Hello:<span>
the equation is : y = ax+b
the slope is a : a×(4/3) = -1......(
perpendicular to a line with a slope of 4/3) because : 4x-3y =8...y=(4/3)x-8/3 ) </span>
<span>
a = -3/4
y=(-3/4)x+b
the line that passes through (3, -2) :- 2 =
(-3/4)(3)+b
b =1/4
<span> the equation is : y = (-3/4)x+1/4</span></span>
First we need to define our conditions. Since the angles are complimentary, the sum of the angles must be equal to 90 degrees.
let x = first angle
y = second angle
x + y = 90
from the second condition in the problem
x + y =72 + x -y
solving for x
x= 54
therefore the shorter side is 90 - 54 = y
y = 36 degree, the shorter side
For the equation F(x) = ax² + bx + c we have:
- maximum value if a<0
- minimum value if a>0
F(x) = -3x² + 18x + 3 ⇒ a = -3, b = 18
a < 0 ⇒ the function has a maximum value
Quadratic function has the maximum value (or minimum) at vertex of its parabola.
The maximum value is k at x=h where: 
and k = F(h)
Therefore:
<h3>
The function has a maximum value of 30 at x = 3</h3>
