Given the differential equation

$5t^3 \frac{dx}{dt} +2x-2=0$

The solution is as follows:

$5t^3 \frac{dx}{dt} +2x-2=0 \\ \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\ \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\ \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\ \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\ \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\ \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\ \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\ \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1$

3 0

3 years ago

Find the mean of the data. give your answer in decimal form

aleksandr82 [10.1K]

Mean = (1+1+2+3+3+3+4+5)/8 = 22/8 = 2.75

answer
mean = 2.75

8 0

3 years ago

A growing farming conglomerate increases its water usage at a rate of 10% every year. If it

Elena-2011 [213]

The quantity of water the farming conglomerate would use 17 years from now is 65,708.11.

<h3>What is the amount of water that would be used 17 years from now?</h3>

The formula for calculating the amount of water that would be needed 1 years from now:

FV = P (1 + r)^N

FV = Future value
P = Present value
R = interest rate
N = number of years

13000 x 1.1^17 = 65,708.11

To learn more about future value, please check: brainly.com/question/18760477

8 0

2 years ago