F(x)=3x/2 for 0≤x≤2
<span>.....=6 - 3x/2 for 2<x≤4 </span>
<span>g(x) = -x/4 + 1 for 0≤x≤4 and g'(x)=-1/4 </span>
<span>so h(x)= f(g(x)) = (3/2)(-¼x+1)=-3x/8 + 3/2 for 0≤x≤2 </span>
<span>for x=1, h'(x)=-3/8 so h'(1)=-3/8 </span>
<span>When x=2, g(2)=1/2 so h'(2)=g'(2)f '(1/2)= -(1/4)(3/2)=-3/8 </span>
<span>When x=3, h(x)=6 - (3/2)(1 - x/4) = 9/2 +3x/8 </span>
<span>h'(x)=3/8 so h'(3) = 3/8</span>
For a hyperbola

where

the directrix is the line

and the focus is at (0, c).
Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
a = √9 = 3
b = √16 = 4
Your hyperbola's constants are ...
a = 3
b = 4
______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
Answer:
A
Step-by-step explanation:
Just took the quiz on edgenuity
A. The number of 10-boards Peter bought is equal to n divided by 10. Then, each of the 10-boxes will get two boxes of nails. The number of boxes of nails that Peter will have after buying n boards will be,
N = (2)(n/10)
Simplifying,
<em> N = n/5</em>
b. If the number of boards are 90 then,
N2 = (90/10)(2)(100 nails/box)
N2 = 1800
Answer: 1800