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
-36
36
5
To find the difference in the distances of Bertha's and Vernon's dives, we can subtract the distance of Vernon's dive from the distance of Bertha's dive: -45 - (-9) = -45 + 9 = -36.
However, since the distance is negative, we need to take the absolute value to find the difference: |-36| = 36.
Therefore, the end of Bertha's dive was 36 times farther than the end of Vernon's dive. The correct answer is 36.
To find the distance between the end of Bertha's dive and the end of Vernon's dive, we need to subtract their ending distances.
End of Bertha's dive: -45 m
End of Vernon's dive: -9 m
Now, subtract the two distances:
-45 m - (-9 m) = -45 m + 9 m = -36 m
The distance between the end of Bertha's dive and the end of Vernon's dive is -36 m.
Since "farther" usually refers to a positive value, we take the absolute value of -36 m to get the positive value:
|-36 m| = 36 m
Therefore, the end of Bertha's dive is 36 m farther than the end of Vernon's dive.
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 ratio between the distances.
The distance of Bertha's dive from the starting platform is -45 m.
The distance of Vernon's dive from the starting platform is -9 m.
To find the ratio, we divide the distance of Bertha's dive by the distance of Vernon's dive:
-45 m / -9 m
When dividing two negative numbers, the result is positive. So the answer is positive, and we get:
45 m / 9 m = 5
Therefore, the end of Bertha's dive was 5 times farther than the end of Vernon's dive. So the answer is 5.