It takes ten identical pieces to form a circular track for a pair of toy racing cars. if the inside arc of each piece is 3.4 inches shorter than the outside arc, what is the width of the track?

 Length of outer track = sum of length of 10 pieces = circumference of the outer circle 

if R is the Radius of outer circle then... 

Circumference of the outer track = 2pi*R 

Similarly the circumference of the inner track (with radius r) = 2pi*r 

length of each outer piece is 3.4 inch more than length of inner piece 

So total outer length is 10*3.4 =34 inches more than the inner length. 

=> Outer Circumference - Inner Circumference = 34 inches 

=> 2pi*R - 2pi*r = 34 

=> 2pi(R -r) = 34 

=> R-r = 34/2pi = 5.41 inches 

=> R-r = Width of the track = 5.41 inches

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

No he does not because he can only cover 12 square feet from 15 square feet which he will still need 3 more. this is because if 3×5=15 square feet and 15-12=3 which thats the answer