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)%0D%0AResponses%0D%0A%0D%0A405%0D%0A405%0D%0A%0D%0A-36%0D%0A-36%0D%0A%0D%0A36%0D%0A36%0D%0A%0D%0A5
To find the difference in the distances, we need to subtract the distance of Vernon's dive from the distance of Bertha's dive: -45 m - (-9 m) = -45 m + 9 m = -36 m.
The end of Bertha's dive was 36 meters farther than the end of Vernon's dive.
To determine how many times farther the end of Bertha's dive is from the end of Vernon's dive, we need to find the ratio between their distances.
First, let's find the absolute values of their distances:
Bertha's distance: |-45| = 45
Vernon's distance: |-9| = 9
Now we can calculate the ratio:
Ratio = Bertha's distance / Vernon's distance
= 45 / 9
= 5
Therefore, the end of Bertha's dive is 5 times farther from the starting platform compared to the end of Vernon's dive.
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 between the two distances.
First, we need to determine the absolute values of the distances, as negative distances don't make sense in this context. The absolute value of -45 is 45, and the absolute value of -9 is 9.
Now we can calculate the ratio by dividing the larger distance by the smaller distance. In this case, 45 is larger than 9, so we divide 45 by 9.
45 / 9 = 5
Therefore, the end of Bertha's dive is 5 times farther than the end of Vernon's dive.