17+8-9=17-1=16
Answer: C)16
B. No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32
Perimeter = 2(l + w)
Perimeter = 64
Assuming: P = 64; w = 11 ; l = ?
64 = 2(l + 11)
64/2 = l + 11
32 - 11 = l
length = 21.
Let x represent the number of type A table and y represent the number of type B tables.
Minimize: C = 265x + 100y
Subject to: x + y ≤ 40
25x + 13y ≥ 760
x ≥ 1, y ≥ 1
From, the graph the corner points are (20, 20), (39, 1), (30, 1)
For (20, 20): C = 265(20) + 100(20) = $7,300
For (39, 1): C = 265(39) + 100 = $10,435
For (30, 1): C = 265(30) + 100 = $8,050
Therefore, for minimum cost, 20 of type A and 20 of type B should be ordered.
Since that f is continuous for all x in R, and f > 0 then
for hence the asymptote is
y = 4
