Tiene pendiente 5 y pasa por el punto Q (-1,-2)
r is the square of the quotint of 2 and 0.5
Find the equation of the line perpendicular to y = 2x + 5 that passes through the point (3, -1) in slope-intercept form.
y equals...
10000×(1−(1+0.10)
−5
)/0.10
use generating functions to solve the recurrence relation an=an+1 +2an-2+2n for n=2,3,4---with initial condition a o=4,a 1=0
10^x=50
13/13/1/3/↑/H
\frac{d}{dt}\left(\sqrt{3t+\sqrt{2+\sqrt{1-t}}}\right)
x→0
+
lim
[
x−3
1
−
log(x−2)
]
