Answer:
a. attached graph; zero real: 2
b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
Step-by-step explanation:
p(x) = x³ + 4x² + 6x - 36
a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.
p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0
b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.
2 | 1 4 6 -36
1 6 18 0
x² + 6x + 18 = 0
Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²
√Δ = 6i
x = -6±6i/2 = 2(-3±3i)/2
x' = -3-3i
x" = -3+3i
p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
465 -(-45) = x
This creates a double negative which becomes a positive and you solution would be 465+45 =X, X = 510
Answer:
false false true false ez
The <em>correct answer</em> is:
|w-20| ≤ 0.12.
Explanation:
We first find the average of the two ends of the inequality:
(19.88+20.12)/2 = 40/2 = 20
This will be the number subtracted from w in the inequality.
Now we find the difference between this value and the ends:
20-19.88 = 0.12
20.12 - 20 = 0.12
This will be what our absolute value inequality ends with; the "answer" part, so to speak.
Since this inequality is written in compact form, it must be an "and" inequality; this means the absolute value inequality must be a "less than or equal to."
This gives us
|w-20| ≤ 0.12
