Find an equation of the line that satisfies the given conditions. through (−2, −6); perpendicular to the line passing through (1, 6) and (5, 4)

The slope of the given line is ∆y/∆x = (4-6)/(5-1) = -2/4 = -1/2. The slope of the desired (perpendicular) line is the negative reciprocal of this, -1/(-1/2) = 2.

The point-slope form of the equation for a line with slope m through point (h, k) is

... y - k = m(x - h)

For slope 2 and the given point, your equation can be

... y + 6 = 2(x + 2)

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !