers:

mestny [16]2 years ago

5 0

Answer:

s = $\frac{149}{10} - \frac{7t}{10}$

t = $\frac{149}{10} -\frac{10s}{7}$

Step-by-step explanation:

(Hope this helps can I pls have brainlist (crown)☺️)

sergeinik [125]2 years ago

5 0

$10s + 7t = 149 \\ 10s + 7t + - 7t = 149 + - 7t \\ 10s = - 7t + 149 \\ \frac{10s}{10} = \frac{ - 7t + 149}{10} \\ \boxed{s = \frac{ - 7}{10} t + \frac{149}{10} }$

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$5s+7t=99 \\ 5s+7t+−7t=99+−7t \\ 5s=−7t+99 \\ \frac{5s}{5} = \frac{ - 7 + 99}{5} \\ \boxed{s = \frac{ - 7}{5} t + \frac{99}{5} }$

correct me if I'm wrong.

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4 years ago

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