Answer:
190 square inches
Step-by-step explanation:
Answer:
no solution
Step-by-step explanation:
no solution x=0
Answer:

Step-by-step explanation:
1) divide each term in the numerator by the term in denominator:
(remember when dividing subtract the smaller exponent from larger exponent
for example:
say a is the larger exponent, do 
x^4/x=x^3
5x^3/x=5x^2
3x^2/x=3x
so
x^3+5x^2+3x
hope this helps!
The solution depends on the value of

. To make things simple, assume

. The homogeneous part of the equation is

and has characteristic equation

which admits the characteristic solution

.
For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be

. Then

So you have


This means


and so the general solution would be