Charlie can bowl 25 games.
20 - 1.25 = 18.75
18.75/.75 = 25 games
i am not 100% sure but...
(4x+3)=47
<u> - 3 - 3 </u>
4x=44
44/4= 11
so x = 11
P(t) = P₀ e^(kt)
<span>Where P₀ is the initial population, </span>
<span>P(t) is the population after "t" time. </span>
<span>t is your rate (can be hours, days, years, etc. in this case, hours) </span>
<span>k is the growth constant for this particular problem. </span>
<span>So using the information given, solve for k: </span>
<span>P₀ = 2000 </span>
<span>P(4) = 2600 </span>
<span>P(t) = P₀ e^(kt) </span>
<span>2600 = 2000e^(k * 4) </span>
<span>1.3 = e^(4k) </span>
<span>Natural log of both sides: </span>
<span>ln(1.3) = 4k </span>
<span>k = ln(1.3) / 4 </span>
<span>Now that we have a value for "k", use that, the same P₀, then solve for P(17): </span>
<span>P(t) = P₀ e^(kt) </span>
<span>P(17) = 2000 e^(17ln(1.3) / 4) </span>
<span>Using a calculator to get ln(1.3) then to simplify from there, we get: </span>
<span>P(17) ≈ 2000 e^(17 * 0.262364 / 4) </span>
<span>P(17) ≈ 2000 e^(4.460188 / 4) </span>
<span>P(17) ≈ 2000 e^(1.115047) </span>
<span>P(17) ≈ 2000 * 3.0497 </span>
<span>P(17) ≈ 6099.4 </span>
<span>Rounded to the nearest unit: </span>
<span>P(17) ≈ 6099 bacteria hope i could help =)))</span>
we are given
(A)
(f×g)(x)=f(x)*g(x)
now, we can plug it
we can simplify it
(B)
Domain:
Firstly, we will find domain of f(x) , g(x) and (fxg)(x)
and then we can find common domain
Domain of f(x):
we know that f(x) is undefined at x=0
so, domain will be
∪
Domain of g(x):
Since, it is polynomial
so, it is defined for all real values of x
now, we can find common domain
so, domain will be
∪..............Answer
Range:
Firstly, we will find range of f(x) , g(x) and (fxg)(x)
and then we can find common range
Range of f(x):
we know that range is all possible values of y for which x is defined
since, horizontal asymptote will be at y=0
so, range is
∪
Range of g(x):
Since, it is quadratic equation
so, its range will be
now, we can find common range
so, range will be
∪.............Answer
