In an interval whose length is z seconds, a body moves (32z + 2z^ 2 ) ft. What is the average speed v of the body in this interval?

This question is incomplete, the complete question is;

In an interval whose length is z seconds, a body moves (32z + 2z²) ft.

What is the average speed v of the body in this interval?

the average speed v of the body in this interval is 32 + 2z ft/s

Step-by-step explanation:

Given that;

Length of travel = z seconds

also the body moves (32z + 2z²) ft

i.e

distance travelled = (32z + 2z²) ft

average speed v of the body in this interval = ?

Now to get the Average Speed

we simply say;

Average Speed V =  (32z + 2z²) / z

Average Speed V = 32 + 2z ft/s

Therefore the average speed v of the body in this interval is 32 + 2z ft/s

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.