A polynomial function of least degree with integral coefficients that has the
given zeros 
Given
Given zeros are 3i, -1 and 0
complex zeros occurs in pairs. 3i is one of the zero
-3i is the other zero
So zeros are 3i, -3i, 0 and -1
Now we write the zeros in factor form
If 'a' is a zero then (x-a) is a factor
the factor form of given zeros
Now we multiply it to get the polynomial
polynomial function of least degree with integral coefficients that has the
given zeros
Learn more : brainly.com/question/7619478
C3=n
N x 2 - 6 + 2
N < (6x8) - 10
Answer:
Step-by-step explanation:
sin 30+tan²(60)+sec²(45)
=1/2+ (√3)^2+(√2)²
=1/2+3+2
=5 1/2
=5.5
Answer:
Explicit tn = 52 + (n - 1)*(-12)
Recursive = tn = t(n - 1) - 12
Step-by-step explanation:
The difference between term n and term n - 1 is can be found by taking the difference between and 2 consecutive terms.
t3 = 28
t2 = 40
d = t3 - t2
d = 28 - 40
d = - 12
Explicit
tn = a1 + (n - 1)*d
a1 = 52
d = - 12
tn = 52 + (n - 1)*(-12)
<em><u>Example</u></em>
Find t5
t5 = 52 + (5 - 1)*(-12)
t5 = 52 + 4 * - 12
t5 = 52 - 48
t5 = 4
Recursive
tn = t(n - 1) - 12
<em><u>Example</u></em>
t5 = t4 - 12
t5 = 16 - 12
t5 = 4 just as it did before.
