数学题。如图求过程。
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a) substitute P(X)=1000,t=10into the equationP(X)=P0*e^0.01t
get 1000=P0*e^0.01*10
1000=P0*e^0.1
P=1000/e^0.1
P=904.837418
P约等于905
b)initial population means when t=0
P(X)=905*e^0.01*0
=905*e^0
=905*1
=905
c)because the initial population is 905,so when t=20 afeter means 20 years
P(X)=905*e^0.01*20
= 905*e^0.2
=1105
d)substitute P(x)=2000 into the equation
P(x)=p0*e^0.01t
2000=905*e0.01t
e^0.01t=2000/905
e^0.01t=2.2099
0.01t*log e=log2.2099
0.01t=0.79296
t=79.29675
so after 79 years the population weill reach 2000.
PS: ^这个符号后是次方的意思,e是e的一次方,你要把e放进计算机才能求喔~~另外,e^0.01t=2.2099这一步你两边都log一下,会比较好求,不知道你老师有没有说过~~好了,希望能帮到你
get 1000=P0*e^0.01*10
1000=P0*e^0.1
P=1000/e^0.1
P=904.837418
P约等于905
b)initial population means when t=0
P(X)=905*e^0.01*0
=905*e^0
=905*1
=905
c)because the initial population is 905,so when t=20 afeter means 20 years
P(X)=905*e^0.01*20
= 905*e^0.2
=1105
d)substitute P(x)=2000 into the equation
P(x)=p0*e^0.01t
2000=905*e0.01t
e^0.01t=2000/905
e^0.01t=2.2099
0.01t*log e=log2.2099
0.01t=0.79296
t=79.29675
so after 79 years the population weill reach 2000.
PS: ^这个符号后是次方的意思,e是e的一次方,你要把e放进计算机才能求喔~~另外,e^0.01t=2.2099这一步你两边都log一下,会比较好求,不知道你老师有没有说过~~好了,希望能帮到你
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