Using the general form; y = mx + b, we get:
y = 4x + 6
Answer:
r = 1/2
c1 = 3/4
c2 = 27/64
c3 = 405/1408
Step-by-step explanation:
Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)
r = 1/2
c1 = 3/4
c2 = 27/64
c3 = 405/1408
The solution is attached.
I must assume that you meant the following:
10(Q-3R)
P = ----------------
R
Multiplying both sides by R, we get PR = 10R(Q-3R), or PR = 10RQ - 30R^2.
Rearranging these terms so that powers of R are in descending order:
30R^2 - 10RQ + PR = 0
Factoring out R, we get
R(30R - 10Q + P) = 0. This has two solutions:
R = 0, and 10Q - P
30R = 10Q - P, so that R = --------------
30
Answer:
C. x²+3x-28
Step-by-step explanation:
f(x) = x +7
g(x) = x -4
f(x) · g(x) = (x-4)(x+7)
= x²+7x-4x-28
= x²+3x-28