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Can you please answer the questions clearly? Simplify the expressions below. s - [}] (b) log 10 (x - 25) (c) log 3813~ 4 Show

Can you please answer the questions clearly?

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Simplify the expressions below. s - [}] (b) log 10 (x - 25) (c) log 3813~ 4 Show that the function is continuous at the given value. f(x)=9x,c=3 The function f{x) is continuous at x = 3 since #(3)= |imHar(x)f(3). and limy_,af(x)=| |implies that Show that the function given below is continuous at x= - 6. 3 +17%-6 fix) = X+B The function f(x} is continuous at x = - & since f( - 6) = u and limy_, gf(x) = Uimplies that limy_ - (%) Elf( -6). @ For the following function, find all points of discontinuity. fix)= x+4 - Select the correct choice below and, if necessary, fill in the answer box to complete your chaice. O A fis discontinuous at | 7 (Simplify your answer. Use a comma to separate answers as needed.) () B. fhas no discontinuity. Logarithms allow for solving equations with unknown exponenis by the fact that x=log 4y is equivalent to y =a*. The natural logarithm has the number as its base and is written In x; that is, log . = In x. Some properties of logarithms are Based on s log 3{xy) = log ax +log 5y log 5 [;J =logax-logay log %' =rlog 4% log 4a\" = x log % a =y @ Which value must you assign to a so that f(x) is continuous at x =37 2 G+x-x Ry PNE it f(x)= x-3 I w a X Ifa= D, then f(x) is continuous at x = 3. @Determine at which points f(x) is discontinuous. f(x)= i X2 -36 ) The function f(x) is discontinuous at x = D (Use a comma to separate answers as needed.) @ A papulation model is used to describe how the rate of reproduction of organisms depends on the amount of nutrients that are available. The model shows how the rate of division of E. coli cells depends upon the amount of sugar added to their growth flask. If r is the rate of reproduction (number of divisions in one hour) and C is the amount of glucose sugar added to the growth medium measured in moles, then the following applies. 1.36C HC)= C+0.24x10 (a) Show that the reproduction rate goes to zero when the sugar level is low; that is lim rcy=o. C0 1.36([ ) = =0 [ ]+024x107* Since r is cont tc=0, Im rc)= = ince r is continuous a b r(C) f(D)

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