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)
Responses
405
405
-36
-36
36
36
5
To find the difference between the distances of Bertha and Vernon's dives, we subtract the two distances:
|-45 - (-9)| = |-45 + 9| = |-36| = 36
Therefore, Bertha's dive ended 36 units farther than Vernon's dive. Answer: 36.
To find out how many times farther the end of Bertha's dive is from the end of Vernon's dive, we need to calculate the ratio of their distances from the starting platform.
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 calculate the ratio, we divide the distance of Bertha's dive by the distance of Vernon's dive:
-45 / -9 = 5
Therefore, the end of Bertha's dive is 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 from the end of Vernon's dive, we need to calculate the ratio between their distances.
The distance of Bertha's dive from the starting platform is -45 m, and the distance of Vernon's dive is -9 m.
To calculate the ratio, we can divide Bertha's distance by Vernon's distance:
Ratio = Bertha's distance / Vernon's distance
Ratio = (-45 m) / (-9 m)
Since we have negative values, we need to remove the negative sign for simplicity.
Ratio = 45 m / 9 m
Now, we can simplify the ratio by dividing both numbers by their greatest common divisor, which is 9 in this case:
Ratio = (45 m / 9 m) / (9 m / 9 m)
Ratio = 5
So, the end of Bertha's dive was 5 times farther from the starting platform than the end of Vernon's dive.
The answer is 5.