Find the angle between each of the following pairs of vectors a⃗ =axi^+ayj^ and b⃗ =bxi^+byj^.. ax1= -3.20, ay1= 4.40; bx1= 2.70, by1= -3.10. i know this is dot prodoct but the calculater just keep saying error

The product point of the two vectors is given by:

Then, the module of both vectors is:

For vector A:

For vector B:

Finally the angle between the vectors is given by:

Substituting values:

The angle between the vectors is:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°