Asked by Axel

Bertha and Vernon are competing in a diving competition. Bertha's dive ended -30 m from the starting platform. Vernon's dive ended -6 m from the starting platform. How many times farther was the end of Bertha's dive than the end of Vernon's dive?

Answers

Answered by Axel
Bertha and Vernon are competing in a diving competition. Bertha's dive ended -30 m from the starting platform. Vernon's dive ended -6 m from the starting platform. How many times farther was the end of Bertha's dive than the end of Vernon's dive?
Answered by Axel
THATS WRONG
Answered by GPT 3.5
The distance of Bertha's dive from the starting platform is -30 m - (-6 m) = -24 m.
Thus, the end of Bertha's dive was 24/6 = <<24/6=4>>4 times farther than the end of Vernon's dive. Answer: \boxed{4}.
Answered by GPT 3.5
The distance of Bertha's dive from the starting platform is -30 m - (-6 m) = -24 m.
Thus, the end of Bertha's dive was |-24 m|/|-6 m|=4 times farther than the end of Vernon's dive.
Therefore the end of Bertha's dive was 4 times farther than the end of Vernon's dive.
Thus, the answer is \boxed{4}.
Answered by GPT 3.5
Apologies for the mistake. Let's try again.

The distance of Bertha's dive from the starting platform is -30 m, and the distance of Vernon's dive from the starting platform is -6 m.

To determine how many times farther Bertha's dive was than Vernon's dive, we can calculate the ratio of their distances.

Ratio = Bertha's distance / Vernon's distance
Ratio = (-30 m) / (-6 m)

Simplifying the ratio, we have:
Ratio = 30 m / 6 m
Ratio = 5

Therefore, the end of Bertha's dive was 5 times farther than the end of Vernon's dive.

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