Solving for the polynomial function of least degree with
integral coefficients whose zeros are -5, 3i
We have:
x = -5
Then x + 5 = 0
Therefore one of the factors of the polynomial function is
(x + 5)
Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0
Therefore one of the factors of the polynomial function is (x^2
+ 9)
The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)
= x^3 + 9x + 5x^2 = 45
Therefore, x^3 + 5x^2 + 9x – 45 = 0
f(x) = x^3 + 5x^2 + 9x – 45
The polynomial function of least degree with integral coefficients
that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45
Answer:
8:4:13
Step-by-step explanation:
Divide each by 3 so you get 8:4:13
Answer:
no 4 is not perfect cube and no 2 is perfect cube
First, we establish
our hypothesis:
<span>Null hypothesis H0: μ = $1.00 </span>
Alternative hypothesis
Ha: μ ≠ $1.00
<span>Let’s say X = the sample average cost of a daily newspaper
= 0.96</span>
u = population mean
cost = 1.00
S = sample standard
deviation = 0.18
Calculating for z
value:
z = (X – u) / S
z = (0.96 – 1) / 0.18
z = – 0.222
From the standard
distribution table at this z value, p-value = 0.4129
Since alpha = 0.01,
the decision therefore is:
<span>Do not reject the null
hypothesis because the p-value is greater than 0.01. There is enough evidence
to support the claim that the mean cost of newspapers is $1. </span>
