Bertha and Vernon are competing in a diving competition. Bertha's dive ended -45 m from the starting platform. Vernon's dive ended -9 m from the starting platform. How many times farther was the end of Bertha's dive than the end of Vernon's dive? (1 point)
To find out how many times farther the end of Bertha's dive was compared to the end of Vernon's dive, we need to calculate the relative difference in distance between the two dives.
First, let's define the distance of Bertha's dive as x and the distance of Vernon's dive as y.
According to the given information, Bertha's dive ended 45 m from the starting platform (x = -45 m) and Vernon's dive ended 9 m from the starting platform (y = -9 m).
To find the relative difference, we can calculate the ratio of the distances:
Relative difference = |x| / |y|
Plugging in the values, we have:
Relative difference = |-45 m| / |-9 m|
Since we are only concerned with the magnitude (positive values) of the distances, we can remove the absolute value signs:
Relative difference = 45 m / 9 m
Reducing the fraction, we get:
Relative difference = 5
Therefore, the end of Bertha's dive was 5 times farther from the starting platform than the end of Vernon's dive.
To find out how many times farther the end of Bertha's dive was compared to Vernon's dive, we need to calculate the ratio of the distances.
The distance of Bertha's dive is -45 m, and the distance of Vernon's dive is -9 m.
The ratio of the distances is calculated by dividing the distance of Bertha's dive by the distance of Vernon's dive:
Ratio = Distance of Bertha's dive / Distance of Vernon's dive
Ratio = -45 m / -9 m
Simplifying the ratio, we get:
Ratio = 5
Therefore, the end of Bertha's dive was 5 times farther than the end of Vernon's dive.
The distance of Bertha's dive compared to Vernon's dive is -45 m - (-9 m) = -45 m + 9 m = -36 m
Bertha's dive was 36/9 = <<36/9=4>>4 times farther than Vernon's dive. Answer: \boxed{4}.