A group of college students developed a solar-powered car and entered it in a race. The car travels at a constant speed of 100 meters per 4 seconds. A solar-powered car on a racetrack. A. What representation can show the distance the car will travel over time?
B. What expression can show the distance the car will travel over time?
C. Compare the representation and the expression. Which shows the distance traveled over time more clearly? Explain.

A) Distance time graph

B) d(t) = 25t

C) The expression shows the distance more clearly.

Step-by-step explanation:

A) A distance time graph as seen in the attachment provides a representation of the distance travelled.

We are told the car travels at a constant speed of 100 meters per 4 seconds. Which means that 100 m for each 4 hours. So, for 200m, it's 8 hours like seen in the graph and for 300m,it's 12 hours as seen in the graph.

B) And expression for the distance is;

d = vt

Where;

d is distance in metres

v is speed in m/s and t is time

We are told that the car travels at a constant speed of 100 meters per 4 seconds.

Thus, v = 100/4 = 25 m/s

Distance travelled over time is;

d(t) = 25t

C) Looking at both A and B above, it's obvious that the expression of the distance shows a more clearer way of getting the distance because once we know the time travelled, we will just plug it into the equation and get the distance. Whereas, for the representation form, one will need to longer graphs if the time spent is very long.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.