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)

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

english, cause i can’t you