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)
405
-36
36
5
The distance of Bertha's dive from the starting platform is 45 m, and the distance of Vernon's dive from the starting platform is 9 m.
To find how many times farther Bertha's dive is, we divide the distance of Bertha's dive by the distance of Vernon's dive:
45/9 = <<45/9=5>>5.
Therefore, the end of Bertha's dive is 5 times farther than the end of Vernon's dive.
The answer is 5.
To find out how many times farther Bertha's dive ended compared to Vernon's dive, we need to calculate the ratio of the distances.
First, let's calculate the absolute values of the distances. Since distance can't be negative, we'll ignore the negative signs.
Bertha's dive ended 45 m from the starting platform (| -45 | = 45).
Vernon's dive ended 9 m from the starting platform (| -9 | = 9).
Now, let's calculate the ratio:
Bertha's distance / Vernon's distance = 45 / 9 = 5.
Therefore, the end of Bertha's dive was 5 times farther than the end of Vernon's dive. So, the correct answer is 5.
To find how many times farther Bertha's dive ended compared to Vernon's dive, we need to divide the distance of Bertha's dive by the distance of Vernon's dive.
The distance of Bertha's dive is -45 m.
The distance of Vernon's dive is -9 m.
Dividing the distance of Bertha's dive by the distance of Vernon's dive:
-45 / -9 = 5
Therefore, the end of Bertha's dive was 5 times farther than the end of Vernon's dive.