Answer:
(I) (-5) is a zero of P(x)
(II) 5 is a zero of P(x)
(III) (-5/2) is a zero of P(x)
Step-by-step explanation:
<h3>
(I) P(x) = x + 5</h3>
Here, P(x) = x + 5
To find the zeroes of P(x)
let P(x) = 0
∴ x + 5 = 0
∴ x = (-5)
Thus, (-5) is a zero of P(x)
<h3>(II) P(x) = x - 5</h3>
Here, P(x) = x - 5
To find the zeroes of P(x)
let P(x) = 0
∴ x - 5 = 0
∴ x = 5
Thus, 5 is a zero of P(x)
<h3>(III) P(x) = 2x + 5</h3>
Here, P(x) = 2x + 5
To find the zeroes of P(x)
let P(x) = 0
∴ 2x + 5 = 0
∴ 2x = -5
∴ x = (-5/2)
Thus, (-5/2) is a zero of P(x)
<u>-</u><u>TheUnknownScientist</u>
Answer:
True
Step-by-step explanation:
When covariance matrices of corresponding class are identical and diagonal matrix and their class probability is same then the class is normally distributed and its discriminant functions are linear.
Step-by-step explanation:
y + 4 = -4(x - 2) = -4x + 8
y - 4 = -4x
-4x - y = -4
4x + y = 4
Answer:
1507.06
.Step-by-step explanation:
The function says
A=A,(1+r)^t
A=1,300×(1+0.03)^(5)
A=1507.06
Hope it helps!
