D. The distributive property That is what jack used
Multiply the original DE by xy:
xy2(1+x2y4+1−−−−−−−√)dx+2x2ydy=0(1)
Let v=xy2, so that dv=y2dx+2xydy. Then (1) becomes
x(y2dx+2xydy)+xy2x2y4+1−−−−−−−√dxxdv+vv2+1−−−−−√dx=0=0
This final equation is easily recognized as separable:
dxxln|x|+CKxvKx2y2−1K2x4y4−2Kx2y2y2=−dvvv2+1−−−−−√=ln∣∣∣v2+1−−−−−√+1v∣∣∣=v2+1−−−−−√+1=x2y4+1−−−−−−−√=x2y4=2KK2x2−1integrate both sides
Numbers of dinners =125 individual portion= 6oz size of can = 32oz
Answer: i have no idea
Step-by-step explanation: i dont know sorry
Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
