When you subtract a positive number from a negative number the difference is always

Mathematics

1 answer:

deff fn [24]3 years ago

8 0

The answer is true.

(+)(+)= +

(-)(-)= +

(-)(+)= -

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

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

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago

Need help with this didn’t understand it

zaharov [31]

Answer:

Ok, the solution to the first one. See below.

Step-by-step explanation:

$1>3-d\\\\\mathrm{Add\:}d\mathrm{\:to\:both\:sides}\\\\1+d>3-d+d\\\\\mathrm{Simplify:}\\\\1+d>3\\\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\\\1+d-1>3-1\\\\\mathrm{Simplify:}\\\\d>2$