Tuesday
December 10, 2013

# Posts by becky

Total # Posts: 470

calculus
based on my question, i retyped... F(v) : (600v^3/2 + 20250)/v F'(v) : (300v^3/2 - 20250)/v^2 F"(v) : (-150v^5/2 + 40500)/v^3 Reiny, isn't your answer suppose to be f(x) = 600 v^(3/2) + 20250 v^-1, instead of f(x) = 600 v^(1/2) + 20250 v^-1 ?? 1/2 power and 3/2 po...

calculus
I do not want the simplified answer by the way...I need the answer as an expression only

calculus
Ok but the power 5/2 is wrong? Should have been power 3/2? I keep on getting the answer 5/2.. I don't know which part of my working is wrong

calculus
Just want to check whether my answer is correct for quotient rule F(v) : (600^3/2 + 20250)/v F'(v) : (300^3/2 - 20250)/v^2 F"(v) : (-150v^5/2 + 40500)/v^3 Can anyone let me know whether my answer is correct for first and second derivative based on F(v) ??

calculus
yes it should be, F(v) : (600^3/2 + 20250)/v F'(v) : (300^3/2 - 20250)/v^2 F"(v) : (-150v^5/2 + 40500)/v^3 for F"(v), this is how i get: F'(v) : (300^3/2 - 20250)/v^2 F"(v) : ((300v^3/2 - 20250)' * v^2) - ((300v^3/2 - 20250) * (v^2)')/v^4 F"...

calculus
nope, it's not a typo. the 5/2 is the power of 5 over 2. it is not 5 divide by 2. is my above answer correct?

calculus
Just want to check whether my answer is correct for quotient rule F(x) : (600^3/2 + 20250)/v F'(x) : (300^3/2 - 20250)/v^2 F"(x) : (-150v^5/2 + 40500)/v^3 Can anyone let me know whether my answer is correct for first and second derivative based on F(x) ?? Thank you!!

calculus
A ship uses \$( 7v^2 ) of fuel per hour when travelling at a constant speed of v km/h. Other expenses in operating the ship (i.e. crew and equipment) amount to \$345 per hour. The ship makes a journey of 100 km. C(v) : (700v^2 + 34500)/v C'(v): (700v^2 - 34500)/v^2 C"(v...

calculus
i got the answer!! thank you!!!

calculus
pls ignore the above question... to make it clearer for the question, it is... f(x): (700v^2 + 34500)/v f'(x) : (700v^2 -34500)/v^2 f''(x) : ???

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