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course Mth173
6/18/13~7:33PM
Text section 1.2Section 1.2 Problems 1-11, 15, 18, 19, 22, 24, 28, 31, 32, 37, 38, 40, 43
1.) The graph shown is concave up
2.) The graph shown is neither concave up or down
3.) The graph is linear and is neither concave up or down
4.) The graph is concave up
5.) The function p=5(1.07)^t is an exponential growth with an initial quantity of 5, growth factor of 1.07, growth rate of .07
????not sure how to tell if it is continuous or not????
6.) P=7.7(0.92)^t is exponential decay with an initial quantity of 7.7, growth factor and rate of 0.92
7.) P=3.2e^.03t is exponential decay with a decay factor of .03 and a initial quantity of 3.2, the decay factor is e (2.73…)
8.) P=15e^-0.06t is exponential decay with a decay factor of .06,
9.) P=P0*e^0.25t where p0 is the initial quantity, e is the growth factor and .25 is the decay rate. This equation represents exponential decay.
10.) P=2*e^-0.5t. this equation is exponential decay. The decay rate is -.5. the decay factor is e. the initial quantity is 2
11.) P=P0*e^.2t. This is exponential growth, with a growth factor of e, initial quantity of e, and a growth rate of .2
12.) P=7e^pi*t. this is exponential growth. Growth rate of pi. Initial quantity of 7. And the growth factor of e
15.) town has initial pop at 1000, t=0
a.) P(t)=1000+50t
b.)P(t)=1000+(1.05)^t
18.)
a.) P=6.7*(1+.012)^t
b.) P(13)=6.7*(1+.012)^13 → 7.82 billion people by 2020
c.)The graph of this function would be concave up at a slow pace over the time interval.
19.)
a.) if a page is reduced to 80% then it must be enlarged by 125% to return it to its original size.
b.) to reduce an original copy to less than 15% it must be copied 9 times
%=1*(.8)^x→ x=9
22.) ????Not sure how to find the possible formula for the graphs of exponetials???(the book does not explain it)
24.)
28.)
a.) prices in Zimbabwe increased by .67% day to day
b.) assuming the same rate of inflation rate all year round → (1+.067)^365
31.) the doubling time is as follows:
p(t)=10,000,000(1.02)^(t)
p(t+tD)=10,000,000(1.02)^(t+tD)
→ 2 P(t) = P(t+tD)
→2(10,000,000(1.02)^(t)) = 10,000,000(1.02)^(t+tD)
→log(2)/log(1.02)→ 35.0027 ~~around 35 years for doubling time
32.) ????dont know how to set this one up???
37.) we can show graphically that the doubling time is around 2.3-2.4 years. At t=0 we have a value around 10,000 by looking at the graph. It takes around 2.3-2.4 years for that value to double.
????Not sure where problems 38, 40, and 43 are, the section problems stop at 37 and im sure I have the right book????
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