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
the answer is C.DAB and BCD
Answer:
Step-by-step explanation:
Answer: 0.0049 hope this is right
Answer:
x = 2
Step-by-step explanation:
(x² logₓ27) log₉x = x + 4
Use change of base formula.
x² (log 27 / log x) (log x / log 9) = x + 4
Simplify.
x² (log 27 / log 9) = x + 4
x² (log 3³ / log 3²) = x + 4
x² (3 log 3 / (2 log 3)) = x + 4
x² (3 / 2) = x + 4
3x² = 2x + 8
3x² − 2x − 8 = 0
(x − 2) (3x + 4) = 0
x = 2 or -4/3
Since x > 0, x = 2.