Answer:
500 inches
Step-by-step explanation:
Let C₁ and n₁ be the circumference and the number of rotations of the larger wheel. Also, let C₂ and n₂ be the circumference and the number of rotations of the smaller wheel. Since the distance moved by both wheels from Ally's house to her neighbor's house is the same, n₁C₁ = n₂C₂. (1)
Also it is given that half the circumference of the larger wheel is 5 inches more than the circumference of one of the smaller wheels.
So, C₁/2 = C₂ + 5
C₁ = 2(C₂ + 5) (2)
Substituting C₁ into (1), we have
n₁C₁ = n₂C₂.
n₁[2(C₂ + 5)] = n₂C₂.
Since n₁ = 10 and n₂ = 25, we have
10[2(C₂ + 5)] = 25C₂
20(C₂ + 5) = 25C₂
expanding the bracket,
20C₂ + 100 = 25C₂
collecting like terms, we have
25C₂ - 20C₂ = 100
5C₂ = 100
dividing both sides by 5, we have
C₂ = 100/5
= 20 inches
So, the distance between Ally's house and her neighbor's house is d = n₂C₂ = 25(20)
= 500 inches
020.99 I think and if not I am sorry
The prime factorisation for 58 is 2 x 29
:)
The answer is 72
You would have to add sixty-five plus seven
