Recent content by stokes

I think I've stated it more then enough: Find the average rate of change in distance as the lighter childs distance changes from 1.5m to 2.5meters.

Thanks.

A child weighs 34 Kg is seated on a seesaw. While a child who weighs 40 kg is situated on the opposite end of the seesaw. The function B(x)= 34x / 40 gives the distance that the 40 kg child must sit from the center of the seesaw when the 34 kg child sits x meters from the center. The seesaw is...

I am really sure I am writing everything down correctly.

1. Homework Statement A child weighs 34 Kg is seated on a seesaw. While a child who weighs 40 kg is situated on the opposite end of the seesaw. The function B(x)= 34x / 40 gives the distance that the 40 kg child must sit from the center of the seesaw when the 34 kg child sits x meters from the...

Please Help.

The thing is the derivative is 34/40. Which equals to 0.85. The prof gave us the answer which is -1.275. I haven't come close to that. I've tried that formula before too... I used... f(x) = f(b) - f(a) / b-a

I still did not get the answer... I am stuck...

Since it is over the interval of 1.5 and 2.5. Couldnt I find B(1.5) and B(2.5) and then find the average?

No it did not... It says find the average rate of change in distance.

Sorry...

I used the quotient rule to fine the instanenous rate of change which turns out to be B'(x)= 68/80. I don't know how to takle the average rate of change.

The seesaw is 9m long find the average rate of change in distance as the lighter childs distance changes from 1.5m to 2.5meters.

Homework Statement A child weighs 34 Kg is seated on a seesaw. While a child who weighs 40 kg is situated on tghe opposite end of the seesaw. The functio B(x)= 34x / 40 gives the distance that the 40 kg child must sit from the center of the seesaw when the 34 kg child sits x meters from the...

Thank you for your help. I don't know why I couldn't put that all together. Thanks again.