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April 1, 2015

April 1, 2015

Posted by **Justin** on Tuesday, September 17, 2013 at 10:17pm.

M'(x) =

3,600x−2 − 1/

(3,600x−1 + x)2.

Estimate M'(10), M'(60), and M'(90). (Round your answers to seven decimal places.)

M'(10) = mpg/mph

M'(60) = mpg/mph

M'(90) = mpg/mph

- Calculus -
**Reiny**, Tuesday, September 17, 2013 at 10:35pmI will read that as

M'(x) = (3600x^-2 - 1)/(3600x^-1 + x)^2

which is

= (3600/x^2 - 1)/(3600/x + x)^2

M'(10) = (3600/100 - 1)/(3600/10+ 10)^2

= 35/(370)^2

= ????

This makes no sense

Either I did not interpret your typing correctly, or there is a typo.

Please use ^ to exponents and brackets to establish where the division is.

e.g. is it

(3600x^-2 - 1)/(3600x^-1 + x)^2

or

3600x^-2 - ( 1/(3600x^-1 + x)^2 ) ?

- Calculus -
**Justin**, Tuesday, September 17, 2013 at 10:48pmok, maybe this is more accurate...

M'(x) = (3600x^-2 -1)/

(3600x^-1 +x)^2

- Calculus -
**evander**, Thursday, March 26, 2015 at 11:06pmyou plug in 10,60,90 for x

so

3600(10)^-1 -1

/

(3600(10)^2 +10)^2 = 0.0002557 round to the seven decimal places also and you do that for each one.

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