52. on page 51 we showed that a one-parameter family of solutions of the rst-order differential equation dyydx 5 xy1/2 is y 5 _ solutions of the rst-order differential equation 1 4 x4 1 c+ solutions of the rst-order differential equation 2 for c $0. each solution in this family is dened on (2`, `). the last statement is not true if we choose c to be negative. for c 5 21, explain why y 5 _ 1 4 x4 2 1+ 2 is not a solution of the de on the interval (2`, `). find an interval of denition i on which y 5 _ 1 4 x4 2 1+ 2 is a solution of the de.

b because it contains 40% of the possible results from the random list.  

0,1,2,3 v 4,5,6,7,8,9  

4 integers to 6 integers  

40% success vs 60% fail

step-by-step explanation:

ιт ιѕ в. 1/4

ιи (fяα¢тισи/єχα¢т fσям) 1/4

ιи (∂є¢ιмαℓ fσям) .25

нσρє ι ¢συℓ∂ нєℓρ уα

8.75

how-to:

first, we need to find what we have to work with.

since i know that we have 7 (years) and 0.8 (inches), we can use these to solve and find the equation!

secondly, what you want to do, is see that we need to know how much growth out of the 0.8 inches he got in only 1 year, instead of 7 years.

let's try to divide.

7 ÷ 0.8 = 8.75