Respuesta :
Answer:
Charles Second jump was [tex]1\frac{5}{6} \ feet[/tex] long.
Step-by-step explanation:
Given:
First Jump marked distance = [tex]16\frac{5}{6} \ feet[/tex]
[tex]16\frac{5}{6} \ feet[/tex] can be Rewritten as [tex]\frac{101}{6} \ feet[/tex]
First Jump distance = [tex]\frac{101}{6} \ feet[/tex]
Second Jump marked distance = [tex]18\frac{2}{3} \ feet[/tex]
[tex]18\frac{2}{3} \ feet[/tex] can be Rewritten as [tex]\frac{56}{3} \ feet[/tex]
Second Jump distance = [tex]\frac{56}{3} \ feet[/tex]
We need to find the how much long is Charles second jump.
To find Length of Second Jump is calculated by subtracting distance of First Jump from Distance of Second jump.
Length of Second Jump = Second Jump distance - First Jump distance
Substituting the values we get.
Length of Second Jump = [tex]\frac{56}{3}-\frac{101}{6}[/tex]
We will make denominator common by taking LCM and then solve for it.
Length of Second Jump = [tex]\frac{56\times 2}{3\times2}-\frac{101\times1}{6\times 1} = \frac{112}{6}-\frac{101}{6}= \frac{112-101}{6}= \frac{11}{6} \ feet[/tex]
Now [tex]\frac{11}{6} \ feet[/tex] can be Rewritten as [tex]1\frac{5}{6} \ feet[/tex]
Length of Second Jump = [tex]1\frac{5}{6} \ feet[/tex]
Hence Charles Second jump was [tex]1\frac{5}{6} \ feet[/tex] long.
