Answer:
1) x < 1, x > 9
2) 2 < x ≤ 6, -4 ≤ x < 0
Step-by-step explanation:
1) lx - 5l > 4
x - 5 = 4
x = 9
-(x - 5) = 4
x = -4 + 5 = 1
x < 1, x > 9
3) 1 < lx-1l ≤ 5
1 < x-1 ≤ 5
2 < x ≤ 6
1 < -(x-1) ≤ 5
-5 ≤ x-1 < -1
-4 ≤ x < 0
Answer:
106
Step-by-step explanation:
Answer:
1) DC and DG
2) D
3) <GDE
4) acute
5) obtuse
Step-by-step explanation:
1) DC and DG
2) D
3) <GDE
4) acute
5) obtuse
F(x) = x² + 2x - 1
f(2) = 2² + 2 · 2 - 1 = 4 + 4 - 1 = 7
f(-2) = (-2)² + 2 · (-2) - 1 = 4 - 4 - 1 = -1
f(3) = 3² + 2 · 3 - 1 = 9 + 6 - 1 = 14
f(-3) = (-3)² + 2 · (-3) - 1 = 9 - 6 - 1 = 2
The range = {-1; 2; 7; 14}
Answer: The average daily inventory is 200 cases.
Step-by-step explanation:
Since we have given that
N(t)=600-20√30t
We need to find the average daily inventory.
![\dfrac{1}{b-a}\int\limits^a_b {600-20\sqrt{30t}} \, dt\\\\=\dfrac{1}{30}\int\limits^{30}_0 {600-20\sqrt{30t}} \, dt \\\\=\dfrac{1}{30}[600t-\dfrac{20(30t)^\frac{3}{2}}{45}|_0^{30}\\\\=\dfrac{1}{30}[18000-\dfrac{20\times 30^3}{45}]\\\\=\dfrac{1}{30}[18000-12000]\\\\=200](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bb-a%7D%5Cint%5Climits%5Ea_b%20%7B600-20%5Csqrt%7B30t%7D%7D%20%5C%2C%20dt%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5Cint%5Climits%5E%7B30%7D_0%20%7B600-20%5Csqrt%7B30t%7D%7D%20%5C%2C%20dt%20%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5B600t-%5Cdfrac%7B20%2830t%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7B45%7D%7C_0%5E%7B30%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5B18000-%5Cdfrac%7B20%5Ctimes%2030%5E3%7D%7B45%7D%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5B18000-12000%5D%5C%5C%5C%5C%3D200)
Hence, the average daily inventory is 200 cases.
