Hello:<span>
the equation is : y = ax+b
the slope is a : a×(-3/2) = -1......(
perpendicular to a line with a slope of -3/2)
a = 2/3
y=(2/3)x+b
the line that passes through (-2, -2) : -2 =
(2/3)(-2)+b
b = -2/3
<span> the equation is : y = (2/3)x-2/3</span></span>
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3
(a +(- b))^3 = (a-b)^2 = a^3 - 3a^2b + 3ab^2- b^3
Hello :
f'(7) = the slope of the tangent line : let A(7,4) B(0,3)<span>
<span>the slope is : (YB - YA)/(XB -XA)= (3-4)/(0-7) = -1/-7 = 1/7=f'(7)
the equation of the tangent line is :
y-3 = 1/7 (x-0)
y = (1/7)x+3
the line tangent and the graph of : f </span></span><span>passes through the point A(7, 4)
x= 7 y=4 so : f(7) =4</span>
