2 1/2 yds = 2.5 x 36 = 90 inches...what she started with
2 ft 8 in = 2(12) + 8 = 32 inches...what she had left
90 - x = 32
90 - 32 = x
x = 58 inches....so she used 58 inches
<h3>Given</h3>
p'(t) = kp²
p(0) = 12; p(10) = 24
<h3>Find</h3>
a) p(t)
b) t such that p(t) = 48
c) the behavior of p(t) after the time of part b
<h3>Solution</h3>
a) The differential equation is separable, so can be solved by separating the variables and integrating.
![\displaystyle\frac{d}{dt}p(t)=k\cdot p(t)^{2}\\\\\int{p^{-2}}\,dp=\int{k}\,dt\\\\-p^{-1}=kt+C\\\\p=\frac{-1}{kt+C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bd%7D%7Bdt%7Dp%28t%29%3Dk%5Ccdot%20p%28t%29%5E%7B2%7D%5C%5C%5C%5C%5Cint%7Bp%5E%7B-2%7D%7D%5C%2Cdp%3D%5Cint%7Bk%7D%5C%2Cdt%5C%5C%5C%5C-p%5E%7B-1%7D%3Dkt%2BC%5C%5C%5C%5Cp%3D%5Cfrac%7B-1%7D%7Bkt%2BC%7D)
Plugging in the given boundary conditions, we can solve for k and C to find
![p(t)=\dfrac{240}{20-t}](https://tex.z-dn.net/?f=p%28t%29%3D%5Cdfrac%7B240%7D%7B20-t%7D)
b) The population doubles when the time to t=20 is cut in half. The first doubling occurred in 10 years; the second one will occur in half that time, 5 years. There will be 48 alligators in the swamp in 2003.
c) The population doubles again in half the time of the previous doubling, so is predicted to be infinite in 2008.
more details,so I can give you an answer
A. (5,7)
b. (5,5)
c. (5,3)
d. (-2,5)
e. (-4,5)
f. (-6,5)
This is just substitution. so 3(2(3)-4(1/2)+3(-2/3)= 3(6-2-2)= 3(2) = 6. Basically you plug in the values they gave you for the variables and then just solve one step at a time