HL congruence theorem is the answer.
Hope this helps :)
<span>the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
The rectangle on the right is 81 and the rectangle on the left is 24. I don't really know about the second one sorry, but I hope this helped.
Answer:
(fg)(x) = 6x³ - 2x² + 12x - 4
Step-by-step explanation:
( f g ) ( x ) = f ( x ) * g ( x )
f(x) = 2x² + 4
g(x) = 3x – 1
( f g ) ( x ) = (2x² + 4) (3x – 1)
= 6x³ - 2x² + 12x - 4