Solve the initial value problem 2yy′+6=y2+6x with y(0)=9. to solve this, we should use the substitution u= y^2+6x (formulas) with this substitution, y= sqrt(u-6x) (formulas) y′= (formulas) enter derivatives using prime notation (e. g., you would enter y′ for dydx). after the substitution from the previous part, we obtain the following linear differential equation in x, u,u′. (equations) the solution to the original initial value problem is described by the following equation in x, y. (equations)

10010010001010010101001001010010100101010010101010010010101010110010001011010010010001

step-by-step explanation:

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2

jfkc

step-by-step explanation:

jhcxzbfzc

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5