#rates question
27 messages · Page 1 of 1 (latest)
for b and c you need to add the wage of the driver to your initial function too
the derivative would be (-1024/x^2)+1
,,f'(x)=0.0043165(\frac{-1024}{x^2}+1)
yule
yule
to derivate this you need to subtract 1 from the power of x then multiply by the result
,,(-1024x^{-2})*(-2)
yule
the site makes it way too complicated i suggest you look into videos explaining the power rule simpler
,,F(x)=(1.69)(0.0043165)(\frac{1024}{x}+x)+(\frac{510}{x})(15)
yule
you should derivate this function and solve for y=0 to find b
,,F(x)=(1.69)(0.0043165)(\frac{1024}{x}+x)+(\frac{350}{x})(24)
yule
this for c
alright again u should look into the power rule
no u cant solve it like that
x in this case is not 32
you have to derivate that function too and solve for y=0
you have to solve all a b and c on their own
here we calculate the money spent in gas while going x mph and the amount we pay driver while he goes x mph thru the road
so essentially you have to find the sweet spot where the sum of those things are minimal
x=32 is only minimal for the gas price
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