Help solve this pls ASAP

Mathematics

1 answer:

daser333 [38]2 years ago

3 0

Answer:

m<h = 100 deg

Step-by-step explanation:

Draw a segment connecting the two parallel lines. Draw that segment near the parallel line symbols (the small arrowheads). Now you have a pentagon. The sum of the measures of the interior angles of a pentagon is

(n - 2)180 = (5 - 2)180 = 3(180) = 540

The angles the new segment makes with the parallel lines are same-side interior angles of parallel lines cut by a transversal, so they are supplementary. Their measures add up to 180 deg.

180 + 98+ 162 + h = 540

h + 440 = 540

h = 100

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