What is the equation in point slope form of the line that passes through the point

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

Answer:

$\large\boxed{y+2=3(x-1)}$

Step-by-step explanation:

$\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\m-slope\\(x_1,\ y_1)-point\\\\\text{We have the slope}\ m=3\ \text{and the point (1, -2). Substitute:}\\\\y-(-2)=3(x-1)\\\\y+2=3(x-1)$

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