okay! 3.(x^7)e^5x 4. A company's toatl cost in millions of dollars is given by C(t)=130-70e^-t where t is the time in years since the start up date. The graph of C(t) is shown to the right. FInd the following. 4(A). the marginal cost, C'(t) (b). C'(0) (c). C'(4) (D) find lim t->infinite C(t) and lim t-->infinite C'(t) will send rest through word document

can't get word document to upload due to subscription running out. i posted some below9)S^a=N=logs(s being lower than log)?=? Convert to logarithmic equation10)use log b 3(b is lower than log) = 1.099 and or log b 8 = 2.079 to find log b 2411) g(x)=e^x(lnx^2) differentiate12) f(x)=ln(e^6x(+6) six being added to e. Differentiate13) solve e^x=5 for x

can't copy paste have to type them.14) a model for consumers’ response to advertising is given by the equation N(a)=1900+500 ln a, where N(a) is number of units sold, a is the amount spent on advertsing in thousands of dollars and a>=1. Complete (a through (d How many units were sold after spending $1000 on advertising? N’(a) and n’(10) Find the maximum value, if it exists. And Find the minimum value if it exists Find lim N’(a) does it make sense to spend more and more dollars on adverstising15) the growth rate for the demand for oil in a particular country is 3% per year. When will the demanded be double that of 2012In 2004 an art collector paid $108,467,000 for a particular painting the same painting sold for $28,000 in 1950 complete parts a through d Find the exponential growth rate k to three decimal places and determine the exponential growth function V, for which V(t) is the painting’s value in dollars, t years after 1950. K=? Predict the value of painting in 2012 What is the doubling time for the value of the painting? How long after 1950 will the value of the painting be $1 billion?

16) of an intial amount of 4000 g of lead-210 how much will remain in 170 years? lead -210 decays at a rate of 3.15%/yr17)The demand D(x), and supply S(x) functions for a certain type of multipurpose printer are as follows. Find equilibrium point. Assume that x is the price in dollars D(x) = q= 490e^-0.003x S(x)=q=150e^0.005x

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18)g(x)= x^6(2.6)^x differentiate19)y = 3^(x^5 +4) four being added to x20) y = log 8 x (8 being lower than log) differentiate21)g(x)= log 47 (47 lower than log) (5x-2)

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