Answer:
C and D
Step-by-step explanation:
Equating the line A and the parabola, we get
-3x + 2 = x² - 3x + 4
0 = x² - 3x + 4 +3x - 2
0 = x² + 2
-2 = x²
which has no real solutions. Then, the line A and the parabola don't intersect each other.
Equating the line B and the parabola, we get
-3x + 3 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 3
0 = x² + 1
-1 = x²
which has no real solutions. Then, the line B and the parabola don't intersect each other.
Equating the line C and the parabola, we get
-3x + 5 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 5
0 = x² - 1
1 = x²
√1 = x
which has 2 solutions, x = 1 and x = -1. Then, the line C and the parabola intersect each other.
Equating the line D and the parabola, we get
-3x + 6 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 6
0 = x² - 2
2 = x²
√2 = x
which has 2 solutions, x ≈ 1.41 and x ≈ -1.41. Then, the line D and the parabola intersect each other.
A. True. We see this by taking the highest order term in each factor:
B. True. Again we look at the leading term's degree and coefficient. f(x) behaves like -3x⁶ when x gets large. The degree is even, so as x goes to either ± ∞, x⁶ will make it positive, but multiplying by -3 will make it negative. So on both sides f(x) approaches -∞.
C. False. f(x) = 0 only for x=0, x = 5, and x = -2.
D. False. Part of this we know from the end behavior discussed in part B. On any closed interval, every polynomial is bounded, so that for any x in [-2, 5], f(x) cannot attain every positive real number.
E. True. x = 0 is a root, so f(0) = 0 and the graph of f(x) passes through (0, 0).
F. False. (0, 2) corresponds to x = 0 and f(x) = 2. But f(0) = 0 ≠ 2.
Answer:
A or 3
Step-by-step explanation:
2/3+2/3=4/3 or 1 1/3
4/3+2/3= 6/3 or 2
Ratio with two equivalent measurements would be 1
Answer:
-2
Step-by-step explanation:
