Respuesta :
The solution to the problem is as follows:
We have 2+.8(2) + .8(.8(2)) + .8(.8(.8(2))) + ... =
2( .8^0 + .8^1 + .8^2 + .8^3 + ... ) =
2(.8^n -1) / (.8-1) . As n-->infinity, .8^n-->0 giving us
2(-1)/(-.2) = 2(5) = 10 meters.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
We have 2+.8(2) + .8(.8(2)) + .8(.8(.8(2))) + ... =
2( .8^0 + .8^1 + .8^2 + .8^3 + ... ) =
2(.8^n -1) / (.8-1) . As n-->infinity, .8^n-->0 giving us
2(-1)/(-.2) = 2(5) = 10 meters.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Answer:
The distance traveled by the ball until it stops bouncing is:
18 meters
Step-by-step explanation:
It is given that:
a ball bounces from a height of 2 meters and returns to 80% of its previous height on each bounce.
This means that the distance traveled by the ball is the distance it travels by going down as well as coming up on bouncing.
and is given as follows:
[tex]\text{Total Distance}=2+0.8\times 2\ up+0.8\times 2\ down+(0.8)^2\times 2\ up+(0.8)^2\times 2\ down+.....\\\\\text{Total distance}=2+4\times (0.8)+4\times (0.8)^2+4\times (0.8)^3+....\\\\\text{Total distance}=2+4\times [0.8+(0.8)^2+(0.8)^3+......]\\\\\text{Total distance}=2+4\times (\dfrac{0.8}{1-0.8})[/tex]
Since, the formula of infinite geometric progression is:
[tex]\sum_{n=1}^{\infty} ar^{n-1}=\dfrac{a}{1-r}[/tex]
i.e.
[tex]\text{Total distance}=2+4\times (\dfrac{0.8}{0.2})\\\\\text{Total distance}=2+4\times 4\\\\\text{Total distance}=2+16\\\\\text{Total distance}=18\ \text{meters}[/tex]
